Calculating variance of poisson distributed random variable












2












$begingroup$


I am calculating variance of a Poisson distributed random variable with mean $lambda$. I am doing it in the following way:



$mathbb{V}(X) = mathbb{E}(X^2) - lambda^2 \
= sum_{xgeq 0} quad x^2frac{e^{-lambda} quadlambda^x}{x!} - lambda^2\
= 0 + lambda e^{-lambda} + sum_{xgeq 2} quad x^2frac{e^{-lambda} quadlambda^x}{x!}-lambda^2\
= lambda e^{-lambda} + e^{-lambda}lambda^2sum_{xgeq 2} quad frac{lambda^{x-2}}{(x-2)!}\
=lambda e^{-lambda} + lambda^2 -lambda^2\
=lambda e^{-lambda}$



The answer is wrong. But I cannot understand what am I doing wrong in breaking the summation on line 3.



Edit: shown step 4 to answer a comment






self-study poisson-distribution arithmetic







share|cite|improve this question











$endgroup$












  • $begingroup$
    How do you get that sum over $x geq 2$ to be equal to $lambda^2$?
    $endgroup$
    – jbowman
    6 hours ago












  • $begingroup$
    I have added a step.
    $endgroup$
    – anotherone
    6 hours ago










  • $begingroup$
    Going from line 3 to line 4, you've assumed $x^2/x! = 1/(x - 2)!$ for $x geq 2$, which is not true. Also, your $-lambda^2$ term disappears on line 4 and then reappears on line 5.
    $endgroup$
    – Alex
    6 hours ago








  • 5




    $begingroup$
    Often times it's easier to start by finding $E[X(X-1)]$. Also, please add the self-study tag, and I'll assist further.
    $endgroup$
    – StatsStudent
    6 hours ago








  • 1




    $begingroup$
    @StatsStudent Yes, I found that just now. Thanks.
    $endgroup$
    – anotherone
    6 hours ago
















2












$begingroup$


I am calculating variance of a Poisson distributed random variable with mean $lambda$. I am doing it in the following way:



$mathbb{V}(X) = mathbb{E}(X^2) - lambda^2 \
= sum_{xgeq 0} quad x^2frac{e^{-lambda} quadlambda^x}{x!} - lambda^2\
= 0 + lambda e^{-lambda} + sum_{xgeq 2} quad x^2frac{e^{-lambda} quadlambda^x}{x!}-lambda^2\
= lambda e^{-lambda} + e^{-lambda}lambda^2sum_{xgeq 2} quad frac{lambda^{x-2}}{(x-2)!}\
=lambda e^{-lambda} + lambda^2 -lambda^2\
=lambda e^{-lambda}$



The answer is wrong. But I cannot understand what am I doing wrong in breaking the summation on line 3.



Edit: shown step 4 to answer a comment






self-study poisson-distribution arithmetic







share|cite|improve this question











$endgroup$












  • $begingroup$
    How do you get that sum over $x geq 2$ to be equal to $lambda^2$?
    $endgroup$
    – jbowman
    6 hours ago












  • $begingroup$
    I have added a step.
    $endgroup$
    – anotherone
    6 hours ago










  • $begingroup$
    Going from line 3 to line 4, you've assumed $x^2/x! = 1/(x - 2)!$ for $x geq 2$, which is not true. Also, your $-lambda^2$ term disappears on line 4 and then reappears on line 5.
    $endgroup$
    – Alex
    6 hours ago








  • 5




    $begingroup$
    Often times it's easier to start by finding $E[X(X-1)]$. Also, please add the self-study tag, and I'll assist further.
    $endgroup$
    – StatsStudent
    6 hours ago








  • 1




    $begingroup$
    @StatsStudent Yes, I found that just now. Thanks.
    $endgroup$
    – anotherone
    6 hours ago














2












2








2


1



$begingroup$


I am calculating variance of a Poisson distributed random variable with mean $lambda$. I am doing it in the following way:



$mathbb{V}(X) = mathbb{E}(X^2) - lambda^2 \
= sum_{xgeq 0} quad x^2frac{e^{-lambda} quadlambda^x}{x!} - lambda^2\
= 0 + lambda e^{-lambda} + sum_{xgeq 2} quad x^2frac{e^{-lambda} quadlambda^x}{x!}-lambda^2\
= lambda e^{-lambda} + e^{-lambda}lambda^2sum_{xgeq 2} quad frac{lambda^{x-2}}{(x-2)!}\
=lambda e^{-lambda} + lambda^2 -lambda^2\
=lambda e^{-lambda}$



The answer is wrong. But I cannot understand what am I doing wrong in breaking the summation on line 3.



