How do we build a confidence interval for the parameter of the exponential distribution?





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EDIT



Let $X_{1},X_{2},ldots,X_{n}$ be a random sample whose distribution is given by $text{Exp}(theta)$, where $theta$ is not known. Precisely, $f(x|theta) = (1/thetaexp)(-x/theta)$ Describe a method to build a confidence interval with confidence coefficient $1 - alpha$ for $theta$.



MY ATTEMPT



Since the distribution in discussion is not normal and I do not know the size of the sample, I think we cannot apply the central limit theorem. One possible approach is to consider the maximum likelihood estimator of $theta$, whose distribution is approximately $mathcal{N}(theta,(nI_{F}(theta)^{-1})$. Another possible approach consists in using the score function, whose distribution is approximately $mathcal{N}(0,nI_{F}(theta))$. However, in both cases, it is assumed the CLT is applicable.



The exercise also provides the following hint: find $c_{1}$ and $c_{2}$ such that
begin{align*}
textbf{P}left(c_{1} < frac{1}{theta}sum_{i=1}^{n} X_{i} < c_{2}right) = 1 -alpha
end{align*}



Can someone help me out? Thanks in advance!










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  • 1




    $begingroup$
    You should clarify which parameterization of the exponential distribution you're using. From the later parts of your post it looks like you're using the scale parameterization rather than the rate parameterization but you should be explicit, not leave it to people to guess.
    $endgroup$
    – Glen_b
    4 hours ago










  • $begingroup$
    Thanks for the comment and sorry for the inconvenience. I edited the question.
    $endgroup$
    – user1337
    4 hours ago






  • 1




    $begingroup$
    Okay, you've defined it as the rate parameterization, which is fine, but then the hint at the end is wrong.
    $endgroup$
    – Glen_b
    3 hours ago












  • $begingroup$
    For rather large $n$ an approach using the CLT might provide a useful approximation. My answer gives an exact CI that works even for small $n.$
    $endgroup$
    – BruceET
    3 hours ago










  • $begingroup$
    There are so many options here because there are different choices of pivots. A C.I. could also be found using $min X_i$ which also has an exp distribution, but this won't be as 'good' as the one based on $sum X_i$.
    $endgroup$
    – StubbornAtom
    18 mins ago


















1












$begingroup$


EDIT



Let $X_{1},X_{2},ldots,X_{n}$ be a random sample whose distribution is given by $text{Exp}(theta)$, where $theta$ is not known. Precisely, $f(x|theta) = (1/thetaexp)(-x/theta)$ Describe a method to build a confidence interval with confidence coefficient $1 - alpha$ for $theta$.



MY ATTEMPT



Since the distribution in discussion is not normal and I do not know the size of the sample, I think we cannot apply the central limit theorem. One possible approach is to consider the maximum likelihood estimator of $theta$, whose distribution is approximately $mathcal{N}(theta,(nI_{F}(theta)^{-1})$. Another possible approach consists in using the score function, whose distribution is approximately $mathcal{N}(0,nI_{F}(theta))$. However, in both cases, it is assumed the CLT is applicable.



The exercise also provides the following hint: find $c_{1}$ and $c_{2}$ such that
begin{align*}
textbf{P}left(c_{1} < frac{1}{theta}sum_{i=1}^{n} X_{i} < c_{2}right) = 1 -alpha
end{align*}



Can someone help me out? Thanks in advance!










share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    You should clarify which parameterization of the exponential distribution you're using. From the later parts of your post it looks like you're using the scale parameterization rather than the rate parameterization but you should be explicit, not leave it to people to guess.
    $endgroup$
    – Glen_b
    4 hours ago










  • $begingroup$
    Thanks for the comment and sorry for the inconvenience. I edited the question.
    $endgroup$
    – user1337
    4 hours ago






  • 1




    $begingroup$
    Okay, you've defined it as the rate parameterization, which is fine, but then the hint at the end is wrong.
    $endgroup$
    – Glen_b
    3 hours ago












  • $begingroup$
    For rather large $n$ an approach using the CLT might provide a useful approximation. My answer gives an exact CI that works even for small $n.$
    $endgroup$
    – BruceET
    3 hours ago










  • $begingroup$
    There are so many options here because there are different choices of pivots. A C.I. could also be found using $min X_i$ which also has an exp distribution, but this won't be as 'good' as the one based on $sum X_i$.
    $endgroup$
    – StubbornAtom
    18 mins ago














1












1








1


0



$begingroup$


EDIT



Let $X_{1},X_{2},ldots,X_{n}$ be a random sample whose distribution is given by $text{Exp}(theta)$, where $theta$ is not known. Precisely, $f(x|theta) = (1/thetaexp)(-x/theta)$ Describe a method to build a confidence interval with confidence coefficient $1 - alpha$ for $theta$.



