1
$\begingroup$

Suppose we have a random sample $X_1, X_2, \ldots, X_n$ from exponential$~(β >0)$ $\text{i.e. }f(x\mid β) = {1/β} ~e^{−x/β}$

and a random sample$~Y_1, Y_2, \ldots, Y_n$ from exponential$~(⍺ >0)$ and assume both sample are independent.

let$~~ θ = P(X_1 < Y_1)$ Find the MVUE of $~θ~~$ for $n=2$.

So, first I calculate the $$θ=\int_0^ ∞\int_x^\infty \frac{1}{β} ~e^{−x/β}~~ \frac{1}{⍺ } ~e^{−y/⍺} \, dy \, dx= \frac\alpha {\alpha+\beta}$$

Then since $f(X,Y) =f(X)\cdot f(Y) =f(X_1)\cdot f(X_2)\cdot f(Y_1)\cdot f(Y_2)$ belongs to exponential family then
$(x_1+x_2, y_1+y_2)$ is complete sufficient statistics.

$x_1+x_2\sim \operatorname{Gamma}(2,\beta)$ and $y_1+y_2\sim \operatorname{Gamma}(2,\alpha).$

Now, I am stuck, any help please.

$\endgroup$
  • $\begingroup$ Im sorry I dont really understand what your are trying to calculate? What is MVUE and is this what you are trying to find? $\endgroup$ – Jesper Hybel Jan 1 at 4:10
  • $\begingroup$ My bad, minimum variance unbiased estimator $\endgroup$ – user0533535412 Jan 1 at 7:19
5
$\begingroup$

Some of your notation is abominable and most unfortunately, you are in good company. You're using the same letter $f$ to refer to several different functions. If instead one writes $f_X$ and $f_Y$ then one can understand the difference between $f_X(3)$ and $f_Y(3),$ and one can understand things like $\Pr(X\le x)$ (where $X$ and $x$ are two different things).

And you should say $X_1+X_2,$ rather than $x_1+x_2,$ has a gamma distribution, and similarly for the other one.

You have $$ f_{X_1,X_2}(x_1,x_2) = \frac 1 {\beta^2} e^{-(x_1+x_2)/\beta} \quad\text{for } x_1,x_2 \ge 0, $$ and the fact that this depends on $(x_1,x_2)$ only through $x_1+x_2$ is sufficient (but not necessary) to establish that $X_1+X_2$ (not $x_1+x_2$) is a sufficient statistic for $\beta.$

Showing completeness is another matter, but before that let's Rao–Blackwellize.

Let $W = \begin{cases} 1 & \text{if } X_1 < Y_1, \\ 0 & \text{otherwise.} \end{cases}$

Then $W$ is an unbiased estimator of $\theta.$ So the Rao–Blackwell estimator is \begin{align} & \operatorname E(W\mid X_1+X_2, Y_1+Y_2) \\[10pt] = {} & \Pr(W=1\mid X_1+X_2, Y_1+Y_2) \\[10pt] = {} & \Pr(X_1<Y_1\mid X_1+X_2, Y_1+Y_2). \end{align} The conditional distribution of $X_1$ given that $X_1+X_2=x$ is uniform on the interval $[0,x]$ because the joint density of $(X_1,X_2)$ is constant on that set. Similarly the conditional distribution of $Y_1$ given $Y_1+Y_2=y$ is uniform on $[0,y].$ Hence the conditional distribution of $(U_1,U_2)=(X_1/x,Y_1/y)$ given $X_1+X_2=x, \, Y_1+Y_2=y$ is uniform in the square $[0,1]\times[0,1].$ We seek $\Pr\left( U_1 < \dfrac y x U_2 \right).$ $$ \Pr\left(U_1 < \frac y x U_2 \right) = \begin{cases} y/(2x) & \text{if } x \ge y \\[8pt] 1 - x/(2y) & \text{if } x \le y. \end{cases} $$ So the Rao–Blackwell estimator is $$ \frac 1 2 \times \begin{cases} \frac{Y_1+Y_2}{X_1+X_2} & \text{if that is} \le 1/2, \\[8pt] 1-\frac{X_1+X_2}{Y_2+Y_2} & \text{if that is} \ge 1/2. \end{cases} $$

That's the UMVUE if we have completeness.

\begin{align} & \operatorname E(g(X_1+X_2)) \\[8pt] = {} & \frac 1 {\Gamma(2)} \int_0^\infty g(x) x^{2-1} e^{-x/\beta} \, \frac{dx} \beta. \end{align} This is the Laplace transform, evaluated at $1/\beta,$ of $x\mapsto xg(x).$ We want it to be $0$ regardless of the value of $\beta.$ That can happen only if $xg(x)$ is $0$ for all values of $x\ge0.$ Thus we have no nontrivial unbiased estimators of zero.

$\endgroup$

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.