finding UMVU estimator of parameter γ(θ)

#1
suppose X_1,X_2,…,X_n,X_{n+1} be a random sample of distribution N(θ,1). if γ(θ)=P(∑Xi>X_{n+1}) (i=1,2,...,n)
how can i find UMVU estimator of parameter γ(θ)
 

BGM

TS Contributor
#2
I guess you will have to used the Lehmann–Scheffé theorem here.

Usually a crude unbiased estimator for the parameter in the form of a probability is the corresponding indicator.

The complete sufficient statistic for \( \theta \) is the sum of the sample.

Now the remaining task is to find the conditional expectation

\( g(t) = E\left[\mathbf{1}\left\{\sum_{i=1}^n X_i > X_{n+1}\right\}\Bigg|
\sum_{i=1}^{n+1} X_i = t\right] \)

Note that the above is just a conditional probability, with the following random variables have a a bivariate normal distribution:

\( \begin{bmatrix} \sum_{i=1}^n X_i - X_{n+1} \\ \sum_{i=1}^{n+1} X_i \end{bmatrix}
\sim \mathcal{N} \left(\begin{bmatrix}(n-1) \theta \\ (n+1)\theta \end{bmatrix},
\begin{bmatrix} n + 1 & n - 1 \\ n - 1 & n + 1 \end{bmatrix} \right)
\)

So the conditional distribution is

\( \sum_{i=1}^n X_i - X_{n+1}\Bigg| \sum_{i=1}^{n+1} X_i = t
\sim \) \( \mathcal{N}\left((n-1) \theta + \frac {n - 1} {n + 1}(t - (n + 1)\theta) = \frac {n - 1} {n + 1}t, \left(1 - \left(\frac {n - 1} {n + 1}\right)^2\right)(n+1)
= \frac {4n} {n+1} \right) \)

And the answer should be very obvious now :)