Estimating confidence Interval for unknown Variance, Normal distribution

I've been stuck with this question for a while:

I've learnt how to use t-distribution to estimate CIs for an unknown variance, but I'm unsure how that applies to this situation.

Any help would be greatly appreciated.

Thank you!

• I'm not sure if it's from a textbook, it's just a practice question from a worksheet that I'm struggling with. – J. Bant Feb 7 '16 at 18:53

$X_1,\ldots,X_n$ are i.i.d. $\mathcal N(\mu,\sigma^2)$ random variables and $\bar X=n^{-1}\sum_{i=1}^nX_i$. The distributions and confidence intervals are as follow.

(a) $$\frac1{\sigma^2}\sum_{i=1}^n(X_i-\bar X)^2\sim\chi_{n-1}^2$$ and $$\Pr\biggl(\frac{\sum_{i=1}^n(X_i-\bar X)^2}{\chi_{n-1,\alpha/2}^2}\le\sigma^2\le\frac{\sum_{i=1}^n(X_i-\bar X)^2}{\chi_{n-1,1-\alpha/2}^2}\biggr)=1-\alpha.$$ (b) $$\frac1{\sigma^2}\sum_{i=1}^n(X_i-\mu)^2\sim\chi_n^2$$ and $$\Pr\biggl(\frac{\sum_{i=1}^n(X_i-\mu)^2}{\chi_{n,\alpha/2}^2}\le\sigma^2\le\frac{\sum_{i=1}^n(X_i-\mu)^2}{\chi_{n,1-\alpha/2}^2}\biggr)=1-\alpha.$$

(c)

$$\frac{n(\bar X-\mu)^2}{\sigma^2}\sim\chi_1^2$$ and

$$\Pr\biggl(\frac{n(\bar X-\mu)^2}{\chi_{1,\alpha/2}^2}\le\sigma^2\le\frac{n(\bar X-\mu)^2}{\chi_{1,1-\alpha/2}^2}\biggr)=1-\alpha.$$

I hope this helps.

• Thank you, but I'm struggling slightly to see how you've related these distributions to Chi squared distributions? I think I'm just being thick here – J. Bant Feb 7 '16 at 19:13
• By definition, $\chi_n^2$ is the distribution of a sum of $n$ squared standard normal random variables (I used this in (b) and (c)). Using Cochran's theorem, we can show that the distribution of $\sigma^{-2}\sum_{i=1}^n(X_i-\bar X)^2$ is $\chi_{n-1}^2$ (see here). – Cm7F7Bb Feb 7 '16 at 19:18
• Ah, very well explained, thank you very much! – J. Bant Feb 8 '16 at 13:59

I believe you have the denominators switched. Also, say $c_1 = \chi^2_{1, 1-\alpha/2}$ and $c_2 =\chi^2_{1, \alpha/2 }$ then for (c), the expected width of the interval is $$\mathbb{E} \left[ n (\bar{X} - \mu )^2 ( \frac{1}{c_2} - \frac{1}{c_1}) \right] = n ( \frac{1}{c_2} - \frac{1}{c_1}) \times \mathbb{E}(\bar{X} - \mu )^2 = n ( \frac{1}{c_2} - \frac{1}{c_1}) \; \; \times \frac{\sigma^2}{n} = ( \frac{1}{c_2} - \frac{1}{c_1}) \sigma^2$$

since $\quad \quad \frac{n}{\sigma^2} ( \bar{X} - \mu ) \sim \chi^2_1$

and thus $\quad \mathbb{E}(\bar{X} - \mu )^2 = \frac{\sigma^2}{n}$

strange the expected width does not depend on $n$...

• "I believe you have the denominators switched" - whom are you talking to? – Alex M. Apr 18 '18 at 15:51