# finding UMVU estimator of parameter γ(θ)

##### New Member
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
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