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 $\sigma_X^2+\sigma_Y^2$?


  • $\begingroup$ I think in the expression for $J$ you want the $+\frac1\theta$ on $y$, not on $x$? $\endgroup$
    – joriki
    Jan 17, 2020 at 16:36
  • $\begingroup$ 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. $\endgroup$
    – MyWorld
    Jan 17, 2020 at 18:11
  • $\begingroup$ 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. $\endgroup$ Jan 17, 2020 at 18:58
  • $\begingroup$ @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? $\endgroup$
    – joriki
    Jan 17, 2020 at 21:53
  • $\begingroup$ 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. $\endgroup$
    – MyWorld
    Jan 18, 2020 at 8:55

1 Answer 1


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.)


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