Finding the Expected value of $\hat{\theta}_n$

Question

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?

Answer 1

Quite simply, define $Y \sim \operatorname{Bernoulli}(1/n)$, hence $$\hat \theta_n = \bar X + nY;$$ consequently, $$\operatorname{E}[\hat \theta_n] = \operatorname{E}[\bar X + nY] = \operatorname{E}[\bar X] + n \operatorname{E}[Y] = \theta + n(1/n) = \theta + 1.$$

Answer 2

I think your logic is correct but there is a very tiny mistake. I think you forgot to take expectation of $\bar{X}$ so it should be

$$E[\hat{\theta}_n] = (1-1/n)E[\bar{X}]+(1/n)(\bar{X}+n) = E[\bar{X}]+1$$