MY ATTEMPT



Since the distribution in discussion is not normal and I do not know the size of the sample, I think we cannot apply the central limit theorem. One possible approach is to consider the maximum likelihood estimator of $theta$, whose distribution is approximately $mathcal{N}(theta,(nI_{F}(theta)^{-1})$. Another possible approach consists in using the score function, whose distribution is approximately $mathcal{N}(0,nI_{F}(theta))$. However, in both cases, it is assumed the CLT is applicable.



The exercise also provides the following hint: find $c_{1}$ and $c_{2}$ such that
begin{align*}
textbf{P}left(c_{1} < frac{1}{theta}sum_{i=1}^{n} X_{i} < c_{2}right) = 1 -alpha
end{align*}



Can someone help me out? Thanks in advance!










share|cite|improve this question











$endgroup$




EDIT



Let $X_{1},X_{2},ldots,X_{n}$ be a random sample whose distribution is given by $text{Exp}(theta)$, where $theta$ is not known. Precisely, $f(x|theta) = (1/thetaexp)(-x/theta)$ Describe a method to build a confidence interval with confidence coefficient $1 - alpha$ for $theta$.



MY ATTEMPT



Since the distribution in discussion is not normal and I do not know the size of the sample, I think we cannot apply the central limit theorem. One possible approach is to consider the maximum likelihood estimator of $theta$, whose distribution is approximately $mathcal{N}(theta,(nI_{F}(theta)^{-1})$. Another possible approach consists in using the score function, whose distribution is approximately $mathcal{N}(0,nI_{F}(theta))$. However, in both cases, it is assumed the CLT is applicable.



The exercise also provides the following hint: find $c_{1}$ and $c_{2}$ such that
begin{align*}
textbf{P}left(c_{1} < frac{1}{theta}sum_{i=1}^{n} X_{i} < c_{2}right) = 1 -alpha
end{align*}



Can someone help me out? Thanks in advance!







self-study confidence-interval exponential-distribution






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edited 1 hour ago







user1337

















asked 7 hours ago









user1337user1337

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  • 1




    $begingroup$
    You should clarify which parameterization of the exponential distribution you're using. From the later parts of your post it looks like you're using the scale parameterization rather than the rate parameterization but you should be explicit, not leave it to people to guess.
    $endgroup$
    – Glen_b
    4 hours ago










  • $begingroup$
    Thanks for the comment and sorry for the inconvenience. I edited the question.
    $endgroup$
    – user1337
    4 hours ago






  • 1




    $begingroup$
    Okay, you've defined it as the rate parameterization, which is fine, but then the hint at the end is wrong.
    $endgroup$
    – Glen_b
    3 hours ago












  • $begingroup$
    For rather large $n$ an approach using the CLT might provide a useful approximation. My answer gives an exact CI that works even for small $n.$
    $endgroup$
    – BruceET
    3 hours ago










  • $begingroup$
    There are so many options here because there are different choices of pivots. A C.I. could also be found using $min X_i$ which also has an exp distribution, but this won't be as 'good' as the one based on $sum X_i$.
    $endgroup$
    – StubbornAtom
    18 mins ago














  • 1




    $begingroup$
    You should clarify which parameterization of the exponential distribution you're using. From the later parts of your post it looks like you're using the scale parameterization rather than the rate parameterization but you should be explicit, not leave it to people to guess.
    $endgroup$
    – Glen_b
    4 hours ago










  • $begingroup$
    Thanks for the comment and sorry for the inconvenience. I edited the question.
    $endgroup$
    – user1337
    4 hours ago






  • 1




    $begingroup$
    Okay, you've defined it as the rate parameterization, which is fine, but then the hint at the end is wrong.
    $endgroup$
    – Glen_b
    3 hours ago












  • $begingroup$
    For rather large $n$ an approach using the CLT might provide a useful approximation. My answer gives an exact CI that works even for small $n.$
    $endgroup$
    – BruceET
    3 hours ago










  • $begingroup$
    There are so many options here because there are different choices of pivots. A C.I. could also be found using $min X_i$ which also has an exp distribution, but this won't be as 'good' as the one based on $sum X_i$.
    $endgroup$
    – StubbornAtom
    18 mins ago








1




1




$begingroup$
You should clarify which parameterization of the exponential distribution you're using. From the later parts of your post it looks like you're using the scale parameterization rather than the rate parameterization but you should be explicit, not leave it to people to guess.
$endgroup$
– Glen_b
4 hours ago




$begingroup$
You should clarify which parameterization of the exponential distribution you're using. From the later parts of your post it looks like you're using the scale parameterization rather than the rate parameterization but you should be explicit, not leave it to people to guess.
$endgroup$
– Glen_b
4 hours ago












