Let $X_1, \dots, X_n$ be mutually independent random variables such that $X_j \sim N(j\theta, \sigma^2)$ for $j = 1,\dots, n \ $ where $\ \sigma^2 > 0 \ $ is known. $$ \bar{X} = \frac{\sum^n_{j = 1}X_j}{n} \ and \ \bar{X^*} = \frac{\Biggl [\sum^n_{j = 1}\frac{X_j}{j} \Biggr ]}{n}. $$

(1) Find a 95% confidence interval for $\theta$ based on $\bar{X}$.

$\bf{My \ thoughts:}$ $E[\bar{X}] = j\theta.$ and $ V[\bar{X}] = \frac{\sigma^2}{n}$ so $\bar{X} \sim N(j\theta,\frac{\sigma^2}{n}).\\$

$$ 0.95 = P[-1.96 \leq Z \leq 1.96]\\ = P[-1.96 \leq \frac{\bar{X} - j\theta}{\frac{\sigma}{\sqrt{n}}} \leq 1.96] \\ = P[-1.96 \bigl(\frac{\sigma}{\sqrt{n}} \bigr) \leq \bar{X} - j\theta \leq 1.96\bigl(\frac{\sigma}{\sqrt{n}} \bigr) ]\\ = P \bigl[-1.96 \bigl(\frac{\sigma}{\sqrt{n}} \bigr)- \bar{X} \leq - j\theta \leq 1.96\bigl(\frac{\sigma}{\sqrt{n}} \bigr) - \bar{X} \bigr]\\ = P \Biggl[\frac{-1.96 \bigl(\frac{\sigma}{\sqrt{n}} \bigr)- \bar{X}}{-j} \leq \theta \leq \frac{1.96 \bigl(\frac{\sigma}{\sqrt{n}} \bigr)- \bar{X}}{-j} \Biggr].$$

Still think that I might be missing something on this one.

(2) Find a 95% confidence interval for $\theta$ based on $\bar{X^*}$.

$\bf{My \ thoughts:}$ $E[\bar{X^*}] = \theta.$ and $ V[\bar{X^*}] = \frac{\sigma^2}{n^2}\sum^n_{j=1}\frac{1}{j^2}$. (From user1551)

So $\bar{X^*} \sim N \biggl(\theta, \frac{\sigma^2}{n^2}\sum^n_{j=1}\frac{1}{j^2} \biggr).$ I started setting this up as

$$0.95 = P[-1.96 \leq Z \leq 1.96]\\ = P[-1.96 \leq \frac{\bar{X^*} - \theta}{\sqrt{\frac{\sigma^2}{n^2}\sum^n_{j=1}\frac{1}{j^2}}} \leq 1.96]$$

I don't think I am right in they way I set up this interval though.

Any help is greatly appreciated.


MLE? What MLE? $\bar{X}^\ast$ is the sum of a number of independent normal distributions, hence it is normally distributed. Its mean and variance are respectively $\mu=\frac1n\sum\limits_{j=1}^n \frac{j\theta}{j}=\theta$ and $v=\frac{\sigma^2}{n^2}\sum\limits_{j=1}^n \frac1{j^2}$. I suppose you know how to construct a $95\%$ confidence interval for $N(\mu,v)$. The case for $\bar{X}$ is even simpler.

  • $\begingroup$ Wow, I was obviously not using my brain when I first looked at this problem. So for part one if have [X bar plus or minus 1.96(sigma/sqrt(n))]/j. Not quite sure if I went about this the right way though. $\endgroup$ – user45185 Mar 10 '13 at 6:15
  • $\begingroup$ Wouldn't the sum of 1/(j^2) be zero since it is a constant and the variance of a constant is zero? $\endgroup$ – user45185 Mar 10 '13 at 6:23
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    $\begingroup$ @user45185 No. Since the $X_j$s are *not* identically distributed (they are independent, though), $\bar{X}$ is the mean of $n$ iid r.v.. Hence the endpoints of the associated 95% CI for $\theta$ are not $\bar{X}\pm1.96\sigma/sqrt{n}$. You still need to evaluate the mean and variance of $\bar{X}$ in this case. The calculations are just simpler than those for $\bar{X}^\ast$. $\endgroup$ – user1551 Mar 10 '13 at 6:56
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    $\begingroup$ And what do you mean by "the variance of a constant is zero"? $\bar{X}^\ast$ is the mean of a number of independent normal random variables (not constants) $X_j$, each having a nonzero variance $\frac1j$. For example, when $j=2$, $\bar{X}^\ast=\frac12(\frac{X_1}1+\frac{X_2}2)=\frac{X_1}2+\frac{X_2}4$, where $X_1\sim N(\theta,\sigma^2)$ and $X_2\sim N(2\theta,\sigma^2)$. So $\frac{X_1}2\sim N(\frac\theta2,\frac{\sigma^2}4)$ and $\frac{X_2}4\sim N(\frac\theta2,\frac{\sigma^2}{16})$. Summing up, we get $\bar{X}^\ast\sim N(\theta,\frac{5\sigma^2}{16})$. $\endgroup$ – user1551 Mar 10 '13 at 7:05
  • $\begingroup$ When finding the variance of $\bar{X^*}$, I have $\frac{1}{n^2}\sum^n_{j=1} V \biggl[\frac{X_j}{j} \biggr]$. What rule can I use to find the variance of what is left inside? $\endgroup$ – user45185 Mar 11 '13 at 17:24

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