Why does Central Limit Theorem break down in my simulation?












18












$begingroup$


Let say I have following numbers:



4,3,5,6,5,3,4,2,5,4,3,6,5


I sample some of them, say, 5 of them, and calculate the sum of 5 samples.
Then I repeat that over and over to get many sums, and I plot the values of sums in a histogram, which will be Gaussian due to the Central Limit Theorem.



But when they are following numbers, I just replaced 4 with some big number:



4,3,5,6,5,3,10000000,2,5,4,3,6,5


Sampling sums of 5 samples from these never becomes Gaussian in histogram, but more like a split and becomes two Gaussians. Why is that?










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$endgroup$












  • $begingroup$
    It won’t do that if you increase it to beyond n = 30 or so ... just my suspicion and more succinct version / restating of the accepted answer below.
    $endgroup$
    – oemb1905
    5 hours ago
















18












$begingroup$


Let say I have following numbers:



4,3,5,6,5,3,4,2,5,4,3,6,5


I sample some of them, say, 5 of them, and calculate the sum of 5 samples.
Then I repeat that over and over to get many sums, and I plot the values of sums in a histogram, which will be Gaussian due to the Central Limit Theorem.



But when they are following numbers, I just replaced 4 with some big number:



4,3,5,6,5,3,10000000,2,5,4,3,6,5


Sampling sums of 5 samples from these never becomes Gaussian in histogram, but more like a split and becomes two Gaussians. Why is that?










share|cite|improve this question











$endgroup$












  • $begingroup$
    It won’t do that if you increase it to beyond n = 30 or so ... just my suspicion and more succinct version / restating of the accepted answer below.
    $endgroup$
    – oemb1905
    5 hours ago














18












18








18


6



$begingroup$


Let say I have following numbers:



4,3,5,6,5,3,4,2,5,4,3,6,5


I sample some of them, say, 5 of them, and calculate the sum of 5 samples.
Then I repeat that over and over to get many sums, and I plot the values of sums in a histogram, which will be Gaussian due to the Central Limit Theorem.



But when they are following numbers, I just replaced 4 with some big number:



4,3,5,6,5,3,10000000,2,5,4,3,6,5


Sampling sums of 5 samples from these never becomes Gaussian in histogram, but more like a split and becomes two Gaussians. Why is that?










share|cite|improve this question











$endgroup$




Let say I have following numbers:



4,3,5,6,5,3,4,2,5,4,3,6,5


I sample some of them, say, 5 of them, and calculate the sum of 5 samples.
Then I repeat that over and over to get many sums, and I plot the values of sums in a histogram, which will be Gaussian due to the Central Limit Theorem.



But when they are following numbers, I just replaced 4 with some big number:



4,3,5,6,5,3,10000000,2,5,4,3,6,5


Sampling sums of 5 samples from these never becomes Gaussian in histogram, but more like a split and becomes two Gaussians. Why is that?







central-limit-theorem






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share|cite|improve this question













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edited 16 hours ago









amoeba

60.8k15205263




60.8k15205263










asked yesterday









JimSDJimSD

1087




1087












  • $begingroup$
    It won’t do that if you increase it to beyond n = 30 or so ... just my suspicion and more succinct version / restating of the accepted answer below.
    $endgroup$
    – oemb1905
    5 hours ago


















  • $begingroup$
    It won’t do that if you increase it to beyond n = 30 or so ... just my suspicion and more succinct version / restating of the accepted answer below.
    $endgroup$
    – oemb1905
    5 hours ago
















$begingroup$
It won’t do that if you increase it to beyond n = 30 or so ... just my suspicion and more succinct version / restating of the accepted answer below.
$endgroup$
– oemb1905
5 hours ago




$begingroup$
It won’t do that if you increase it to beyond n = 30 or so ... just my suspicion and more succinct version / restating of the accepted answer below.
$endgroup$
– oemb1905
5 hours ago










4 Answers
4






active

oldest

votes


















13












$begingroup$

Let's recall, precisely, what the central limit theorem says.




If $X_1, X_2, cdots, X_k$ are independent and identically distributed random variables, then $frac{X_1 + X_2 + cdots + X_k}{k}$ converges in distribution to a normal distribution (*).




When we have a static list of numbers like



4,3,5,6,5,3,10000000,2,5,4,3,6,5


and we are sampling by taking a number at random from this list, to apply the central limit theorem we need to be sure that our sampling scheme satisfies these two conditions of independence and identically distributed.




  • Identically distributed is no problem: each number in the list is equally likely to be chosen.

  • Independent is more subtle, and depends on our sampling scheme. If we are sampling without replacement, then we violate independence. It is only when we sample with replacement that the central limit theorem is applicable.


So, if we use with replacement sampling in your scheme, then we should be able to apply the central limit theorem. At the same time, you are right, if our sample is of size 5, then we are going to see very different behaviour depending on if the very large number is chosen, or not chosen in our sample.



So what's the rub? Well, the rate of convergence to a normal distribution is very dependent on the shape of the population we are sampling from, in particular, if our population is very skew, we expect it to take a long time to converge to the normal. This is the case in our example, so we should not expect that a sample of size 5 is sufficient to show the normal structure.



Three Normal Distributions



Above I repeated your experiment (with replacement sampling) for samples of size 5, 100, and 1000. You can see that the normal structure is emergent for very large samples.



(*) Note there are some technical conditions needed here, like finite mean and variance. They are easily verified to be true in our sampling from a list example.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thank you for a very quick and perfect answer. Idea of CLT, replacement, the need for more samples when data distribution is skewed,... It is very clear now. My original intention of question is, just as you mentioned, the case when one large number is included without replacement and the number of sampling is fixed. It behaves very differently, and therefore we need to consider "conditional" CLT for the case a large number is sampled and the case not sampled. I wonder if there is any research or prior work for that.. But thank you anyway.
    $endgroup$
    – JimSD
    22 hours ago






  • 6




    $begingroup$
    "f we are sampling with replacement, then we violate independence." I think you meant to write "without replacement" here? Especially as you then go on to sample with replacement to invoke the CLT.
    $endgroup$
    – syntonym
    17 hours ago










  • $begingroup$
    @syntonym Yup! Thanks for the edit!
    $endgroup$
    – Matthew Drury
    15 hours ago










  • $begingroup$
    " It is only when we sample without replacement that the central limit theorem is applicable". I think you wanted to write sample with replacement in the above line.
    $endgroup$
    – naive
    14 hours ago












  • $begingroup$
    Ahg! So many typos! Fixed.
    $endgroup$
    – Matthew Drury
    14 hours ago



















