Let $X \sim Gamma(\alpha,1)$ and $Y|X=x \sim Exp(\frac{1}{\theta x}), \alpha >1$ and $\theta >0$ are unknown. Let $\tau=E(Y)$. Suppose that based on the random sample $Y_1,...,Y_n$, we have MLEs, $\hat{\alpha}$ and $\hat{\theta}$. Use these MLEs to develop an asymptotic $1-\alpha$ confidence interval for $\tau$.

my work:

First, I need to find $\tau=E(Y)=E(\frac{1}{\theta x})=\frac{1}{\theta}E(\frac{1}{x})$. We use a transformation of $T=\frac{1}{X}$, where $f_T(t)=\frac{1}{\Gamma(\alpha)t^{\alpha+1}}e^{-1/t},t>0$. However, I am having trouble evaluating $E(T)=\int^\infty_0\frac{1}{\Gamma(\alpha)t^{\alpha}}e^{-1/t}dt$.

Assuming we have $\tau$, we can get the asymptotic $1-\alpha$ CI by using the asymptotic property of MLE. We know that $\hat{\alpha}\sim AN(\alpha,\frac{1}{ni(\alpha)})$ and $\hat{\theta} \sim AN(\theta,\frac{1}{ni(\theta)})$. However, I am failing to see how I can obtain the asymptotic CI for $\tau$.

updated work:

Thanks to Oriol, I get that $\tau=\frac{1}{\theta}\frac{\Gamma(\alpha-1)}{\Gamma(\alpha)}$. I now see that through the invariance property of MLE, $\hat{\tau}=\frac{1}{\hat{\theta}}\frac{\Gamma(\hat{\alpha}-1)}{\Gamma(\hat{\alpha})}$.

We can get our asymptotic $1-\alpha$ confidence interval for $\tau$ with

$\hat{\tau} \pm z_{\alpha/2}\frac{1}{\sqrt{ni(\hat{\tau})}} \implies \hat{\tau} \pm z_{\alpha/2}\sqrt{\hat{Var}(\tau(\hat{\theta},\hat{\alpha})|\theta,\alpha)}$.

To be quite honest, I do not see how to derive either $\sqrt{ni(\hat{\tau})}$ or $\sqrt{\hat{Var}(\tau(\hat{\theta},\hat{\alpha})|\theta,\alpha)}=\sqrt{\hat{V}}$. Up to this point, since regularity conditions hold, I have been using $i(\tau)$ to denote the Fisher information for a single observation and would prefer to see a solution using this form of Fisher information. Regarding the variance term, I know that

$\hat{V} \approx \frac{(\tau'(\hat{\theta},\hat{\alpha}))^2}{-\frac{\partial^2}{\partial \theta \partial \alpha}logL(\hat{\theta},\hat{\alpha}|X)}$,

but I do not know how to derive this term.

  • 1
    $\begingroup$ What are the pdfs of $Exp(a)$ and $Gamma(a,1)$? $\endgroup$ Commented Apr 18, 2020 at 6:04
  • $\begingroup$ @StubbornAtom $f(y)=\frac{1}{1/(\theta x)}exp(-\frac{y}{1/(\theta x)}), y>0$ and $f(x|\alpha)=\frac{1}{\Gamma(\alpha)}x^{\alpha-1}e^{-x},x>0$, using the scale parametrization for the Gamma distribution $\endgroup$
    – Ron Snow
    Commented Apr 18, 2020 at 21:33

1 Answer 1


I see you are computing $\tau$ using the law of total expectations. Using linearity of expectations and $Y\sim Exp\left(\theta x\right)$ (I see in your comment that this is the pdf of $Y | X=x$. Usually the notation is different) we get: $$\tau = \mathbb{E}_Y[Y]=\mathbb{E}_Y[\mathbb{E}_X[Y|X]]=\mathbb{E}_X[\mathbb{E}_Y[Y|X]]=\mathbb{E}_X\left[\frac{1}{\theta X}\right]$$

but your mistake is that as you see the expectation is taken with respect to $X$ (not its inverse). So you can compute it as:

\begin{align}\tau&=\mathbb{E}_X\left[\frac{1}{\theta X}\right]=\frac{1}{\theta}\int_0^\infty \frac{1}{x} \frac{1}{\Gamma(\alpha)} x^{\alpha - 1} e^{-x}dx \\&= \frac{1}{\theta}\frac{\Gamma(\alpha-1)}{\Gamma(\alpha)}\int_0^\infty \frac{1}{\Gamma(\alpha - 1)} x^{(\alpha - 1) - 1} e^{-x}dx=\frac{1}{\theta}\frac{\Gamma(\alpha-1)}{\Gamma(\alpha)}\end{align} where the last step follows since we are integrating a pdf. Now you can take $\hat{\tau}=\frac{1}{\hat{\theta}}\frac{\Gamma(\hat{\alpha}-1)}{\Gamma(\hat{\alpha})}$.

You can further simplify the latter expression if $\alpha \in \mathbb{N}$, in which case $\Gamma(\alpha)=(\alpha-1)!$ and hence $\hat{\tau}=\frac{1}{\hat{\theta}(\hat{\alpha} - 1)}$.

You can compute the distribution of $Y$ using the law of total probability:

\begin{align} f_Y(y;\alpha,\theta)&=\int_0^\infty f_{Y|X=x}(y) f_X(x) dx = \int_0^\infty \theta x e^{-\theta x y}\frac{1}{\Gamma(\alpha)} x^{\alpha - 1} e^{-x} dx \\ &=\frac{\Gamma(\alpha + 1)\theta}{\Gamma(\alpha)(1+\theta y)^{\alpha + 1}} \int_0^\infty \frac{(1+\theta y)^{\alpha + 1}}{\Gamma(\alpha + 1)} x^{(\alpha + 1) - 1} e^{-x(1 + \theta y)} dx \\ &=\frac{\Gamma(\alpha + 1)\theta}{\Gamma(\alpha)(1+\theta y)^{\alpha + 1}}\end{align}

where I used the same trick of integrating over a pdf, in this case $Ga(\alpha + 1 , (1+\theta y))$. You can check that the result is non-negative and integrates to 1 so it's actually a pdf (recall $y\in[0,\infty)$).

Now you have to derive the MLEs of $\alpha$ and $\theta$ so you can compute the confidence intervals. To do so, you first need to compute the log-likelihood of your sample, which by independence is

\begin{align} \ell(\alpha,\theta)&=\sum_{i=1}^n \log f(y_i ; \alpha, \theta)\\ &=\sum_{i=1}^n \left[ \log \Gamma(\alpha + 1) + \log \theta - \log \Gamma(\alpha) - (\alpha + 1 )\log(1+\theta y_i) \right] \end{align}

Hope you can continue from here.

  • $\begingroup$ Thank you. I see how you derived $\hat{\tau}$. Unfortunately, I cannot assume $\alpha \in N$. I have added updated work to my post to get closer to finding the asymptotic $1-\alpha$ confidence interval, but I do not know where to go from there. $\endgroup$
    – Ron Snow
    Commented Apr 20, 2020 at 16:56
  • $\begingroup$ Okay- got it. I'll find a way to use the Fisher information matrix from here! $\endgroup$
    – Ron Snow
    Commented Apr 20, 2020 at 18:34

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