Btukfyl

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

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 $θ=int_0^ ∞int_x^infty frac{1}{β} ~e^{−x/β}~~ frac{1}{⍺ } ~e^{−y/⍺} , dy , dx= fracalpha {alpha+beta}$

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.

$x_1+x_2sim operatorname{Gamma}(2,beta)$ and $y_1+y_2sim operatorname{Gamma}(2,alpha).$

Now, I am stuck, any help please.

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 $Pr(Xle x)$ (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
$$
f_{X_1,X_2}(x_1,x_2) = frac 1 {beta^2} e^{-(x_1+x_2)/beta} quadtext{for } x_1,x_2 ge 0,
$$
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(Wmid X_1+X_2, Y_1+Y_2) \[10pt]
= {} & Pr(W=1mid X_1+X_2, Y_1+Y_2) \[10pt]
= {} & Pr(X_1<Y_1mid 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 $Prleft( U_1 < dfrac y x U_2 right).$
$$
Prleft(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 UMVU 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 $xmapsto xg(x).$ 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 $xge0.$ Thus we have no nontrivial unbiased estimators of zero.