8












$begingroup$

In general, the size of each sample should be more than $5$ for the CLT approximation to be good. A rule of thumb is a sample of size $30$ or more. But, with the population of your first example, $5$ is OK.



pop <- c(4, 3, 5, 6, 5, 3, 4, 2, 5, 4, 3, 6, 5)
N <- 10^5
n <- 5
x <- matrix(sample(pop, size = N*n, replace = TRUE), nrow = N)
x_bar <- rowMeans(x)
hist(x_bar, freq = FALSE, col = "cyan")
f <- function(t) dnorm(t, mean = mean(pop), sd = sd(pop)/sqrt(n))
curve(f, add = TRUE, lwd = 2, col = "red")


enter image description here



In your second example, because of the shape of the population distribution (it's too much skewed; see guy's comment bellow), even samples of size $30$ won't give you a good approximation for the distribution of the sample mean using the CLT.



pop <- c(4, 3, 5, 6, 5, 3, 10000000, 2, 5, 4, 3, 6, 5)
N <- 10^5
n <- 30
x <- matrix(sample(pop, size = N*n, replace = TRUE), nrow = N)
x_bar <- rowMeans(x)
hist(x_bar, freq = FALSE, col = "cyan")
f <- function(t) dnorm(t, mean = mean(pop), sd = sd(pop)/sqrt(n))
curve(f, add = TRUE, lwd = 2, col = "red")


enter image description here



But, with this second population, samples of, say, size $100$ are fine.



pop <- c(4, 3, 5, 6, 5, 3, 10000000, 2, 5, 4, 3, 6, 5)
N <- 10^5
n <- 100
x <- matrix(sample(pop, size = N*n, replace = TRUE), nrow = N)
x_bar <- rowMeans(x)
hist(x_bar, freq = FALSE, col = "cyan")
f <- function(t) dnorm(t, mean = mean(pop), sd = sd(pop)/sqrt(n))
curve(f, add = TRUE, lwd = 2, col = "red")


enter image description here






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$endgroup$









  • 3




    $begingroup$
    It’s not the variance that is problem. One way of getting rigorous control is using the ratio of the third central moment to the standard deviation cubed, as in the Berry-Esseen theorem.
    $endgroup$
    – guy
    yesterday












  • $begingroup$
    Perfect. Added. Tks.
    $endgroup$
    – Paulo C. Marques F.
    yesterday










  • $begingroup$
    Thank you for a quick, visual, and perfect answer with a code. I was very surprised how quick it was! I was not aware of the appropriate number of sampling. I was thinking of the case where the number of sampling is fixed.
    $endgroup$
    – JimSD
    22 hours ago










  • $begingroup$
    @guy, Thank you for the that. I didn't know the idea of "the ratio of the third central moment to the standard deviation cubed in Berry-Esseen theorem". I just wish to tackle the case where there is one large number like outlier is included in distribution. And that kind of distribution can be refereed to as you mentioned, I suppose. If if you know any prior work dealing with that kind of distribution, let me know, thank you.
    $endgroup$
    – JimSD
    22 hours ago



















3












$begingroup$

I'd just like to explain, using complex cumulant-generating functions, why everyone keeps blaming this on skew.



Let's write the random variable you're sampling as $mu+sigma Z$, where $mu$ is the mean and $sigma$ the standard deviation so $Z$ has mean $0$ and variance $1$. The cumulant-generating function of $Z$ is $-frac{1}{2}t^2-frac{igamma_1}{6}t^3+o(t^3)$. Here $gamma_1$ denotes the skew of $Z$; we could write it in terms of the skew $kappa_3$ of the original variable $mu+sigma Z$, viz. $gamma_1=sigma^{-3}kappa_3$.



If we divide the sum of $n$ samples of $Z$'s distribution by $sqrt{n}$, the result has cgf $$nleft(-frac{1}{2}left(frac{t}{sqrt{n}}right)^2-frac{igamma_1}{6}left(frac{t}{sqrt{n}}right)^3right)+o(t^3)=-frac{1}{2}t^2-frac{igamma_1}{6sqrt{n}}t^3+o(t^3).$$For a Normal approximation to be valid at large enough $t$ for the graph to look right, we need sufficiently large $n$. This calculation motivates $nproptogamma_1^2$. The two samples you considered have very different values of $gamma_1$.






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    -1












    $begingroup$

    Short answer is, you don't have a big enough sample to make central limit theorem apply.






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    • 1




      $begingroup$
      That this cannot be a valid explanation is evident from the observation that the CLT gives a good approximation for the first set of data in the question, which is equally small.
      $endgroup$
      – whuber
      14 hours ago










    • $begingroup$
      @whuber: I think you are saying that the normal distribution gives a reasonably good approximation for a sample of five from the first set. Since there are only an finite number of values for the sums (13 possible values without replacement and 21 possible values with replacement), the approximation does not get much better with a large number of samples of five, and the initial approximation is more due to the initial pattern...
      $endgroup$
      – Henry
      13 hours ago










    • $begingroup$
      @whuber Since the distribution of the first set looks left skewed, I would expect the sum of five also to be left skewed, in a less extreme way than I would expect the sum of five from the second set to be right skewed. To get the skewness to reduce further, I would have thought that you would need a larger sample size
      $endgroup$
      – Henry
      13 hours ago






    • 1




      $begingroup$
      @Henry Thank you for your comments. I wasn't making a remark about these particular circumstances, but only about the logic of this answer, in the hope that it could be explained further.
      $endgroup$
      – whuber
      11 hours ago











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    4 Answers
    4






    active

    oldest

    votes








    4 Answers
    4






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    13












    $begingroup$

    Let's recall, precisely, what the central limit theorem says.




    If $X_1, X_2, cdots, X_k$ are independent and identically distributed random variables, then $frac{X_1 + X_2 + cdots + X_k}{k}$ converges in distribution to a normal distribution (*).




    When we have a static list of numbers like



    4,3,5,6,5,3,10000000,2,5,4,3,6,5


    and we are sampling by taking a number at random from this list, to apply the central limit theorem we need to be sure that our sampling scheme satisfies these two conditions of independence and identically distributed.




    • Identically distributed is no problem: each number in the list is equally likely to be chosen.

    • Independent is more subtle, and depends on our sampling scheme. If we are sampling without replacement, then we violate independence. It is only when we sample with replacement that the central limit theorem is applicable.


    So, if we use with replacement sampling in your scheme, then we should be able to apply the central limit theorem. At the same time, you are right, if our sample is of size 5, then we are going to see very different behaviour depending on if the very large number is chosen, or not chosen in our sample.