Edit: shown step 4 to answer a comment






self-study poisson-distribution arithmetic







share|cite|improve this question











$endgroup$




I am calculating variance of a Poisson distributed random variable with mean $lambda$. I am doing it in the following way:



$mathbb{V}(X) = mathbb{E}(X^2) - lambda^2 \
= sum_{xgeq 0} quad x^2frac{e^{-lambda} quadlambda^x}{x!} - lambda^2\
= 0 + lambda e^{-lambda} + sum_{xgeq 2} quad x^2frac{e^{-lambda} quadlambda^x}{x!}-lambda^2\
= lambda e^{-lambda} + e^{-lambda}lambda^2sum_{xgeq 2} quad frac{lambda^{x-2}}{(x-2)!}\
=lambda e^{-lambda} + lambda^2 -lambda^2\
=lambda e^{-lambda}$



The answer is wrong. But I cannot understand what am I doing wrong in breaking the summation on line 3.



Edit: shown step 4 to answer a comment






self-study poisson-distribution arithmetic




self-study poisson-distribution arithmetic






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 5 hours ago









whuber♦

203k33443813




203k33443813










asked 6 hours ago









anotheroneanotherone

1206




1206












  • $begingroup$
    How do you get that sum over $x geq 2$ to be equal to $lambda^2$?
    $endgroup$
    – jbowman
    6 hours ago












  • $begingroup$
    I have added a step.
    $endgroup$
    – anotherone
    6 hours ago










  • $begingroup$
    Going from line 3 to line 4, you've assumed $x^2/x! = 1/(x - 2)!$ for $x geq 2$, which is not true. Also, your $-lambda^2$ term disappears on line 4 and then reappears on line 5.
    $endgroup$
    – Alex
    6 hours ago








  • 5




    $begingroup$
    Often times it's easier to start by finding $E[X(X-1)]$. Also, please add the self-study tag, and I'll assist further.
    $endgroup$
    – StatsStudent
    6 hours ago








  • 1




    $begingroup$
    @StatsStudent Yes, I found that just now. Thanks.
    $endgroup$
    – anotherone
    6 hours ago


















  • $begingroup$
    How do you get that sum over $x geq 2$ to be equal to $lambda^2$?
    $endgroup$
    – jbowman
    6 hours ago












  • $begingroup$
    I have added a step.
    $endgroup$
    – anotherone
    6 hours ago










  • $begingroup$
    Going from line 3 to line 4, you've assumed $x^2/x! = 1/(x - 2)!$ for $x geq 2$, which is not true. Also, your $-lambda^2$ term disappears on line 4 and then reappears on line 5.
    $endgroup$
    – Alex
    6 hours ago








  • 5




    $begingroup$
    Often times it's easier to start by finding $E[X(X-1)]$. Also, please add the self-study tag, and I'll assist further.
    $endgroup$
    – StatsStudent
    6 hours ago








  • 1




    $begingroup$
    @StatsStudent Yes, I found that just now. Thanks.
    $endgroup$
    – anotherone
    6 hours ago
















$begingroup$
How do you get that sum over $x geq 2$ to be equal to $lambda^2$?
$endgroup$
– jbowman
6 hours ago






$begingroup$
How do you get that sum over $x geq 2$ to be equal to $lambda^2$?
$endgroup$
– jbowman
6 hours ago














$begingroup$
I have added a step.
$endgroup$
– anotherone
6 hours ago




$begingroup$
I have added a step.
$endgroup$
– anotherone
6 hours ago












$begingroup$
Going from line 3 to line 4, you've assumed $x^2/x! = 1/(x - 2)!$ for $x geq 2$, which is not true. Also, your $-lambda^2$ term disappears on line 4 and then reappears on line 5.
$endgroup$
– Alex
6 hours ago






$begingroup$
Going from line 3 to line 4, you've assumed $x^2/x! = 1/(x - 2)!$ for $x geq 2$, which is not true. Also, your $-lambda^2$ term disappears on line 4 and then reappears on line 5.
$endgroup$
– Alex
6 hours ago






5




5




$begingroup$
Often times it's easier to start by finding $E[X(X-1)]$. Also, please add the self-study tag, and I'll assist further.
$endgroup$
– StatsStudent
6 hours ago






