I have a random sample drawn from a $N(\theta,\sigma^2)$ distribution with $\sigma^2$ known. I am trying to estimate $\theta$.

I need to calculate the efficiency of the unbiased estimator, $\bar{X}^2-\frac{\sigma^2}{n}$. To do so, I have been trying to find the pdf of the estimator by using the MGF. However, I have been having some difficulty in doing so.

My thought process thus far:

$M_{\frac{1}{n^2} \sum_{i=1}^{n} X_{i}^{2} -\frac{\sigma^2}{n} (t)} =e^{\frac{-\sigma^2}{n}t}M_{\sum_{i=1}^{n}X_{i}^2}(\frac{1}{n^2}t)$

Can we say that $X_{i}^2=\chi_{1}^2$? If so:

$e^{-\frac{\sigma^2}{n}t}[M_{X_{i}^2}(\frac{1}{n^2}t)]^n =e^{-\frac{\sigma^2}{n}t}[(\frac{1}{1-2(\frac{1}{n^2}t)})^{\frac{1}{2}}]^n =e^{-\frac{\sigma^2}{n}t}[\frac{1}{1-\frac{2}{n^2}}t]^{\frac{n}{2}}$

Do we instead need to take into account that $X\sim N(\theta,\sigma^2)$? Does that mean we need to make $\frac{(X_{i}-\theta)^2}{\sigma^2}$ appear in our equation? If so, how would I do that? Does the following make any sense?

$\frac{1}{n^2}\displaystyle \sum_{i=1}^{n}X_{i}^2-\frac{\sigma^2}{n} =\sigma^2 \frac{1}{\sigma^2}\frac{1}{n^2}\displaystyle \sum_{i=1}^{n}X_{i}^2-\frac{\sigma^2}{n} =\sigma^2\frac{1}{\sigma^2}\frac{1}{n^2}\displaystyle \sum_{i=1}^nX_{i}^2-2\theta X_i+\theta^2+2\theta X_i-\theta^2-\frac{\sigma^2}{n} =\sigma^2\frac{1}{n^2}\displaystyle \sum_{i=1}^{n}\frac{(X_i-\theta)^2}{\sigma^2}+\frac{1}{n^2}\displaystyle \sum_{i=1}^n 2\theta X_i-\frac{1}{n^2}\displaystyle \sum_{i=1}^n \theta^2-\frac{\sigma^2}{n} =\frac{\sigma^2}{n^2}\chi^2_n+\frac{2\theta}{n^2}\displaystyle \sum_{i=1}^{n}X_i-\frac{\theta^2}{n}-\frac{\sigma^2}{n}$

*We'd then take the MGF of the above expression.

Are we now dealing with the sum of a chi-squared and normal distribution? Am I overthinking this?

I'd appreciate any guidance! As you can see, I am quite confused as to what to do. Thank you in advance!

  • $\begingroup$ Oh, I'm sorry! I should have been more clear. The random sample is drawn from a $N(\theta, \sigma^2)$ distribution, where $\sigma^2$ is known. We are estimating $\theta$. $\endgroup$ – Jess Feb 5 '13 at 16:30
  • $\begingroup$ $\overline{X}$ has perfect efficiency, i.e., it saturates the Cramer-Rao bound. $\endgroup$ – zyx Feb 5 '13 at 23:49
  • $\begingroup$ I think you are estimating $\theta^2$ rather than $\theta$? Your estimator is not even consistent for $\theta$. $\endgroup$ – Yining Wang Jun 19 '17 at 16:34

In order to proof ordinary least squares estimator as BLUE (best linear unbiased estimator) it is sufficient to proof Gauss Markov theorem. I guess you could probably try to go in that direction. A proof if the Gauss Markov theorem can be found under



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