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Let $(X_1, Y_1),\ldots, (X_n, Y_n)$ be a random sample of size $n$ from the continuous distribution with joint pdf:

$$f_{X, Y} (x, y\mid\theta) = \frac{1}{\theta y}\exp\left(\frac{-x}{\theta y}\right)I_{(0,\infty)}(x) I_{(0,1)}(y) I_{(0,\infty)}(\theta)$$

(a) Find a complete and sufficient statistic for $\theta$.

(b) Find the maximum likelihood estimator (MLE) for $\theta$.

(c) Find the MLE for $P(X < Y)$.

(d) Let $V= \frac{X}{\theta}$ and $W = Y$. Show that the joint distribution of $(V, W)$ does not depend on $\theta$.

(e) Define the statistic $S((X_1, Y_1),\ldots,(X_n, Y_n))$ by:

$$S((X_1, Y_1),\ldots,(X_n, Y_n)) = \frac{\sum\limits_{i=1}^n X_i Y_i} {\sum\limits_{j=1}^n X_j}$$

Show that $S((X_1, Y_1),\ldots,(X_n, Y_n))$ is an ancillary statistic for the model $f_{X,Y}(x,y\mid \theta)$. Note that you cannot claim the it is part of the scale parameter family when the pdf is a joint distribution. Thus, to show that S is ancillary, I will have to show that its distribution does not depend on $\theta$.

I have completed parts a, b, and d, but I am stuck on how to start the last two parts. Any help would be greatly appreciated. I thought for part c to utilize the invariance property of MLE's, but I wasn't sure how to go about using it. Then for part e, I have no idea how to start it.

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I'm pretty sure the reason part (d) is there is that you're supposed to do part (e) by using the result of part (d). Notice that $\dfrac{\sum_i V_i W_i}{\sum_i V_i}= \dfrac{\sum_i X_i Y_i}{\sum_i X_i}$.

For part (c), you need to find a function of $\theta$: $$ g(\theta) = \Pr(X<Y). $$ If you have the MLE $\hat\theta$ of $\theta$, then $g(\hat\theta)$ is the MLE for $\Pr(X<Y)$.

After a hasty computation I get $\Pr(X<Y)= 1 - e^{-1/\theta}$, so its MLE should be $1-e^{-1/\hat\theta}$.

....and for the MLE for $\theta$ I get $\displaystyle\frac1n\sum_{i=1}^n \dfrac{X_i}{Y_i}$, so the MLE for $\Pr(X<Y)$ should be $$ 1 - \exp\left( \frac{-n}{\sum_{i=1}^n (X_i/Y_i)} \right). $$

Later addendum: The joint distribution of $(X,Y)$ is \begin{align} & \phantom{{}=} \frac1y\exp\left(\frac{-x}{\theta y}\right)\,\frac{dx}{\theta}\, dy\qquad\text{on }x>0,\ 0<y<1. \\[8pt] & = \frac1y\exp\left(\frac {-v}{y}\right)\,dv\,dy\qquad\text{on }v>0,\ 0<y<1. \end{align} We want $\Pr(X<Y)$. That is $$ \int_0^1\int_0^y\frac1y\exp\left(\frac{-x}{\theta y}\right)\,\frac{dx}{\theta}\, dy. $$ In the inner integral, $x$ goes from $0$ to $y$. Then \begin{align} & \phantom{{}=} \int_0^1 \int_0^y\frac1y\exp\left(\frac{-x}{\theta y}\right)\,\frac{dx}{\theta}\, dy = \int_0^1 \int_0^{y/\theta} \frac1y\exp\left(\frac {-v}y\right) \, dv\,dy \\[8pt] & = \int_0^\infty \left[ -\exp\left(\frac {-v}y\right)\right]_{v=0}^{v=y/\theta} \, dy = \int_0^1 (1 - e^{-1/\theta}) \, dy = 1 - e^{-1/\theta}. \end{align}

Second later addendum: The likelihood function is $$ L(\theta) = \prod_{i=1}^n \frac{1}{\theta y_i}\exp\left(\frac{-x_i}{\theta y_i}\right) = \frac{1}{\theta^n \prod_{i=1}^n y_i}\exp\left( \frac{-1}\theta \sum_{i=1}^n \frac{x_i}{y_i} \right) $$ So $$ \ell(\theta)=\log L(\theta) = -n\log\theta -\frac1\theta\sum_{i=1}^n\frac{x_i}{y_i}+\text{constant} $$ and then $$ \ell\;'(\theta) = \frac{-n}\theta + \frac{1}{\theta^2}\sum_{i=1}^n\frac{x_i}{y_i} =\frac{1}{\theta^2}\left( -n\theta+\sum_{i=1}^n\frac{x_i}{y_i} \right). $$ Finally we get $$ \begin{cases} \ell\;'(\theta) > 0 & \text{if } 0\le\theta<\frac1n\sum_{i=1}^n\frac{x_i}{y_i}, \\[10pt] \ell\;'(\theta) = 0 & \text{if } \theta=\frac1n\sum_{i=1}^n\frac{x_i}{y_i}, \\[10pt] \ell\;'(\theta) < 0 & \text{if } \theta>\frac1n\sum_{i=1}^n\frac{x_i}{y_i}. \end{cases} $$

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Thank you for the help. But I was wondering about your MLE for $\theta$. When you took the ln of the likelihood function of $\theta$ given X and Y, the ln of $e^{\sum\limits_{i=1}^n\frac{X_{i}}{Y_{i}*\theta}}$ you can take the $\theta$ out of the summation, but when you add it, then doesn't it become $\frac{n}{\theta}$? So, then when you take the derivative it is now $\frac{n}{\theta^{2}}$, right? So then when you solved for $\theta$, how did you get the n there? – user61752 Feb 11 '13 at 4:44
I guess I am also a little confused on how you came up with your $g(\theta) = Pr(X<Y)$ function. Any hints on how you computed that? Please and thank you! – user61752 Feb 11 '13 at 4:53
I've added an explanation of how I found $\Pr(X<Y)$. – Michael Hardy Feb 11 '13 at 14:53
. . . . . and now I've added an explanation of how I found the MLE. – Michael Hardy Feb 11 '13 at 15:25
Thank you for the help! It is greatly appreciated! – user61752 Feb 11 '13 at 20:23

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