    So what's the rub? Well, the rate of convergence to a normal distribution is very dependent on the shape of the population we are sampling from, in particular, if our population is very skew, we expect it to take a long time to converge to the normal. This is the case in our example, so we should not expect that a sample of size 5 is sufficient to show the normal structure.



    Three Normal Distributions



    Above I repeated your experiment (with replacement sampling) for samples of size 5, 100, and 1000. You can see that the normal structure is emergent for very large samples.



    (*) Note there are some technical conditions needed here, like finite mean and variance. They are easily verified to be true in our sampling from a list example.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      Thank you for a very quick and perfect answer. Idea of CLT, replacement, the need for more samples when data distribution is skewed,... It is very clear now. My original intention of question is, just as you mentioned, the case when one large number is included without replacement and the number of sampling is fixed. It behaves very differently, and therefore we need to consider "conditional" CLT for the case a large number is sampled and the case not sampled. I wonder if there is any research or prior work for that.. But thank you anyway.
      $endgroup$
      – JimSD
      22 hours ago






    • 6




      $begingroup$
      "f we are sampling with replacement, then we violate independence." I think you meant to write "without replacement" here? Especially as you then go on to sample with replacement to invoke the CLT.
      $endgroup$
      – syntonym
      17 hours ago










    • $begingroup$
      @syntonym Yup! Thanks for the edit!
      $endgroup$
      – Matthew Drury
      15 hours ago










    • $begingroup$
      " It is only when we sample without replacement that the central limit theorem is applicable". I think you wanted to write sample with replacement in the above line.
      $endgroup$
      – naive
      14 hours ago












    • $begingroup$
      Ahg! So many typos! Fixed.
      $endgroup$
      – Matthew Drury
      14 hours ago
















    13












    $begingroup$

    Let's recall, precisely, what the central limit theorem says.




    If $X_1, X_2, cdots, X_k$ are independent and identically distributed random variables, then $frac{X_1 + X_2 + cdots + X_k}{k}$ converges in distribution to a normal distribution (*).




    When we have a static list of numbers like



    4,3,5,6,5,3,10000000,2,5,4,3,6,5


    and we are sampling by taking a number at random from this list, to apply the central limit theorem we need to be sure that our sampling scheme satisfies these two conditions of independence and identically distributed.




    • Identically distributed is no problem: each number in the list is equally likely to be chosen.

    • Independent is more subtle, and depends on our sampling scheme. If we are sampling without replacement, then we violate independence. It is only when we sample with replacement that the central limit theorem is applicable.


    So, if we use with replacement sampling in your scheme, then we should be able to apply the central limit theorem. At the same time, you are right, if our sample is of size 5, then we are going to see very different behaviour depending on if the very large number is chosen, or not chosen in our sample.



    So what's the rub? Well, the rate of convergence to a normal distribution is very dependent on the shape of the population we are sampling from, in particular, if our population is very skew, we expect it to take a long time to converge to the normal. This is the case in our example, so we should not expect that a sample of size 5 is sufficient to show the normal structure.



    Three Normal Distributions



    Above I repeated your experiment (with replacement sampling) for samples of size 5, 100, and 1000. You can see that the normal structure is emergent for very large samples.



    (*) Note there are some technical conditions needed here, like finite mean and variance. They are easily verified to be true in our sampling from a list example.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      Thank you for a very quick and perfect answer. Idea of CLT, replacement, the need for more samples when data distribution is skewed,... It is very clear now. My original intention of question is, just as you mentioned, the case when one large number is included without replacement and the number of sampling is fixed. It behaves very differently, and therefore we need to consider "conditional" CLT for the case a large number is sampled and the case not sampled. I wonder if there is any research or prior work for that.. But thank you anyway.
      $endgroup$
      – JimSD
      22 hours ago






    • 6




      $begingroup$
      "f we are sampling with replacement, then we violate independence." I think you meant to write "without replacement" here? Especially as you then go on to sample with replacement to invoke the CLT.
      $endgroup$
      – syntonym
      17 hours ago










    • $begingroup$
      @syntonym Yup! Thanks for the edit!
      $endgroup$
      – Matthew Drury
      15 hours ago










    • $begingroup$
      " It is only when we sample without replacement that the central limit theorem is applicable". I think you wanted to write sample with replacement in the above line.
      $endgroup$
      – naive
      14 hours ago












    • $begingroup$
      Ahg! So many typos! Fixed.
      $endgroup$
      – Matthew Drury
      14 hours ago














    13












    13








    13





    $begingroup$

    Let's recall, precisely, what the central limit theorem says.




    If $X_1, X_2, cdots, X_k$ are independent and identically distributed random variables, then $frac{X_1 + X_2 + cdots + X_k}{k}$ converges in distribution to a normal distribution (*).




    When we have a static list of numbers like



    4,3,5,6,5,3,10000000,2,5,4,3,6,5


    and we are sampling by taking a number at random from this list, to apply the central limit theorem we need to be sure that our sampling scheme satisfies these two conditions of independence and identically distributed.




    • Identically distributed is no problem: each number in the list is equally likely to be chosen.

    • Independent is more subtle, and depends on our sampling scheme. If we are sampling without replacement, then we violate independence. It is only when we sample with replacement that the central limit theorem is applicable.


    So, if we use with replacement sampling in your scheme, then we should be able to apply the central limit theorem. At the same time, you are right, if our sample is of size 5, then we are going to see very different behaviour depending on if the very large number is chosen, or not chosen in our sample.



    So what's the rub? Well, the rate of convergence to a normal distribution is very dependent on the shape of the population we are sampling from, in particular, if our population is very skew, we expect it to take a long time to converge to the normal. This is the case in our example, so we should not expect that a sample of size 5 is sufficient to show the normal structure.



    Three Normal Distributions



    Above I repeated your experiment (with replacement sampling) for samples of size 5, 100, and 1000. You can see that the normal structure is emergent for very large samples.



    (*) Note there are some technical conditions needed here, like finite mean and variance. They are easily verified to be true in our sampling from a list example.






    share|cite|improve this answer











    $endgroup$



    Let's recall, precisely, what the central limit theorem says.




    If $X_1, X_2, cdots, X_k$ are independent and identically distributed random variables, then $frac{X_1 + X_2 + cdots + X_k}{k}$ converges in distribution to a normal distribution (*).




    When we have a static list of numbers like



    4,3,5,6,5,3,10000000,2,5,4,3,6,5


    and we are sampling by taking a number at random from this list, to apply the central limit theorem we need to be sure that our sampling scheme satisfies these two conditions of independence and identically distributed.




    • Identically distributed is no problem: each number in the list is equally likely to be chosen.