$begingroup$
Thanks for the comment and sorry for the inconvenience. I edited the question.
$endgroup$
– user1337
4 hours ago




$begingroup$
Thanks for the comment and sorry for the inconvenience. I edited the question.
$endgroup$
– user1337
4 hours ago




1




1




$begingroup$
Okay, you've defined it as the rate parameterization, which is fine, but then the hint at the end is wrong.
$endgroup$
– Glen_b
3 hours ago






$begingroup$
Okay, you've defined it as the rate parameterization, which is fine, but then the hint at the end is wrong.
$endgroup$
– Glen_b
3 hours ago














$begingroup$
For rather large $n$ an approach using the CLT might provide a useful approximation. My answer gives an exact CI that works even for small $n.$
$endgroup$
– BruceET
3 hours ago




$begingroup$
For rather large $n$ an approach using the CLT might provide a useful approximation. My answer gives an exact CI that works even for small $n.$
$endgroup$
– BruceET
3 hours ago












$begingroup$
There are so many options here because there are different choices of pivots. A C.I. could also be found using $min X_i$ which also has an exp distribution, but this won't be as 'good' as the one based on $sum X_i$.
$endgroup$
– StubbornAtom
18 mins ago




$begingroup$
There are so many options here because there are different choices of pivots. A C.I. could also be found using $min X_i$ which also has an exp distribution, but this won't be as 'good' as the one based on $sum X_i$.
$endgroup$
– StubbornAtom
18 mins ago










2 Answers
2






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$begingroup$

You don't say how the exponential distribution is
parameterized. Two parameterizations are in common use--mean and rate.



Let $E(X_i) = mu.$ Then one
can show that $$frac 1 mu sum_{i=1}^n X_i sim
mathsf{Gamma}(text{shape} = n, text{rate=scale} = 1).$$



In R statistical software the exponential distribution is parameterized according rate $lambda = 1/mu.$ Let $n = 10$ and $lambda = 1/5,$ so that $mu = 5.$ The following program simulates $m = 10^6$ samples of size $n = 10$ from $mathsf{Exp}(text{rate} = lambda = 1/5),$ finds $$Q = frac 1 mu sum_{i=1}^n X_i =
lambda sum_{i=1}^n X_i$$
for each sample, and plots the histogram of the one million $Q$'s, The figure
illustrates that $Q sim mathsf{Gamma}(10, 1).$
(Use MGFs for a formal proof.)



set.seed(414)   # for reproducibility
q = replicate(10^5, sum(rexp(10, 1/5))/5)
lbl = "Simulated Dist'n of Q with Density of GAMMA(10, 1)"
hist(q, prob=T, br=30, col="skyblue2", main=lbl)
curve(dgamma(x,10,1), col="red", add=T)


enter image description here



Thus, for $n = 10$ the constants $c_1 = 4.975$ and
$c_2 = 17.084$ for
a 95% confidence interval are quantiles 0.025 and 0.975, respectively, of $Q sim mathsf{Gamma}(10, 1).$



qgamma(c(.025, .975), 10, 1)
[1] 4.795389 17.084803


In particular, for the exponential sample shown below (second row),
a 95% confidence interval is $(2.224, 7.922).$ Notice the reversal of the quantiles in 'pivoting' $Q,$ which
has $mu$ in the denominator.



set.seed(1234); x = sort(round(rexp(10, 1/5), 2)); x
[1] 0.03 0.45 1.01 1.23 1.94 3.80 4.12 4.19 8.71 12.51
t = sum(x); t
[1] 37.99
t/qgamma(c(.975, .025), 10, 1)
[1] 2.223614 7.922194


Note: Because the chi-squared distribution is a member of the gamma family, it is possible to find endpoints for such a confidence interval in terms of a chi-squared distribution.



See Wikipedia on exponential distributions under 'confidence intervals'. (That discussion uses rate parameter $lambda$ for the exponential distribution, instead of $mu.)$






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    1












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    Taking $theta$ as the scale parameter, it can be shown that:



    $$frac{n bar{X}}{theta} sim text{Ga}(n,1).$$



    To form a confidence interval we choose any critical points $c_1 < c_2$ from the $text{Ga}(n,1)$ distribution such that these points contain probability $1-alpha$ of the distribution. Using the above pivotal quantity we then have:



    $$mathbb{P} Bigg( c_1 leqslant frac{n bar{X}}{theta} leqslant c_2 Bigg) = 1-alpha
    quad quad quad quad quad
    int limits_{c_1}^{c_2} text{Ga}(r|n,1) dr = 1 - alpha.$$



    Re-arranging the inequality in this probability statement and substituting the observed sample mean gives the confidence interval:



    $$text{CI}_theta(1-alpha) = Bigg[ frac{n bar{x}}{c_2} , frac{n bar{x}}{c_1} Bigg].$$



    This confidence interval is valid for any choice of $c_1<c_2$ so long as it obeys the required integral condition. For simplicity, many analysts use the symmetric critical points. However, it is possible to optimise the confidence interval by minimising its length, which we show below.