$begingroup$
Often times it's easier to start by finding $E[X(X-1)]$. Also, please add the self-study tag, and I'll assist further.
$endgroup$
– StatsStudent
6 hours ago






1




1




$begingroup$
@StatsStudent Yes, I found that just now. Thanks.
$endgroup$
– anotherone
6 hours ago




$begingroup$
@StatsStudent Yes, I found that just now. Thanks.
$endgroup$
– anotherone
6 hours ago










1 Answer
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The Poisson distribution is one of those distributions that involves a factorial denominator. This this type of distribution, the simplest way to find the raw and central moments is to first find the expected values of the falling factorials:



$$mathbb{E}((X)_r) = mathbb{E} bigg( X(X-1) cdots (X-r+1) bigg) = sum_{x=0}^infty x(x-1) cdots (x-r+1) cdot frac{lambda^x}{x!} e^{-lambda}.$$



Have a go at solving this infinite sum (it is relatively simple) and then you will have a general form for the expected value of the falling factorials of $X$. These are all polynomials in $X$, so they can be used to find any of the raw moments or central moments by appropriate arithmetic. In your particular case you want the variance, so you will want to look at the expected values of the falling factorials up to $r=2$.






share|cite|improve this answer









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    $begingroup$

    The Poisson distribution is one of those distributions that involves a factorial denominator. This this type of distribution, the simplest way to find the raw and central moments is to first find the expected values of the falling factorials:



    $$mathbb{E}((X)_r) = mathbb{E} bigg( X(X-1) cdots (X-r+1) bigg) = sum_{x=0}^infty x(x-1) cdots (x-r+1) cdot frac{lambda^x}{x!} e^{-lambda}.$$



    Have a go at solving this infinite sum (it is relatively simple) and then you will have a general form for the expected value of the falling factorials of $X$. These are all polynomials in $X$, so they can be used to find any of the raw moments or central moments by appropriate arithmetic. In your particular case you want the variance, so you will want to look at the expected values of the falling factorials up to $r=2$.






    share|cite|improve this answer









    $endgroup$


















      2












      $begingroup$

      The Poisson distribution is one of those distributions that involves a factorial denominator. This this type of distribution, the simplest way to find the raw and central moments is to first find the expected values of the falling factorials:



      $$mathbb{E}((X)_r) = mathbb{E} bigg( X(X-1) cdots (X-r+1) bigg) = sum_{x=0}^infty x(x-1) cdots (x-r+1) cdot frac{lambda^x}{x!} e^{-lambda}.$$



      Have a go at solving this infinite sum (it is relatively simple) and then you will have a general form for the expected value of the falling factorials of $X$. These are all polynomials in $X$, so they can be used to find any of the raw moments or central moments by appropriate arithmetic. In your particular case you want the variance, so you will want to look at the expected values of the falling factorials up to $r=2$.






      share|cite|improve this answer









      $endgroup$
















        2












        2








        2





        $begingroup$

        The Poisson distribution is one of those distributions that involves a factorial denominator. This this type of distribution, the simplest way to find the raw and central moments is to first find the expected values of the falling factorials:



        $$mathbb{E}((X)_r) = mathbb{E} bigg( X(X-1) cdots (X-r+1) bigg) = sum_{x=0}^infty x(x-1) cdots (x-r+1) cdot frac{lambda^x}{x!} e^{-lambda}.$$



        Have a go at solving this infinite sum (it is relatively simple) and then you will have a general form for the expected value of the falling factorials of $X$. These are all polynomials in $X$, so they can be used to find any of the raw moments or central moments by appropriate arithmetic. In your particular case you want the variance, so you will want to look at the expected values of the falling factorials up to $r=2$.






        share|cite|improve this answer









        $endgroup$



        The Poisson distribution is one of those distributions that involves a factorial denominator. This this type of distribution, the simplest way to find the raw and central moments is to first find the expected values of the falling factorials:



        $$mathbb{E}((X)_r) = mathbb{E} bigg( X(X-1) cdots (X-r+1) bigg) = sum_{x=0}^infty x(x-1) cdots (x-r+1) cdot frac{lambda^x}{x!} e^{-lambda}.$$



        Have a go at solving this infinite sum (it is relatively simple) and then you will have a general form for the expected value of the falling factorials of $X$. These are all polynomials in $X$, so they can be used to find any of the raw moments or central moments by appropriate arithmetic. In your particular case you want the variance, so you will want to look at the expected values of the falling factorials up to $r=2$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered 5 hours ago









        BenBen

        24.6k226117




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