    • Independent is more subtle, and depends on our sampling scheme. If we are sampling without replacement, then we violate independence. It is only when we sample with replacement that the central limit theorem is applicable.


    So, if we use with replacement sampling in your scheme, then we should be able to apply the central limit theorem. At the same time, you are right, if our sample is of size 5, then we are going to see very different behaviour depending on if the very large number is chosen, or not chosen in our sample.



    So what's the rub? Well, the rate of convergence to a normal distribution is very dependent on the shape of the population we are sampling from, in particular, if our population is very skew, we expect it to take a long time to converge to the normal. This is the case in our example, so we should not expect that a sample of size 5 is sufficient to show the normal structure.



    Three Normal Distributions



    Above I repeated your experiment (with replacement sampling) for samples of size 5, 100, and 1000. You can see that the normal structure is emergent for very large samples.



    (*) Note there are some technical conditions needed here, like finite mean and variance. They are easily verified to be true in our sampling from a list example.







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited 14 hours ago

























    answered yesterday









    Matthew DruryMatthew Drury

    26.1k264106




    26.1k264106












    • $begingroup$
      Thank you for a very quick and perfect answer. Idea of CLT, replacement, the need for more samples when data distribution is skewed,... It is very clear now. My original intention of question is, just as you mentioned, the case when one large number is included without replacement and the number of sampling is fixed. It behaves very differently, and therefore we need to consider "conditional" CLT for the case a large number is sampled and the case not sampled. I wonder if there is any research or prior work for that.. But thank you anyway.
      $endgroup$
      – JimSD
      22 hours ago






    • 6




      $begingroup$
      "f we are sampling with replacement, then we violate independence." I think you meant to write "without replacement" here? Especially as you then go on to sample with replacement to invoke the CLT.
      $endgroup$
      – syntonym
      17 hours ago










    • $begingroup$
      @syntonym Yup! Thanks for the edit!
      $endgroup$
      – Matthew Drury
      15 hours ago










    • $begingroup$
      " It is only when we sample without replacement that the central limit theorem is applicable". I think you wanted to write sample with replacement in the above line.
      $endgroup$
      – naive
      14 hours ago












    • $begingroup$
      Ahg! So many typos! Fixed.
      $endgroup$
      – Matthew Drury
      14 hours ago


















    • $begingroup$
      Thank you for a very quick and perfect answer. Idea of CLT, replacement, the need for more samples when data distribution is skewed,... It is very clear now. My original intention of question is, just as you mentioned, the case when one large number is included without replacement and the number of sampling is fixed. It behaves very differently, and therefore we need to consider "conditional" CLT for the case a large number is sampled and the case not sampled. I wonder if there is any research or prior work for that.. But thank you anyway.
      $endgroup$
      – JimSD
      22 hours ago






    • 6




      $begingroup$
      "f we are sampling with replacement, then we violate independence." I think you meant to write "without replacement" here? Especially as you then go on to sample with replacement to invoke the CLT.
      $endgroup$
      – syntonym
      17 hours ago










    • $begingroup$
      @syntonym Yup! Thanks for the edit!
      $endgroup$
      – Matthew Drury
      15 hours ago










    • $begingroup$
      " It is only when we sample without replacement that the central limit theorem is applicable". I think you wanted to write sample with replacement in the above line.
      $endgroup$
      – naive
      14 hours ago












    • $begingroup$
      Ahg! So many typos! Fixed.
      $endgroup$
      – Matthew Drury
      14 hours ago
















    $begingroup$
    Thank you for a very quick and perfect answer. Idea of CLT, replacement, the need for more samples when data distribution is skewed,... It is very clear now. My original intention of question is, just as you mentioned, the case when one large number is included without replacement and the number of sampling is fixed. It behaves very differently, and therefore we need to consider "conditional" CLT for the case a large number is sampled and the case not sampled. I wonder if there is any research or prior work for that.. But thank you anyway.
    $endgroup$
    – JimSD
    22 hours ago




    $begingroup$
    Thank you for a very quick and perfect answer. Idea of CLT, replacement, the need for more samples when data distribution is skewed,... It is very clear now. My original intention of question is, just as you mentioned, the case when one large number is included without replacement and the number of sampling is fixed. It behaves very differently, and therefore we need to consider "conditional" CLT for the case a large number is sampled and the case not sampled. I wonder if there is any research or prior work for that.. But thank you anyway.
    $endgroup$
    – JimSD
    22 hours ago




    6




    6




    $begingroup$
    "f we are sampling with replacement, then we violate independence." I think you meant to write "without replacement" here? Especially as you then go on to sample with replacement to invoke the CLT.
    $endgroup$
    – syntonym
    17 hours ago




    $begingroup$
    "f we are sampling with replacement, then we violate independence." I think you meant to write "without replacement" here? Especially as you then go on to sample with replacement to invoke the CLT.
    $endgroup$
    – syntonym
    17 hours ago












    $begingroup$
    @syntonym Yup! Thanks for the edit!
    $endgroup$
    – Matthew Drury
    15 hours ago




    $begingroup$
    @syntonym Yup! Thanks for the edit!
    $endgroup$
    – Matthew Drury
    15 hours ago












    $begingroup$
    " It is only when we sample without replacement that the central limit theorem is applicable". I think you wanted to write sample with replacement in the above line.
    $endgroup$
    – naive
    14 hours ago






    $begingroup$
    " It is only when we sample without replacement that the central limit theorem is applicable". I think you wanted to write sample with replacement in the above line.
    $endgroup$
    – naive
    14 hours ago














    $begingroup$
    Ahg! So many typos! Fixed.
    $endgroup$
    – Matthew Drury
    14 hours ago




    $begingroup$
    Ahg! So many typos! Fixed.
    $endgroup$
    – Matthew Drury
    14 hours ago













    8












    $begingroup$

    In general, the size of each sample should be more than $5$ for the CLT approximation to be good. A rule of thumb is a sample of size $30$ or more. But, with the population of your first example, $5$ is OK.



    pop <- c(4, 3, 5, 6, 5, 3, 4, 2, 5, 4, 3, 6, 5)
    N <- 10^5
    n <- 5
    x <- matrix(sample(pop, size = N*n, replace = TRUE), nrow = N)
    x_bar <- rowMeans(x)
    hist(x_bar, freq = FALSE, col = "cyan")
    f <- function(t) dnorm(t, mean = mean(pop), sd = sd(pop)/sqrt(n))
    curve(f, add = TRUE, lwd = 2, col = "red")


    enter image description here



    In your second example, because of the shape of the population distribution (it's too much skewed; see guy's comment bellow), even samples of size $30$ won't give you a good approximation for the distribution of the sample mean using the CLT.