    Optimising the confidence interval: The length of this confidence interval is proportional to $1/c_1-1/c_2$, and so we minimise the length of the interval by choosing the critical points to minimise this distance. This can be done using the nlm function in R. In the following code we give a function for the minimum-length confidence interval for this problem, which we apply to some simulated data.



    #Set the objective function for minimisation
    OBJECTIVE <- function(c1, n, alpha) {
    pp <- pgamma(c1, n, 1, lower.tail = TRUE);
    c2 <- qgamma(1 - alpha + pp, n, 1, lower.tail = TRUE);
    1/c1 - 1/c2; }

    #Find the minimum-length confidence interval
    CONF_INT <- function(n, alpha, xbar) {
    START_c1 <- qgamma(alpha/2, n, 1, lower.tail = TRUE);
    MINIMISE <- nlm(f = OBJECTIVE, p = START_c1, n = n, alpha = alpha);
    c1 <- MINIMISE$estimate;
    pp <- pgamma(c1, n, 1, lower.tail = TRUE);
    c2 <- qgamma(1 - alpha + pp, n, 1, lower.tail = TRUE);
    c(n*xbar/c2, n*xbar/c1); }

    #Generate simulation data
    set.seed(921730198);
    n <- 300;
    scale <- 25.4;
    DATA <- rexp(n, rate = 1/scale);

    #Application of confidence interval to simulated data
    n <- length(DATA);
    xbar <- mean(DATA);
    alpha <- 0.05;

    CONF_INT(n, alpha, xbar);

    [1] 23.32040 29.24858





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      2 Answers
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      2 Answers
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      active

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      2












      $begingroup$

      You don't say how the exponential distribution is
      parameterized. Two parameterizations are in common use--mean and rate.



      Let $E(X_i) = mu.$ Then one
      can show that $$frac 1 mu sum_{i=1}^n X_i sim
      mathsf{Gamma}(text{shape} = n, text{rate=scale} = 1).$$



      In R statistical software the exponential distribution is parameterized according rate $lambda = 1/mu.$ Let $n = 10$ and $lambda = 1/5,$ so that $mu = 5.$ The following program simulates $m = 10^6$ samples of size $n = 10$ from $mathsf{Exp}(text{rate} = lambda = 1/5),$ finds $$Q = frac 1 mu sum_{i=1}^n X_i =
      lambda sum_{i=1}^n X_i$$
      for each sample, and plots the histogram of the one million $Q$'s, The figure
      illustrates that $Q sim mathsf{Gamma}(10, 1).$
      (Use MGFs for a formal proof.)



      set.seed(414)   # for reproducibility
      q = replicate(10^5, sum(rexp(10, 1/5))/5)
      lbl = "Simulated Dist'n of Q with Density of GAMMA(10, 1)"
      hist(q, prob=T, br=30, col="skyblue2", main=lbl)
      curve(dgamma(x,10,1), col="red", add=T)


      enter image description here



      Thus, for $n = 10$ the constants $c_1 = 4.975$ and
      $c_2 = 17.084$ for
      a 95% confidence interval are quantiles 0.025 and 0.975, respectively, of $Q sim mathsf{Gamma}(10, 1).$



      qgamma(c(.025, .975), 10, 1)
      [1] 4.795389 17.084803


      In particular, for the exponential sample shown below (second row),
      a 95% confidence interval is $(2.224, 7.922).$ Notice the reversal of the quantiles in 'pivoting' $Q,$ which
      has $mu$ in the denominator.



      set.seed(1234); x = sort(round(rexp(10, 1/5), 2)); x
      [1] 0.03 0.45 1.01 1.23 1.94 3.80 4.12 4.19 8.71 12.51
      t = sum(x); t
      [1] 37.99
      t/qgamma(c(.975, .025), 10, 1)
      [1] 2.223614 7.922194


      Note: Because the chi-squared distribution is a member of the gamma family, it is possible to find endpoints for such a confidence interval in terms of a chi-squared distribution.



      See Wikipedia on exponential distributions under 'confidence intervals'. (That discussion uses rate parameter $lambda$ for the exponential distribution, instead of $mu.)$






      share|cite|improve this answer











      $endgroup$


















        2












        $begingroup$

        You don't say how the exponential distribution is
        parameterized. Two parameterizations are in common use--mean and rate.



