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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?

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  • $\begingroup$ What is $x(ba)$? Is that supposed to be $\bar x$ (ie, $\bar x$)? $\endgroup$ – gung May 20 '15 at 17:24
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    $\begingroup$ It seems your variance calculation (at "The variance is..."), if carried out with $0.1$ replaced by $0.0$, would tell you that the variance of $\hat\theta_1$ is $0$. Is that really plausible? Certainly not. Therefore there must be an error in that calculation. $\endgroup$ – whuber May 20 '15 at 17:25

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