    pop <- c(4, 3, 5, 6, 5, 3, 10000000, 2, 5, 4, 3, 6, 5)
    N <- 10^5
    n <- 30
    x <- matrix(sample(pop, size = N*n, replace = TRUE), nrow = N)
    x_bar <- rowMeans(x)
    hist(x_bar, freq = FALSE, col = "cyan")
    f <- function(t) dnorm(t, mean = mean(pop), sd = sd(pop)/sqrt(n))
    curve(f, add = TRUE, lwd = 2, col = "red")


    enter image description here



    But, with this second population, samples of, say, size $100$ are fine.



    pop <- c(4, 3, 5, 6, 5, 3, 10000000, 2, 5, 4, 3, 6, 5)
    N <- 10^5
    n <- 100
    x <- matrix(sample(pop, size = N*n, replace = TRUE), nrow = N)
    x_bar <- rowMeans(x)
    hist(x_bar, freq = FALSE, col = "cyan")
    f <- function(t) dnorm(t, mean = mean(pop), sd = sd(pop)/sqrt(n))
    curve(f, add = TRUE, lwd = 2, col = "red")


    enter image description here






    share|cite|improve this answer











    $endgroup$









    • 3




      $begingroup$
      It’s not the variance that is problem. One way of getting rigorous control is using the ratio of the third central moment to the standard deviation cubed, as in the Berry-Esseen theorem.
      $endgroup$
      – guy
      yesterday












    • $begingroup$
      Perfect. Added. Tks.
      $endgroup$
      – Paulo C. Marques F.
      yesterday










    • $begingroup$
      Thank you for a quick, visual, and perfect answer with a code. I was very surprised how quick it was! I was not aware of the appropriate number of sampling. I was thinking of the case where the number of sampling is fixed.
      $endgroup$
      – JimSD
      22 hours ago










    • $begingroup$
      @guy, Thank you for the that. I didn't know the idea of "the ratio of the third central moment to the standard deviation cubed in Berry-Esseen theorem". I just wish to tackle the case where there is one large number like outlier is included in distribution. And that kind of distribution can be refereed to as you mentioned, I suppose. If if you know any prior work dealing with that kind of distribution, let me know, thank you.
      $endgroup$
      – JimSD
      22 hours ago
















    8












    $begingroup$

    In general, the size of each sample should be more than $5$ for the CLT approximation to be good. A rule of thumb is a sample of size $30$ or more. But, with the population of your first example, $5$ is OK.



    pop <- c(4, 3, 5, 6, 5, 3, 4, 2, 5, 4, 3, 6, 5)
    N <- 10^5
    n <- 5
    x <- matrix(sample(pop, size = N*n, replace = TRUE), nrow = N)
    x_bar <- rowMeans(x)
    hist(x_bar, freq = FALSE, col = "cyan")
    f <- function(t) dnorm(t, mean = mean(pop), sd = sd(pop)/sqrt(n))
    curve(f, add = TRUE, lwd = 2, col = "red")


    enter image description here



    In your second example, because of the shape of the population distribution (it's too much skewed; see guy's comment bellow), even samples of size $30$ won't give you a good approximation for the distribution of the sample mean using the CLT.



    pop <- c(4, 3, 5, 6, 5, 3, 10000000, 2, 5, 4, 3, 6, 5)
    N <- 10^5
    n <- 30
    x <- matrix(sample(pop, size = N*n, replace = TRUE), nrow = N)
    x_bar <- rowMeans(x)
    hist(x_bar, freq = FALSE, col = "cyan")
    f <- function(t) dnorm(t, mean = mean(pop), sd = sd(pop)/sqrt(n))
    curve(f, add = TRUE, lwd = 2, col = "red")


    enter image description here



    But, with this second population, samples of, say, size $100$ are fine.



    pop <- c(4, 3, 5, 6, 5, 3, 10000000, 2, 5, 4, 3, 6, 5)
    N <- 10^5
    n <- 100
    x <- matrix(sample(pop, size = N*n, replace = TRUE), nrow = N)
    x_bar <- rowMeans(x)
    hist(x_bar, freq = FALSE, col = "cyan")
    f <- function(t) dnorm(t, mean = mean(pop), sd = sd(pop)/sqrt(n))
    curve(f, add = TRUE, lwd = 2, col = "red")


    enter image description here






    share|cite|improve this answer











    $endgroup$









    • 3




      $begingroup$
      It’s not the variance that is problem. One way of getting rigorous control is using the ratio of the third central moment to the standard deviation cubed, as in the Berry-Esseen theorem.
      $endgroup$
      – guy
      yesterday












    • $begingroup$
      Perfect. Added. Tks.
      $endgroup$
      – Paulo C. Marques F.
      yesterday










    • $begingroup$
      Thank you for a quick, visual, and perfect answer with a code. I was very surprised how quick it was! I was not aware of the appropriate number of sampling. I was thinking of the case where the number of sampling is fixed.
      $endgroup$
      – JimSD
      22 hours ago










    • $begingroup$
      @guy, Thank you for the that. I didn't know the idea of "the ratio of the third central moment to the standard deviation cubed in Berry-Esseen theorem". I just wish to tackle the case where there is one large number like outlier is included in distribution. And that kind of distribution can be refereed to as you mentioned, I suppose. If if you know any prior work dealing with that kind of distribution, let me know, thank you.
      $endgroup$
      – JimSD
      22 hours ago














    8












    8








    8





    $begingroup$

    In general, the size of each sample should be more than $5$ for the CLT approximation to be good. A rule of thumb is a sample of size $30$ or more. But, with the population of your first example, $5$ is OK.



    pop <- c(4, 3, 5, 6, 5, 3, 4, 2, 5, 4, 3, 6, 5)
    N <- 10^5
    n <- 5
    x <- matrix(sample(pop, size = N*n, replace = TRUE), nrow = N)
    x_bar <- rowMeans(x)
    hist(x_bar, freq = FALSE, col = "cyan")
    f <- function(t) dnorm(t, mean = mean(pop), sd = sd(pop)/sqrt(n))
    curve(f, add = TRUE, lwd = 2, col = "red")


    enter image description here



    In your second example, because of the shape of the population distribution (it's too much skewed; see guy's comment bellow), even samples of size $30$ won't give you a good approximation for the distribution of the sample mean using the CLT.



    pop <- c(4, 3, 5, 6, 5, 3, 10000000, 2, 5, 4, 3, 6, 5)
    N <- 10^5
    n <- 30
    x <- matrix(sample(pop, size = N*n, replace = TRUE), nrow = N)
    x_bar <- rowMeans(x)
    hist(x_bar, freq = FALSE, col = "cyan")
    f <- function(t) dnorm(t, mean = mean(pop), sd = sd(pop)/sqrt(n))
    curve(f, add = TRUE, lwd = 2, col = "red")


    enter image description here



    But, with this second population, samples of, say, size $100$ are fine.