        Let $E(X_i) = mu.$ Then one
        can show that $$frac 1 mu sum_{i=1}^n X_i sim
        mathsf{Gamma}(text{shape} = n, text{rate=scale} = 1).$$



        In R statistical software the exponential distribution is parameterized according rate $lambda = 1/mu.$ Let $n = 10$ and $lambda = 1/5,$ so that $mu = 5.$ The following program simulates $m = 10^6$ samples of size $n = 10$ from $mathsf{Exp}(text{rate} = lambda = 1/5),$ finds $$Q = frac 1 mu sum_{i=1}^n X_i =
        lambda sum_{i=1}^n X_i$$
        for each sample, and plots the histogram of the one million $Q$'s, The figure
        illustrates that $Q sim mathsf{Gamma}(10, 1).$
        (Use MGFs for a formal proof.)



        set.seed(414)   # for reproducibility
        q = replicate(10^5, sum(rexp(10, 1/5))/5)
        lbl = "Simulated Dist'n of Q with Density of GAMMA(10, 1)"
        hist(q, prob=T, br=30, col="skyblue2", main=lbl)
        curve(dgamma(x,10,1), col="red", add=T)


        enter image description here



        Thus, for $n = 10$ the constants $c_1 = 4.975$ and
        $c_2 = 17.084$ for
        a 95% confidence interval are quantiles 0.025 and 0.975, respectively, of $Q sim mathsf{Gamma}(10, 1).$



        qgamma(c(.025, .975), 10, 1)
        [1] 4.795389 17.084803


        In particular, for the exponential sample shown below (second row),
        a 95% confidence interval is $(2.224, 7.922).$ Notice the reversal of the quantiles in 'pivoting' $Q,$ which
        has $mu$ in the denominator.



        set.seed(1234); x = sort(round(rexp(10, 1/5), 2)); x
        [1] 0.03 0.45 1.01 1.23 1.94 3.80 4.12 4.19 8.71 12.51
        t = sum(x); t
        [1] 37.99
        t/qgamma(c(.975, .025), 10, 1)
        [1] 2.223614 7.922194


        Note: Because the chi-squared distribution is a member of the gamma family, it is possible to find endpoints for such a confidence interval in terms of a chi-squared distribution.



        See Wikipedia on exponential distributions under 'confidence intervals'. (That discussion uses rate parameter $lambda$ for the exponential distribution, instead of $mu.)$






        share|cite|improve this answer











        $endgroup$
















          2












          2








          2





          $begingroup$

          You don't say how the exponential distribution is
          parameterized. Two parameterizations are in common use--mean and rate.



          Let $E(X_i) = mu.$ Then one
          can show that $$frac 1 mu sum_{i=1}^n X_i sim
          mathsf{Gamma}(text{shape} = n, text{rate=scale} = 1).$$



          In R statistical software the exponential distribution is parameterized according rate $lambda = 1/mu.$ Let $n = 10$ and $lambda = 1/5,$ so that $mu = 5.$ The following program simulates $m = 10^6$ samples of size $n = 10$ from $mathsf{Exp}(text{rate} = lambda = 1/5),$ finds $$Q = frac 1 mu sum_{i=1}^n X_i =
          lambda sum_{i=1}^n X_i$$
          for each sample, and plots the histogram of the one million $Q$'s, The figure
          illustrates that $Q sim mathsf{Gamma}(10, 1).$
          (Use MGFs for a formal proof.)



          set.seed(414)   # for reproducibility
          q = replicate(10^5, sum(rexp(10, 1/5))/5)
          lbl = "Simulated Dist'n of Q with Density of GAMMA(10, 1)"
          hist(q, prob=T, br=30, col="skyblue2", main=lbl)
          curve(dgamma(x,10,1), col="red", add=T)


          enter image description here



          Thus, for $n = 10$ the constants $c_1 = 4.975$ and
          $c_2 = 17.084$ for
          a 95% confidence interval are quantiles 0.025 and 0.975, respectively, of $Q sim mathsf{Gamma}(10, 1).$



          qgamma(c(.025, .975), 10, 1)
          [1] 4.795389 17.084803


          In particular, for the exponential sample shown below (second row),
          a 95% confidence interval is $(2.224, 7.922).$ Notice the reversal of the quantiles in 'pivoting' $Q,$ which
          has $mu$ in the denominator.



          set.seed(1234); x = sort(round(rexp(10, 1/5), 2)); x
          [1] 0.03 0.45 1.01 1.23 1.94 3.80 4.12 4.19 8.71 12.51
          t = sum(x); t
          [1] 37.99
          t/qgamma(c(.975, .025), 10, 1)
          [1] 2.223614 7.922194


          Note: Because the chi-squared distribution is a member of the gamma family, it is possible to find endpoints for such a confidence interval in terms of a chi-squared distribution.



