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I am little bit confused about how we get the upper bound for CI for variance: We have $\frac{(n-1) s^2}{\sigma^{2}}\sim X^2(n-1)$

$$P\left(\frac{(n-1)s^2}{\sigma^{2}}>X^2_{1-\alpha,n-1}\right)=1-\alpha$$ and if we solve for $\sigma^{2}$ then we should get $$P\left(\frac{1}{\sigma^{2}}>\frac{X^2_{1-\alpha,n-1}}{(n-1)s^2}\right)=1-α\Longrightarrow P\left(\sigma^{2}>\frac{(n-1)s^2}{X^2_{1-\alpha,n-1}}\right)=1-α\\\Longrightarrow CI=\left[\frac{(n-1)s^2}{X^2_{1-\alpha,n-1}},+\infty\right].$$ But I do not know how the upper bound CI come out to be $\left[0,\dfrac{(n-1)s^2}{X^2_{1-\alpha,n-1}}\right]$???

F.O
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1 Answers1

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You want $\Pr(U <\sigma^2<V) = 1-\alpha$ where $U$ and $V$ are observable random variables. And "observable" means you can compute them if you know $s^2$ and don't know $\sigma^2.$

Find $A$ and $B$ so that $$ \Pr\left(\frac{(n-1)s^2}{\sigma^2}>A\right)=\frac \alpha 2 \text{ and } \Pr\left(\frac{(n-1)s^2}{\sigma^2}<B\right)=\frac \alpha 2. $$ Then you have $\alpha/2$ in each of the two tails and therefore $1-\alpha$ in the middle. So $$ \Pr\left( B < \frac{(n-1)s^2}{\sigma^2} < A \right) = 1-\alpha. $$

\begin{align} & \frac{(n-1)s^2}{\sigma^2} < A \text{ is equivalent to } \frac{(n-1)s^2} A < \sigma^2, \text{ and} \\[10pt] & \frac{(n-1)s^2}{\sigma^2} > B \text{ is equivalent to } \frac{(n-1)s^2} B > \sigma^2. \\[10pt] \text{The } & \text{conjunction of those two is equivalent to} \\[10pt] & \frac{(n-1)s^2} A < \sigma^2 < \frac{(n-1)s^2} B. \end{align}

Note that $$ \frac 1 {\sigma^2}>\frac{X^2_{1-\alpha,n-1}}{(n-1)s^2} $$ is equivalent to $$ \sigma^2 < \frac{(n-1)s^2}{X^2_{1-\alpha,n-1}}. $$

Michael Hardy
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