I am having difficulty with this. My procedure for solving it is that
$$ E(\theta)= \frac 1 2 E(X-0.1) + \frac 1 2 E(X+0.1) = \frac 1 2 $$ So, $E(theta)\frac 1 2 - (\theta)\frac 1 2 = 0$, which means it is unbiased.
The variance is
$$ E(\theta^2)-E(\theta)^2 = \frac 1 2 E((X-0.1)^2) + \frac 1 2 E((x+0.1))^2)-(E(x))^2 = 0.01 $$ And $V(x(ba)) = 1/12$
So, I thought that since the variance is smaller than $x(ba)$, I should choose it.
So far, am I doing it right?
The next one is that
$$ E(\theta) = \frac 1 2 E(x) + \frac 1 2 E(x+0.1) = 0.55 $$ $0.55 - 0.5$ is $0.05$, which means it's biased and the bias is $0.05$. $$ V(\theta) = E(\theta^2)-E(\theta)^2 = E(x^2) + 0.1E(x) - (0.55)^2 $$ It is biased so, I should choose $x(ba)$ as an estimator.
I'm not sure about my answer; can someone please tell me whether I am wrong or not?