          See Wikipedia on exponential distributions under 'confidence intervals'. (That discussion uses rate parameter $lambda$ for the exponential distribution, instead of $mu.)$






          share|cite|improve this answer











          $endgroup$



          You don't say how the exponential distribution is
          parameterized. Two parameterizations are in common use--mean and rate.



          Let $E(X_i) = mu.$ Then one
          can show that $$frac 1 mu sum_{i=1}^n X_i sim
          mathsf{Gamma}(text{shape} = n, text{rate=scale} = 1).$$



          In R statistical software the exponential distribution is parameterized according rate $lambda = 1/mu.$ Let $n = 10$ and $lambda = 1/5,$ so that $mu = 5.$ The following program simulates $m = 10^6$ samples of size $n = 10$ from $mathsf{Exp}(text{rate} = lambda = 1/5),$ finds $$Q = frac 1 mu sum_{i=1}^n X_i =
          lambda sum_{i=1}^n X_i$$
          for each sample, and plots the histogram of the one million $Q$'s, The figure
          illustrates that $Q sim mathsf{Gamma}(10, 1).$
          (Use MGFs for a formal proof.)



          set.seed(414)   # for reproducibility
          q = replicate(10^5, sum(rexp(10, 1/5))/5)
          lbl = "Simulated Dist'n of Q with Density of GAMMA(10, 1)"
          hist(q, prob=T, br=30, col="skyblue2", main=lbl)
          curve(dgamma(x,10,1), col="red", add=T)


          enter image description here



          Thus, for $n = 10$ the constants $c_1 = 4.975$ and
          $c_2 = 17.084$ for
          a 95% confidence interval are quantiles 0.025 and 0.975, respectively, of $Q sim mathsf{Gamma}(10, 1).$



          qgamma(c(.025, .975), 10, 1)
          [1] 4.795389 17.084803


          In particular, for the exponential sample shown below (second row),
          a 95% confidence interval is $(2.224, 7.922).$ Notice the reversal of the quantiles in 'pivoting' $Q,$ which
          has $mu$ in the denominator.



          set.seed(1234); x = sort(round(rexp(10, 1/5), 2)); x
          [1] 0.03 0.45 1.01 1.23 1.94 3.80 4.12 4.19 8.71 12.51
          t = sum(x); t
          [1] 37.99
          t/qgamma(c(.975, .025), 10, 1)
          [1] 2.223614 7.922194


          Note: Because the chi-squared distribution is a member of the gamma family, it is possible to find endpoints for such a confidence interval in terms of a chi-squared distribution.



          See Wikipedia on exponential distributions under 'confidence intervals'. (That discussion uses rate parameter $lambda$ for the exponential distribution, instead of $mu.)$







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited 4 hours ago

























          answered 4 hours ago









          BruceETBruceET

          6,6331721




          6,6331721

























              1












              $begingroup$

              Taking $theta$ as the scale parameter, it can be shown that:



              $$frac{n bar{X}}{theta} sim text{Ga}(n,1).$$



              To form a confidence interval we choose any critical points $c_1 < c_2$ from the $text{Ga}(n,1)$ distribution such that these points contain probability $1-alpha$ of the distribution. Using the above pivotal quantity we then have:



              $$mathbb{P} Bigg( c_1 leqslant frac{n bar{X}}{theta} leqslant c_2 Bigg) = 1-alpha
              quad quad quad quad quad
              int limits_{c_1}^{c_2} text{Ga}(r|n,1) dr = 1 - alpha.$$



              Re-arranging the inequality in this probability statement and substituting the observed sample mean gives the confidence interval:



              $$text{CI}_theta(1-alpha) = Bigg[ frac{n bar{x}}{c_2} , frac{n bar{x}}{c_1} Bigg].$$



              This confidence interval is valid for any choice of $c_1<c_2$ so long as it obeys the required integral condition. For simplicity, many analysts use the symmetric critical points. However, it is possible to optimise the confidence interval by minimising its length, which we show below.





              Optimising the confidence interval: The length of this confidence interval is proportional to $1/c_1-1/c_2$, and so we minimise the length of the interval by choosing the critical points to minimise this distance. This can be done using the nlm function in R. In the following code we give a function for the minimum-length confidence interval for this problem, which we apply to some simulated data.



