# MLE of simultaneous exponential distributions

Given the $$X_i\sim \text{exp}({\theta})$$ and $$Y_i\sim \text{exp}(\frac{1}{\theta})$$, where $$X_i$$ and $$Y_i$$ are indpendent, with the same $$\theta>0$$. I have to find the MLE and its distribution.

I find the simultaneous distribution to be: $$f_{X_i,Y_i}(x,y)=e^{-\left(\frac{x_i}{\theta }+\theta y_i\right)}1_{(0,\infty )}(x_i)1_{(0,\infty )}(y_i),$$ and the Fisherinformation: $$i_n(\theta)=\frac{2 n}{\theta ^2}$$

As it's convex, the maximum of the loglikelihood must be a minima, so I find the MLE: $$\hat{\theta}=\left(\frac{X_{\bullet }}{Y_{\bullet }}\right)^{0.5},$$ where $$X_{\bullet }=\sum _{i=1}^n x_i$$ and $$Y_{\bullet }=\sum _{i=1}^n y_i$$. So the asymptotic distribution is: $$\hat{\theta}\sim\mathcal{N}\left(\theta ,\frac{\theta^2}{2n^2}\right),$$ as $$\hat{\theta}\approx\mathcal{N}\left(\theta,\frac{1}{ni_n(\theta)}\right),$$ where $$i_\theta$$ is the Fisherinformation.

I am in doubt whether this is the right procedure.

EDIT

It is the right procedure - my professor provided the correct solution.

For a sample of $$n$$ observations on $$(X_i,Y_i)$$, the likelihood function given $$(x_1,y_1),\ldots,(x_n,y_n)$$ is

\begin{align} L(\theta)&=\prod_{i=1}^n e^{-\left(x_i/\theta+\theta y_i\right)}\mathbf1_{x_i>0,y_i>0} \\&=\exp\left[{-\sum_{i=1}^n\left(\frac{x_i}{\theta}+\theta y_i\right)}\right]\mathbf1_{x_1,\ldots,x_n,y_1,\ldots,y_n>0} \\&=\exp\left[-\frac{n\bar x}{\theta}-\theta n\bar y\right]\mathbf1_{x_{(1)},y_{(1)}>0}\quad,\,\theta>0 \end{align}

This, as you say, yields the MLE $$\hat\theta(\mathbf X,\mathbf Y)=\sqrt{\frac{\overline X}{\overline Y}}$$, where $$\overline X,\overline Y$$ are the sample means.

Now $$X_i$$'s are i.i.d $$\mathsf{Exp}$$ with mean $$\theta$$, and $$Y_i$$'s are i.i.d $$\mathsf{Exp}$$ with mean $$1/\theta$$, where $$X_i,Y_i$$ are independent.

Or, $$\frac{2}{\theta}X_i\stackrel{\text{ i.i.d }}\sim\chi^2_2\quad,\text{ independent of }\quad2\theta Y_i\stackrel{\text{ i.i.d }}\sim\chi^2_2$$

Therefore summing over the observations, we have

$$\frac{2}{\theta}n\overline X\sim\chi^2_{2n}\quad,\text{ independent of }\quad 2\theta n\overline Y\sim\chi^2_{2n}$$

Taking ratio of the statistics gives $$\frac{\overline X}{\theta^2 \overline Y}\sim F_{2n,2n}$$

So the (exact) distribution of the MLE $$\hat\theta$$ can be expressed as $$\frac{\hat\theta}{\theta}\stackrel{d}{=} \sqrt F\quad,\text{ where }F\sim F_{2n,2n}$$

I don't think the approach with asymptotic distribution is correct.

Since by (weak) law of large numbers $$\frac{1}{n}\sum\limits_{i=1}^n X_i\stackrel{P}\longrightarrow E(X_1)=\theta$$ and $$\frac{1}{n}\sum\limits_{i=1}^n Y_i\stackrel{P}\longrightarrow E(Y_1)=\frac{1}{\theta}$$,

$$\hat\theta^2=\frac{\frac{1}{n}\sum\limits_{i=1}^n X_i}{\frac{1}{n}\sum\limits_{i=1}^n Y_i}\stackrel{P}\longrightarrow \frac{E(X_1)}{E(Y_1)}=\theta^2$$

And by continuous mapping theorem, $$\hat\theta \stackrel{P}\longrightarrow \theta$$

As a result, asymptotic distribution of the MLE $$\hat\theta$$ is degenerate at $$\theta$$.

Also see this relevant post.

• Thank you, @StubbornAtom. Am I correct if I say, that the asymptotic distribution for the $\hat{\theta}$ is correct? Have you any advice in regards to knowing when the asymptotic approach is not best? – Frederik Apr 20 at 2:54
• @Frederik Why look for asymptotic distribution when exact distribution is easily available? I don't understand how you obtained the asymptotic distribution in your post. The MLE has an asymptotic normal distribution (under some conditions) typically when there is a single sample of observations. Here you have two (independent) samples. – StubbornAtom Apr 20 at 7:54
• I understand your concern, however, the reason I am looking for the asymptotic distribution is that I am solving an old exam set. In this set, they want me to find MLE, as I have, and the asymptotic distribution hereof - even though finding the exact distribution would be better. I have found the asymptotic distribution with formula: $$\hat{\theta}\approx\mathcal{N}\left(\theta,\hat{\text{se}}^2\right)=\mathcal{N}\left(\theta,\frac{1}{I_n(\theta)}\right),$$ from "All of Statistics" on p. 129. – Frederik Apr 20 at 8:18
• @Frederik Have a look at my edit. – StubbornAtom Apr 20 at 14:23
• Thank you, @StubbornAtom. I will try to look into it, as I'm not that familiar with the understanding of the way that MLE can degenerate. – Frederik Apr 21 at 8:35