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