              #Set the objective function for minimisation
              OBJECTIVE <- function(c1, n, alpha) {
              pp <- pgamma(c1, n, 1, lower.tail = TRUE);
              c2 <- qgamma(1 - alpha + pp, n, 1, lower.tail = TRUE);
              1/c1 - 1/c2; }

              #Find the minimum-length confidence interval
              CONF_INT <- function(n, alpha, xbar) {
              START_c1 <- qgamma(alpha/2, n, 1, lower.tail = TRUE);
              MINIMISE <- nlm(f = OBJECTIVE, p = START_c1, n = n, alpha = alpha);
              c1 <- MINIMISE$estimate;
              pp <- pgamma(c1, n, 1, lower.tail = TRUE);
              c2 <- qgamma(1 - alpha + pp, n, 1, lower.tail = TRUE);
              c(n*xbar/c2, n*xbar/c1); }

              #Generate simulation data
              set.seed(921730198);
              n <- 300;
              scale <- 25.4;
              DATA <- rexp(n, rate = 1/scale);

              #Application of confidence interval to simulated data
              n <- length(DATA);
              xbar <- mean(DATA);
              alpha <- 0.05;

              CONF_INT(n, alpha, xbar);

              [1] 23.32040 29.24858





              share|cite|improve this answer











              $endgroup$


















                1












                $begingroup$

                Taking $theta$ as the scale parameter, it can be shown that:



                $$frac{n bar{X}}{theta} sim text{Ga}(n,1).$$



                To form a confidence interval we choose any critical points $c_1 < c_2$ from the $text{Ga}(n,1)$ distribution such that these points contain probability $1-alpha$ of the distribution. Using the above pivotal quantity we then have:



                $$mathbb{P} Bigg( c_1 leqslant frac{n bar{X}}{theta} leqslant c_2 Bigg) = 1-alpha
                quad quad quad quad quad
                int limits_{c_1}^{c_2} text{Ga}(r|n,1) dr = 1 - alpha.$$



                Re-arranging the inequality in this probability statement and substituting the observed sample mean gives the confidence interval:



                $$text{CI}_theta(1-alpha) = Bigg[ frac{n bar{x}}{c_2} , frac{n bar{x}}{c_1} Bigg].$$



                This confidence interval is valid for any choice of $c_1<c_2$ so long as it obeys the required integral condition. For simplicity, many analysts use the symmetric critical points. However, it is possible to optimise the confidence interval by minimising its length, which we show below.





                Optimising the confidence interval: The length of this confidence interval is proportional to $1/c_1-1/c_2$, and so we minimise the length of the interval by choosing the critical points to minimise this distance. This can be done using the nlm function in R. In the following code we give a function for the minimum-length confidence interval for this problem, which we apply to some simulated data.



                #Set the objective function for minimisation
                OBJECTIVE <- function(c1, n, alpha) {
                pp <- pgamma(c1, n, 1, lower.tail = TRUE);
                c2 <- qgamma(1 - alpha + pp, n, 1, lower.tail = TRUE);
                1/c1 - 1/c2; }

                #Find the minimum-length confidence interval
                CONF_INT <- function(n, alpha, xbar) {
                START_c1 <- qgamma(alpha/2, n, 1, lower.tail = TRUE);
                MINIMISE <- nlm(f = OBJECTIVE, p = START_c1, n = n, alpha = alpha);
                c1 <- MINIMISE$estimate;
                pp <- pgamma(c1, n, 1, lower.tail = TRUE);
                c2 <- qgamma(1 - alpha + pp, n, 1, lower.tail = TRUE);
                c(n*xbar/c2, n*xbar/c1); }

                #Generate simulation data
                set.seed(921730198);
                n <- 300;
                scale <- 25.4;
                DATA <- rexp(n, rate = 1/scale);

                #Application of confidence interval to simulated data
                n <- length(DATA);
                xbar <- mean(DATA);
                alpha <- 0.05;

                CONF_INT(n, alpha, xbar);

                [1] 23.32040 29.24858





                share|cite|improve this answer











                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  Taking $theta$ as the scale parameter, it can be shown that:



                  $$frac{n bar{X}}{theta} sim text{Ga}(n,1).$$



                  To form a confidence interval we choose any critical points $c_1 < c_2$ from the $text{Ga}(n,1)$ distribution such that these points contain probability $1-alpha$ of the distribution. Using the above pivotal quantity we then have:



                  $$mathbb{P} Bigg( c_1 leqslant frac{n bar{X}}{theta} leqslant c_2 Bigg) = 1-alpha
                  quad quad quad quad quad
                  int limits_{c_1}^{c_2} text{Ga}(r|n,1) dr = 1 - alpha.$$



                  Re-arranging the inequality in this probability statement and substituting the observed sample mean gives the confidence interval:



                  $$text{CI}_theta(1-alpha) = Bigg[ frac{n bar{x}}{c_2} , frac{n bar{x}}{c_1} Bigg].$$



                  This confidence interval is valid for any choice of $c_1<c_2$ so long as it obeys the required integral condition. For simplicity, many analysts use the symmetric critical points. However, it is possible to optimise the confidence interval by minimising its length, which we show below.





