# Asymptotic distribution of MLE given conditional density function

$$\newcommand{\Var}{\operatorname{Var}}\newcommand{\Cov}{\operatorname{Cov}}$$

Consider $$f_{Y|X=x}(y)=\theta e^{-\theta(y-x)} I(y\ge x)$$. Find the MLE of $$\theta$$ based on $$n$$ realisations of the pair $$(X,Y)$$ and determine its asymptotic distribution.

Note: There is no additional information about $$X$$. It is not given that $$X$$ and $$Y$$ are independent.

My attempt:

I found $$\hat{\theta}= \frac 1{\bar{Y}_n-\bar{X}_n}$$. I calculated $$E(Y),\Var(Y)$$ with the law of total expectation/variation: $$E(Y)=E[E(Y|X)]=\cdots=E(X)+\frac 1 {\theta},\quad \Var(Y)=\frac 1 {\theta^2}.$$

Then $$\sqrt n \left( \begin{pmatrix} \bar{Y}_n \\ \bar{X}_n \end{pmatrix}- \begin{pmatrix} E(Y) \\ E(X) \end{pmatrix}\right)\to_d N\left(\mathbf{0},\begin{pmatrix} \Var(Y)&\Cov(Y,X)\\ \Cov(Y,X) & \Var(X)\end{pmatrix}\right).$$

I'm not sure how to find the covariance, but let's introduce symbols for now: $$\Var(Y)=\sigma_Y^2, \Var(X)=\sigma_X^2, \Cov(Y,X)=\rho$$.

Define $$g(x,y)=\frac 1{x-y}$$, then $$J = \begin{pmatrix} \frac{-1}{(x-y)^2} && \frac{1}{(x-y)^2} \end{pmatrix}_{x=E(X)+\frac 1{\theta}, y=E(X)} = (-\theta^2 \quad \theta^2).$$

After some calculations we get $$\sqrt{n}(\hat{\theta}-\theta)\to_d N(0,(\sigma^2_X+\sigma_Y^2)\theta^4-2\rho \theta^4).$$

Now I was wondering if

• this approach is correct;
• $$\Cov(Y,X)$$ can be calculated
• it is possible to simplify $$\sigma_X^2+\sigma_Y^2$$?

Thanks.

• I think in the expression for $J$ you want the $+\frac1\theta$ on $y$, not on $x$? Jan 17, 2020 at 16:36
• I defined $g$ as a function of $(x,y)$, but in the CLT-equation the roles of $X$ and $Y$ are reversed. It's a little confusing indeed. Jan 17, 2020 at 18:11
• The given information makes it clear that $Y-X$ is independent of $X.$ But if the distribution of $X$ also depends on $\theta$ then there would be something to say about that. Jan 17, 2020 at 18:58
• @Zachary: It's indeed quite confusing; I still don't understand it. What are $g$ and $J$? You introduce them but then don't seem to use them. Are you working with some given form of the central limit theorem that you didn't include in the post? Jan 17, 2020 at 21:53
• It's the multivariate version of the central limit theorem. I then introduced the function $g$ for the multivariate delta method and $J$ is used to find the final variance: $J M J^T$, where $M$ is the covariance matrix. Jan 18, 2020 at 8:55

Since the conditional distribution of $$Y-X$$ given $$X$$ is the same regardless of the value of $$X,$$ you can conclude that $$X$$ and $$Y-X$$ are independent. For that reason you don't need the full joint distribution in order to find the asymptotic distribution of $$1/(\,\overline Y_n - \overline X_n\,).$$
(You can conclude that $$\operatorname{cov}(Y,X) = \operatorname{var}(X),$$ but as stated above, your purpose doesn't require that.)