Calculate variance of statistical function

Question

We have a sample $X_1,..,X_n$ from $N(0,θ)$. Ι can show the statistic $T(X)=\frac{1}{n}\sum_{i=1}^n X_i^2$ to be an unbiased estimator of $θ$. How can I prove its variance is $\dfrac{2θ^2}{n}?$

Answer 1

Note that $\frac{X_i}{θ}=Z_i \sim N(0,1)$. Hence, $Z^2_i \sim \chi^2_1$, whose mean and variance is 1 and 2, respectively (see here for more details).

Another way is to compute the variance of $Z^2_i$ directly, which is $\mathbb E (Z^4_i) - \mathbb E (Z^2_i)^2=3-1=2$ (see here for more details on the raw moments of normal distribution).

Thus, $\text{var}(X^2_i)=2θ^2$, which gives the result you want.

Answer 2

Hi: What Amir did was fine but another way to do it is to take the expectation of the expression directly.

$E(\frac{1}{n} \sum_{i=1}^{n} X_{i}^2) = \frac{1}{n} \sum_{i=1}^{n} E(X_{i}^2) = \frac{1}{n} n \theta = \theta$.

Note: $E(X_{i}^2) = Var(X_{i}) = \theta$ because the mean, $\mu$, is zero.