    pop <- c(4, 3, 5, 6, 5, 3, 10000000, 2, 5, 4, 3, 6, 5)
    N <- 10^5
    n <- 100
    x <- matrix(sample(pop, size = N*n, replace = TRUE), nrow = N)
    x_bar <- rowMeans(x)
    hist(x_bar, freq = FALSE, col = "cyan")
    f <- function(t) dnorm(t, mean = mean(pop), sd = sd(pop)/sqrt(n))
    curve(f, add = TRUE, lwd = 2, col = "red")


    enter image description here






    share|cite|improve this answer











    $endgroup$



    In general, the size of each sample should be more than $5$ for the CLT approximation to be good. A rule of thumb is a sample of size $30$ or more. But, with the population of your first example, $5$ is OK.



    pop <- c(4, 3, 5, 6, 5, 3, 4, 2, 5, 4, 3, 6, 5)
    N <- 10^5
    n <- 5
    x <- matrix(sample(pop, size = N*n, replace = TRUE), nrow = N)
    x_bar <- rowMeans(x)
    hist(x_bar, freq = FALSE, col = "cyan")
    f <- function(t) dnorm(t, mean = mean(pop), sd = sd(pop)/sqrt(n))
    curve(f, add = TRUE, lwd = 2, col = "red")


    enter image description here



    In your second example, because of the shape of the population distribution (it's too much skewed; see guy's comment bellow), even samples of size $30$ won't give you a good approximation for the distribution of the sample mean using the CLT.



    pop <- c(4, 3, 5, 6, 5, 3, 10000000, 2, 5, 4, 3, 6, 5)
    N <- 10^5
    n <- 30
    x <- matrix(sample(pop, size = N*n, replace = TRUE), nrow = N)
    x_bar <- rowMeans(x)
    hist(x_bar, freq = FALSE, col = "cyan")
    f <- function(t) dnorm(t, mean = mean(pop), sd = sd(pop)/sqrt(n))
    curve(f, add = TRUE, lwd = 2, col = "red")


    enter image description here



    But, with this second population, samples of, say, size $100$ are fine.



    pop <- c(4, 3, 5, 6, 5, 3, 10000000, 2, 5, 4, 3, 6, 5)
    N <- 10^5
    n <- 100
    x <- matrix(sample(pop, size = N*n, replace = TRUE), nrow = N)
    x_bar <- rowMeans(x)
    hist(x_bar, freq = FALSE, col = "cyan")
    f <- function(t) dnorm(t, mean = mean(pop), sd = sd(pop)/sqrt(n))
    curve(f, add = TRUE, lwd = 2, col = "red")


    enter image description here







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited 12 hours ago

























    answered yesterday









    Paulo C. Marques F.Paulo C. Marques F.

    17.2k35497




    17.2k35497








    • 3




      $begingroup$
      It’s not the variance that is problem. One way of getting rigorous control is using the ratio of the third central moment to the standard deviation cubed, as in the Berry-Esseen theorem.
      $endgroup$
      – guy
      yesterday












    • $begingroup$
      Perfect. Added. Tks.
      $endgroup$
      – Paulo C. Marques F.
      yesterday










    • $begingroup$
      Thank you for a quick, visual, and perfect answer with a code. I was very surprised how quick it was! I was not aware of the appropriate number of sampling. I was thinking of the case where the number of sampling is fixed.
      $endgroup$
      – JimSD
      22 hours ago










    • $begingroup$
      @guy, Thank you for the that. I didn't know the idea of "the ratio of the third central moment to the standard deviation cubed in Berry-Esseen theorem". I just wish to tackle the case where there is one large number like outlier is included in distribution. And that kind of distribution can be refereed to as you mentioned, I suppose. If if you know any prior work dealing with that kind of distribution, let me know, thank you.
      $endgroup$
      – JimSD
      22 hours ago














    • 3




      $begingroup$
      It’s not the variance that is problem. One way of getting rigorous control is using the ratio of the third central moment to the standard deviation cubed, as in the Berry-Esseen theorem.
      $endgroup$
      – guy
      yesterday












    • $begingroup$
      Perfect. Added. Tks.
      $endgroup$
      – Paulo C. Marques F.
      yesterday










    • $begingroup$
      Thank you for a quick, visual, and perfect answer with a code. I was very surprised how quick it was! I was not aware of the appropriate number of sampling. I was thinking of the case where the number of sampling is fixed.
      $endgroup$
      – JimSD
      22 hours ago










    • $begingroup$
      @guy, Thank you for the that. I didn't know the idea of "the ratio of the third central moment to the standard deviation cubed in Berry-Esseen theorem". I just wish to tackle the case where there is one large number like outlier is included in distribution. And that kind of distribution can be refereed to as you mentioned, I suppose. If if you know any prior work dealing with that kind of distribution, let me know, thank you.
      $endgroup$
      – JimSD
      22 hours ago








    3




    3




    $begingroup$
    It’s not the variance that is problem. One way of getting rigorous control is using the ratio of the third central moment to the standard deviation cubed, as in the Berry-Esseen theorem.
    $endgroup$
    – guy
    yesterday






    $begingroup$
    It’s not the variance that is problem. One way of getting rigorous control is using the ratio of the third central moment to the standard deviation cubed, as in the Berry-Esseen theorem.
    $endgroup$
    – guy
    yesterday














    $begingroup$
    Perfect. Added. Tks.
    $endgroup$
    – Paulo C. Marques F.
    yesterday




    $begingroup$
    Perfect. Added. Tks.
    $endgroup$
    – Paulo C. Marques F.
    yesterday












    $begingroup$
    Thank you for a quick, visual, and perfect answer with a code. I was very surprised how quick it was! I was not aware of the appropriate number of sampling. I was thinking of the case where the number of sampling is fixed.
    $endgroup$
    – JimSD
    22 hours ago




    $begingroup$
    Thank you for a quick, visual, and perfect answer with a code. I was very surprised how quick it was! I was not aware of the appropriate number of sampling. I was thinking of the case where the number of sampling is fixed.
    $endgroup$
    – JimSD
    22 hours ago












    $begingroup$
    @guy, Thank you for the that. I didn't know the idea of "the ratio of the third central moment to the standard deviation cubed in Berry-Esseen theorem". I just wish to tackle the case where there is one large number like outlier is included in distribution. And that kind of distribution can be refereed to as you mentioned, I suppose. If if you know any prior work dealing with that kind of distribution, let me know, thank you.
    $endgroup$
    – JimSD
    22 hours ago




    $begingroup$
    @guy, Thank you for the that. I didn't know the idea of "the ratio of the third central moment to the standard deviation cubed in Berry-Esseen theorem". I just wish to tackle the case where there is one large number like outlier is included in distribution. And that kind of distribution can be refereed to as you mentioned, I suppose. If if you know any prior work dealing with that kind of distribution, let me know, thank you.
    $endgroup$
    – JimSD
    22 hours ago











    3












    $begingroup$

    I'd just like to explain, using complex cumulant-generating functions, why everyone keeps blaming this on skew.



