Let $X_1,X_2,\ldots,X_n$ be a random sample from $\mathcal{N}(\theta,1)$. Consider the following (randomized) estimator of $\theta$ given a sample of size $n$:
$$
\hat{\theta}_n = \bar{X} + \begin{cases}
0 & \text{with probability } 1−1/n,\\
n & \text{with probability } 1/n.
\end{cases}
$$
In this problem, what is expected value, that is, E[$\hat{\theta}_n$]?
I thought of $E[\hat{\theta}_n]$=($1-1/n$)$E[\bar{X}]$ + $(1/n)$($E[\bar{X}]+n$)=$E[\bar{X}]$+$1$
However is it correct?
Best Answer
Quite simply, define $Y \sim \operatorname{Bernoulli}(1/n)$, hence $$\hat \theta_n = \bar X + nY;$$ consequently, $$\operatorname{E}[\hat \theta_n] = \operatorname{E}[\bar X + nY] = \operatorname{E}[\bar X] + n \operatorname{E}[Y] = \theta + n(1/n) = \theta + 1.$$