                  Optimising the confidence interval: The length of this confidence interval is proportional to $1/c_1-1/c_2$, and so we minimise the length of the interval by choosing the critical points to minimise this distance. This can be done using the nlm function in R. In the following code we give a function for the minimum-length confidence interval for this problem, which we apply to some simulated data.



                  #Set the objective function for minimisation
                  OBJECTIVE <- function(c1, n, alpha) {
                  pp <- pgamma(c1, n, 1, lower.tail = TRUE);
                  c2 <- qgamma(1 - alpha + pp, n, 1, lower.tail = TRUE);
                  1/c1 - 1/c2; }

                  #Find the minimum-length confidence interval
                  CONF_INT <- function(n, alpha, xbar) {
                  START_c1 <- qgamma(alpha/2, n, 1, lower.tail = TRUE);
                  MINIMISE <- nlm(f = OBJECTIVE, p = START_c1, n = n, alpha = alpha);
                  c1 <- MINIMISE$estimate;
                  pp <- pgamma(c1, n, 1, lower.tail = TRUE);
                  c2 <- qgamma(1 - alpha + pp, n, 1, lower.tail = TRUE);
                  c(n*xbar/c2, n*xbar/c1); }

                  #Generate simulation data
                  set.seed(921730198);
                  n <- 300;
                  scale <- 25.4;
                  DATA <- rexp(n, rate = 1/scale);

                  #Application of confidence interval to simulated data
                  n <- length(DATA);
                  xbar <- mean(DATA);
                  alpha <- 0.05;

                  CONF_INT(n, alpha, xbar);

                  [1] 23.32040 29.24858





                  share|cite|improve this answer











                  $endgroup$



                  Taking $theta$ as the scale parameter, it can be shown that:



                  $$frac{n bar{X}}{theta} sim text{Ga}(n,1).$$



                  To form a confidence interval we choose any critical points $c_1 < c_2$ from the $text{Ga}(n,1)$ distribution such that these points contain probability $1-alpha$ of the distribution. Using the above pivotal quantity we then have:



                  $$mathbb{P} Bigg( c_1 leqslant frac{n bar{X}}{theta} leqslant c_2 Bigg) = 1-alpha
                  quad quad quad quad quad
                  int limits_{c_1}^{c_2} text{Ga}(r|n,1) dr = 1 - alpha.$$



                  Re-arranging the inequality in this probability statement and substituting the observed sample mean gives the confidence interval:



                  $$text{CI}_theta(1-alpha) = Bigg[ frac{n bar{x}}{c_2} , frac{n bar{x}}{c_1} Bigg].$$



                  This confidence interval is valid for any choice of $c_1<c_2$ so long as it obeys the required integral condition. For simplicity, many analysts use the symmetric critical points. However, it is possible to optimise the confidence interval by minimising its length, which we show below.





                  Optimising the confidence interval: The length of this confidence interval is proportional to $1/c_1-1/c_2$, and so we minimise the length of the interval by choosing the critical points to minimise this distance. This can be done using the nlm function in R. In the following code we give a function for the minimum-length confidence interval for this problem, which we apply to some simulated data.



                  #Set the objective function for minimisation
                  OBJECTIVE <- function(c1, n, alpha) {
                  pp <- pgamma(c1, n, 1, lower.tail = TRUE);
                  c2 <- qgamma(1 - alpha + pp, n, 1, lower.tail = TRUE);
                  1/c1 - 1/c2; }

                  #Find the minimum-length confidence interval
                  CONF_INT <- function(n, alpha, xbar) {
                  START_c1 <- qgamma(alpha/2, n, 1, lower.tail = TRUE);
                  MINIMISE <- nlm(f = OBJECTIVE, p = START_c1, n = n, alpha = alpha);
                  c1 <- MINIMISE$estimate;
                  pp <- pgamma(c1, n, 1, lower.tail = TRUE);
                  c2 <- qgamma(1 - alpha + pp, n, 1, lower.tail = TRUE);
                  c(n*xbar/c2, n*xbar/c1); }

                  #Generate simulation data
                  set.seed(921730198);
                  n <- 300;
                  scale <- 25.4;
                  DATA <- rexp(n, rate = 1/scale);

                  #Application of confidence interval to simulated data
                  n <- length(DATA);
                  xbar <- mean(DATA);
                  alpha <- 0.05;

                  CONF_INT(n, alpha, xbar);

                  [1] 23.32040 29.24858






                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited 1 hour ago

























                  answered 1 hour ago









                  BenBen

                  28.3k233128




                  28.3k233128






























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