Solving Variance Problem: Computing E(\hat{\theta}) and E(\hat{\theta}^2)

[mnb96]

Hello,
we are given N independent random variables $z_i$ defined as follows:

$$z_i = \theta + v_i$$

where the r.v. $v_i$ are zero-mean normal distributions $v_i \sim N(0,\sigma^2)$.

I want to compute the variance of the estimator

$$\hat{\theta}=\frac{1}{n}\sum_{i=1}^n z_i$$

However I can't seem to get the right result (I get always zero).

I computed first $E(\hat{\theta})^2=\theta^2$.

Then I try to compute $E(\hat{\theta^2})$ as follows:

$$\frac{1}{n^2}E\left[\left(\sum_{i=1}^n z_i \right)^2\right] = \frac{1}{n^2}E\left[\left(\sum_{i=1}^n (z_i)^2 \right) + \left(\sum_{i=1}^n z_i \right) \left( \sum_{j\neq i} z_j \right) \right]$$

$$= \frac{1}{n^2}E\left[\sum_{i=1}^n z_i^2 \right] + \frac{n-1}{n}\theta^2$$

$$= \frac{1}{n^2}E\left[\sum_{i=1}^n (\theta^2 +2\theta v_i + v_i^2) \right] + \frac{n-1}{n}\theta^2$$

$$= \frac{1}{n}\theta^2 + \frac{n-1}{n}\theta^2$$

From this we get $$Var(\hat{\theta})=E(\hat{\theta}^2)-E(\hat{\theta})^2 = 0$$.
Where is the mistake?

 

[jbunniii]

$$\frac{1}{n^2}E\left[\sum_{i=1}^n (\theta^2 +2\theta v_i + v_i^2) \right] + \frac{n-1}{n}\theta^2$$

$$= \frac{1}{n}\theta^2 + \frac{n-1}{n}\theta^2$$

This is wrong. You dropped the $v_i^2$ term.

 

[mnb96]

uh...that's true! Thanks.

$$E(v_i^2) = \frac{1}{\sqrt{2\pi \sigma^2}}\int_\mathbb{R} x^2 e^\frac{-x^2}{2\sigma^2} dx$$

This integral is indeed $$> \\ 0$$

 

[jbunniii]

uh...that's true! Thanks.

$$E(v_i^2) = \frac{1}{\sqrt{2\pi \sigma^2}}\int_\mathbb{R} x^2 e^\frac{-x^2}{2\sigma^2} dx$$

This integral is indeed $$> \\ 0$$

In fact, $E(v_i^2) = \sigma^2$.