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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.

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  • $\begingroup$ Im sorry I dont really understand what your are trying to calculate? What is MVUE and is this what you are trying to find? $\endgroup$ Jan 1, 2019 at 4:10
  • $\begingroup$ My bad, minimum variance unbiased estimator $\endgroup$
    – user161381
    Jan 1, 2019 at 7:19

1 Answer 1

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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.

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