    Let's write the random variable you're sampling as $mu+sigma Z$, where $mu$ is the mean and $sigma$ the standard deviation so $Z$ has mean $0$ and variance $1$. The cumulant-generating function of $Z$ is $-frac{1}{2}t^2-frac{igamma_1}{6}t^3+o(t^3)$. Here $gamma_1$ denotes the skew of $Z$; we could write it in terms of the skew $kappa_3$ of the original variable $mu+sigma Z$, viz. $gamma_1=sigma^{-3}kappa_3$.



    If we divide the sum of $n$ samples of $Z$'s distribution by $sqrt{n}$, the result has cgf $$nleft(-frac{1}{2}left(frac{t}{sqrt{n}}right)^2-frac{igamma_1}{6}left(frac{t}{sqrt{n}}right)^3right)+o(t^3)=-frac{1}{2}t^2-frac{igamma_1}{6sqrt{n}}t^3+o(t^3).$$For a Normal approximation to be valid at large enough $t$ for the graph to look right, we need sufficiently large $n$. This calculation motivates $nproptogamma_1^2$. The two samples you considered have very different values of $gamma_1$.






    share|cite|improve this answer









    $endgroup$


















      3












      $begingroup$

      I'd just like to explain, using complex cumulant-generating functions, why everyone keeps blaming this on skew.



      Let's write the random variable you're sampling as $mu+sigma Z$, where $mu$ is the mean and $sigma$ the standard deviation so $Z$ has mean $0$ and variance $1$. The cumulant-generating function of $Z$ is $-frac{1}{2}t^2-frac{igamma_1}{6}t^3+o(t^3)$. Here $gamma_1$ denotes the skew of $Z$; we could write it in terms of the skew $kappa_3$ of the original variable $mu+sigma Z$, viz. $gamma_1=sigma^{-3}kappa_3$.



      If we divide the sum of $n$ samples of $Z$'s distribution by $sqrt{n}$, the result has cgf $$nleft(-frac{1}{2}left(frac{t}{sqrt{n}}right)^2-frac{igamma_1}{6}left(frac{t}{sqrt{n}}right)^3right)+o(t^3)=-frac{1}{2}t^2-frac{igamma_1}{6sqrt{n}}t^3+o(t^3).$$For a Normal approximation to be valid at large enough $t$ for the graph to look right, we need sufficiently large $n$. This calculation motivates $nproptogamma_1^2$. The two samples you considered have very different values of $gamma_1$.






      share|cite|improve this answer









      $endgroup$
















        3












        3








        3





        $begingroup$

        I'd just like to explain, using complex cumulant-generating functions, why everyone keeps blaming this on skew.



        Let's write the random variable you're sampling as $mu+sigma Z$, where $mu$ is the mean and $sigma$ the standard deviation so $Z$ has mean $0$ and variance $1$. The cumulant-generating function of $Z$ is $-frac{1}{2}t^2-frac{igamma_1}{6}t^3+o(t^3)$. Here $gamma_1$ denotes the skew of $Z$; we could write it in terms of the skew $kappa_3$ of the original variable $mu+sigma Z$, viz. $gamma_1=sigma^{-3}kappa_3$.



        If we divide the sum of $n$ samples of $Z$'s distribution by $sqrt{n}$, the result has cgf $$nleft(-frac{1}{2}left(frac{t}{sqrt{n}}right)^2-frac{igamma_1}{6}left(frac{t}{sqrt{n}}right)^3right)+o(t^3)=-frac{1}{2}t^2-frac{igamma_1}{6sqrt{n}}t^3+o(t^3).$$For a Normal approximation to be valid at large enough $t$ for the graph to look right, we need sufficiently large $n$. This calculation motivates $nproptogamma_1^2$. The two samples you considered have very different values of $gamma_1$.






        share|cite|improve this answer









        $endgroup$



        I'd just like to explain, using complex cumulant-generating functions, why everyone keeps blaming this on skew.



        Let's write the random variable you're sampling as $mu+sigma Z$, where $mu$ is the mean and $sigma$ the standard deviation so $Z$ has mean $0$ and variance $1$. The cumulant-generating function of $Z$ is $-frac{1}{2}t^2-frac{igamma_1}{6}t^3+o(t^3)$. Here $gamma_1$ denotes the skew of $Z$; we could write it in terms of the skew $kappa_3$ of the original variable $mu+sigma Z$, viz. $gamma_1=sigma^{-3}kappa_3$.



        If we divide the sum of $n$ samples of $Z$'s distribution by $sqrt{n}$, the result has cgf $$nleft(-frac{1}{2}left(frac{t}{sqrt{n}}right)^2-frac{igamma_1}{6}left(frac{t}{sqrt{n}}right)^3right)+o(t^3)=-frac{1}{2}t^2-frac{igamma_1}{6sqrt{n}}t^3+o(t^3).$$For a Normal approximation to be valid at large enough $t$ for the graph to look right, we need sufficiently large $n$. This calculation motivates $nproptogamma_1^2$. The two samples you considered have very different values of $gamma_1$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered 14 hours ago









        J.G.J.G.

        20616




        20616























            -1












            $begingroup$

            Short answer is, you don't have a big enough sample to make central limit theorem apply.






            share|cite|improve this answer








            New contributor




            feynman is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
            Check out our Code of Conduct.






            $endgroup$









            • 1




              $begingroup$
              That this cannot be a valid explanation is evident from the observation that the CLT gives a good approximation for the first set of data in the question, which is equally small.
              $endgroup$
              – whuber
              14 hours ago










            • $begingroup$
              @whuber: I think you are saying that the normal distribution gives a reasonably good approximation for a sample of five from the first set. Since there are only an finite number of values for the sums (13 possible values without replacement and 21 possible values with replacement), the approximation does not get much better with a large number of samples of five, and the initial approximation is more due to the initial pattern...
              $endgroup$
              – Henry
              13 hours ago










            • $begingroup$
              @whuber Since the distribution of the first set looks left skewed, I would expect the sum of five also to be left skewed, in a less extreme way than I would expect the sum of five from the second set to be right skewed. To get the skewness to reduce further, I would have thought that you would need a larger sample size
              $endgroup$
              – Henry
              13 hours ago






            • 1




              $begingroup$
              @Henry Thank you for your comments. I wasn't making a remark about these particular circumstances, but only about the logic of this answer, in the hope that it could be explained further.
              $endgroup$
              – whuber
              11 hours ago
















            -1












            $begingroup$

            Short answer is, you don't have a big enough sample to make central limit theorem apply.






            share|cite|improve this answer








            New contributor




            feynman is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
            Check out our Code of Conduct.






