# Finding the MVUE from two independent random samples

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.

• Im sorry I dont really understand what your are trying to calculate? What is MVUE and is this what you are trying to find? Jan 1, 2019 at 4:10
• My bad, minimum variance unbiased estimator Jan 1, 2019 at 7:19

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 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.