calculating variance problem

Question

When calculate variance, It's 1/2E((x-0.1)^2)+1/2E((x+0.1)^2)-E(x)^2

Is it right? So, the answer is 0.01

Am I missing something?

And whenever I calculate the vatiance of the second estimator

There left E(x^2)

What's wrong with that?

Answer 1

The mean of the sample, $\bar X$, is a constant.   It's expected to be what it is.   $\mathsf E(\bar X)=\bar X$

The expectation of $\hat\theta$ is half the sum of its values (as they are equi-probable). Likewise the expectation of the square.

$$\begin{align} \mathsf {Var}(\hat\theta_1) & = \mathsf E(\hat{\theta_1^2})-\mathsf E(\hat\theta_1)^2 \\ & = \Big(\tfrac 1 2 (\bar X-0.1)^2+\tfrac 1 2(\bar X+0.1)^2\Big) - {\Big(\tfrac 1 2 (\bar X-0.1)+\tfrac 1 2(\bar X+0.1)\Big)}^2 \\ & = \Big(\bar X^2+0.01\Big) - {\Big(\bar X\Big)}^2 \\ & = 0.01 \end{align}$$

Then likewise:

$$\begin{align} \mathsf {Var}(\hat\theta_2) & = \mathsf E(\hat{\theta_2^2})-\mathsf E(\hat\theta_2)^2 \\ & = \Big(\tfrac 1 2 (\bar X)^2+\tfrac 1 2(\bar X+0.1)^2\Big) - {\Big(\tfrac 1 2 (\bar X)+\tfrac 1 2(\bar X+0.1)\Big)}^2 \\ & = \Big(\bar X^2+0.1\bar X+0.005\Big) - {\Big(\bar X+0.05\Big)}^2 \\ & = \Big(\bar X^2+0.1\bar X+0.005\Big) - \Big(\bar X^2+0.1\bar X+0.0025\Big) \\ & = 0.0025 \end{align}$$

That is all.