            $endgroup$









            • 1




              $begingroup$
              That this cannot be a valid explanation is evident from the observation that the CLT gives a good approximation for the first set of data in the question, which is equally small.
              $endgroup$
              – whuber
              14 hours ago










            • $begingroup$
              @whuber: I think you are saying that the normal distribution gives a reasonably good approximation for a sample of five from the first set. Since there are only an finite number of values for the sums (13 possible values without replacement and 21 possible values with replacement), the approximation does not get much better with a large number of samples of five, and the initial approximation is more due to the initial pattern...
              $endgroup$
              – Henry
              13 hours ago










            • $begingroup$
              @whuber Since the distribution of the first set looks left skewed, I would expect the sum of five also to be left skewed, in a less extreme way than I would expect the sum of five from the second set to be right skewed. To get the skewness to reduce further, I would have thought that you would need a larger sample size
              $endgroup$
              – Henry
              13 hours ago






            • 1




              $begingroup$
              @Henry Thank you for your comments. I wasn't making a remark about these particular circumstances, but only about the logic of this answer, in the hope that it could be explained further.
              $endgroup$
              – whuber
              11 hours ago














            -1












            -1








            -1





            $begingroup$

            Short answer is, you don't have a big enough sample to make central limit theorem apply.






            share|cite|improve this answer








            New contributor




            feynman is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
            Check out our Code of Conduct.






            $endgroup$



            Short answer is, you don't have a big enough sample to make central limit theorem apply.







            share|cite|improve this answer








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            answered 16 hours ago









            feynmanfeynman

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            • 1




              $begingroup$
              That this cannot be a valid explanation is evident from the observation that the CLT gives a good approximation for the first set of data in the question, which is equally small.
              $endgroup$
              – whuber
              14 hours ago










            • $begingroup$
              @whuber: I think you are saying that the normal distribution gives a reasonably good approximation for a sample of five from the first set. Since there are only an finite number of values for the sums (13 possible values without replacement and 21 possible values with replacement), the approximation does not get much better with a large number of samples of five, and the initial approximation is more due to the initial pattern...
              $endgroup$
              – Henry
              13 hours ago










            • $begingroup$
              @whuber Since the distribution of the first set looks left skewed, I would expect the sum of five also to be left skewed, in a less extreme way than I would expect the sum of five from the second set to be right skewed. To get the skewness to reduce further, I would have thought that you would need a larger sample size
              $endgroup$
              – Henry
              13 hours ago






            • 1




              $begingroup$
              @Henry Thank you for your comments. I wasn't making a remark about these particular circumstances, but only about the logic of this answer, in the hope that it could be explained further.
              $endgroup$
              – whuber
              11 hours ago














            • 1




              $begingroup$
              That this cannot be a valid explanation is evident from the observation that the CLT gives a good approximation for the first set of data in the question, which is equally small.
              $endgroup$
              – whuber
              14 hours ago










            • $begingroup$
              @whuber: I think you are saying that the normal distribution gives a reasonably good approximation for a sample of five from the first set. Since there are only an finite number of values for the sums (13 possible values without replacement and 21 possible values with replacement), the approximation does not get much better with a large number of samples of five, and the initial approximation is more due to the initial pattern...
              $endgroup$
              – Henry
              13 hours ago










            • $begingroup$
              @whuber Since the distribution of the first set looks left skewed, I would expect the sum of five also to be left skewed, in a less extreme way than I would expect the sum of five from the second set to be right skewed. To get the skewness to reduce further, I would have thought that you would need a larger sample size
              $endgroup$
              – Henry
              13 hours ago






            • 1




              $begingroup$
              @Henry Thank you for your comments. I wasn't making a remark about these particular circumstances, but only about the logic of this answer, in the hope that it could be explained further.
              $endgroup$
              – whuber
              11 hours ago








            1




            1




            $begingroup$
            That this cannot be a valid explanation is evident from the observation that the CLT gives a good approximation for the first set of data in the question, which is equally small.
            $endgroup$
            – whuber
            14 hours ago




            $begingroup$
            That this cannot be a valid explanation is evident from the observation that the CLT gives a good approximation for the first set of data in the question, which is equally small.
            $endgroup$
            – whuber
            14 hours ago












            $begingroup$
            @whuber: I think you are saying that the normal distribution gives a reasonably good approximation for a sample of five from the first set. Since there are only an finite number of values for the sums (13 possible values without replacement and 21 possible values with replacement), the approximation does not get much better with a large number of samples of five, and the initial approximation is more due to the initial pattern...
            $endgroup$
            – Henry
            13 hours ago




            $begingroup$
            @whuber: I think you are saying that the normal distribution gives a reasonably good approximation for a sample of five from the first set. Since there are only an finite number of values for the sums (13 possible values without replacement and 21 possible values with replacement), the approximation does not get much better with a large number of samples of five, and the initial approximation is more due to the initial pattern...
            $endgroup$
            – Henry
            13 hours ago












            $begingroup$
            @whuber Since the distribution of the first set looks left skewed, I would expect the sum of five also to be left skewed, in a less extreme way than I would expect the sum of five from the second set to be right skewed. To get the skewness to reduce further, I would have thought that you would need a larger sample size
            $endgroup$
            – Henry
            13 hours ago




            $begingroup$
            @whuber Since the distribution of the first set looks left skewed, I would expect the sum of five also to be left skewed, in a less extreme way than I would expect the sum of five from the second set to be right skewed. To get the skewness to reduce further, I would have thought that you would need a larger sample size
            $endgroup$
            – Henry
            13 hours ago




            1




            1




            $begingroup$
            @Henry Thank you for your comments. I wasn't making a remark about these particular circumstances, but only about the logic of this answer, in the hope that it could be explained further.
            $endgroup$
            – whuber
            11 hours ago




            $begingroup$
            @Henry Thank you for your comments. I wasn't making a remark about these particular circumstances, but only about the logic of this answer, in the hope that it could be explained further.
            $endgroup$
            – whuber
            11 hours ago


















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