Square of the sample mean from a Normal distribution

Question

Let $X$ be the mean of a random sample of size $n$ from a $N(θ,σ^2)$ distribution, $−∞ < θ < ∞$ and $σ^2 > 0$. Assume that $σ^2$ is known.

Show that $X^2 - (σ^2)/n$ is an unbiased estimator of $θ$, and find its efficiency.

Some hints about the distribution of $X^2$?

Thanks.

Answer 1

This is only true when $\theta = 0$ or $\theta =1$. Assuming $X_1,\ldots, X_n\stackrel{\mathrm{i.i.d.}}\sim N(\theta,\sigma^2)$, the Bienaymé formula yields \begin{align} \operatorname{Var}(\bar X_n) &= \operatorname{Var}\left(\frac 1n\sum_{i=1}^n X_i\right)\\ &= \frac1{n^2} \sum_{i=1}^n \operatorname{Var}(X_i)\\ &= \frac1{n^2} n\sigma^2\\ &= \frac{\sigma^2}n. \end{align} Since the second moment of the sample mean is given by \begin{align} \mathbb E[\bar X_n^2] &= \operatorname{Var}(\bar X_n) + \mathbb E[\bar X_n]^2\\ &= \frac{\sigma^2}n + \mathbb E\left[\frac1n\sum_{i=1}^n X_i\right]^2\\ &= \frac{\sigma^2}n +\frac1{n^2}\cdot(n\theta)^2\\ &= \frac{\sigma^2}n + \theta^2, \end{align} it is clear that $$ \mathbb E\left[\bar X_n^2 -\frac{\sigma^2}n \right] = \frac{\sigma^2}n + \theta^2 - \frac{\sigma^2}n = \theta^2, $$ and $\theta^2 = \theta$ implies that $\theta\in\{0,1\}$.