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Assume that $X_1,X_2,\ldots,X_N\sim N(\mu,2^2)$ and $Y_1,Y_2,\ldots,Y_M\sim N(0,\sigma^2)$.

a)Find the Cramer-Rao Lower Bound (CRLB) for the variance of the unbiased estimators of $\mu$.

b)Find the CRLB for the variances of the unbiased estimators of $\mu^2$.

c)Is the MLE, $\hat{\mu}$, a uniformly minimum variance unbiased estimator (UMVUE) of $\mu$?

so for part a) I got $\dfrac{4}{n}$ and for part b) I got $\dfrac{\sigma^2}{n}$. Are these answers correct? Just want to know if I'm on the right track.

Lastly for part C, can anyone give me some guidance on where to start? Kind of lost haha

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So what have the $Y_i$ variables got to do with it?

Anyway, your answer to part (a) looks fine. Your answer to part (b) does not. You should be able to derive part (b) from part (a). In particular, if the CRLB on $\theta$ is say $CRLB(\theta)$, and if $g(\theta)$ is some differentiable function of $\theta$ that we are interested in estimating, then, subject to some regularity conditions, the CRLB on $g(\theta)$ is:

$$\left(\frac{\partial g(\theta) }{\partial \theta }\right)^2 CRLB(\theta) $$

In this case, $g(\theta)$ = $\mu^2$, and $\theta$ is $\mu$, so the rest is left to u.


And since he has now solved it :) (see comment below), one can also check one's work on such things with a computer algebra system. For this problem, part (b) would simply be:

$$\text{CRLB}=\frac{1}{n *\text{FisherInformation}\left[\mu ^2,f\right]}$$

where FisherInformation is a mathStatica funkeh monkeh, and $f$ is the $N(\mu,\sigma^2)$ pdf, and which returns, as output:

$\frac{4 \mu ^2 \sigma ^2}{n}$

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  • $\begingroup$ Hi wolfies thanks for your input. What you're saying makes sense to me. After doing some calculations I get $\frac{16\theta^2}{n}$. Is this correct? $\endgroup$
    – Tim
    May 16, 2013 at 20:37
  • $\begingroup$ Yes - if you replace $\theta$ with $\mu$ :) $\endgroup$
    – wolfies
    May 16, 2013 at 20:39
  • $\begingroup$ sweet thanks wolfies :) I feel so accomplished haha but without your help I would not have been able to get part b thanks $\endgroup$
    – Tim
    May 16, 2013 at 20:42
  • $\begingroup$ lol funkeh monkeh haha thanks. I don't have mathstatica but I know what you're doing. Thanks wolfies! $\endgroup$
    – Tim
    May 16, 2013 at 20:48

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