Chi-squared confidence interval for variance When constructing, for example, a $90\%$ confidence interval for the population variance using the chi-squared distribution, we have:
\begin{align}
& P\left(a<\frac{(n-1)S^2}{\sigma^2}<b\right) \\
= {} & P\left(\frac{\sum(\bar{X}-X_i)^2}{b}<\sigma^2<\frac{\sum(\bar{X}-X_i)^2}{a}\right)=0.9, \\ & \text{ where } \frac{(n-1)S^2}{\sigma^2}\sim\chi_{n-1}^2.
\end{align}
In my course, we then find $a, b$ such that $$P(\chi_{n-1}^{2}<a)=P(\chi_{n-1}^{2}>b)=0.05.$$
My question is, given that the chi-squared distribution is asymmetric for small $n,$ why do we pick $0.05$ for both sides? Surely we’d have a shorter confidence interval if we weighted more of the $0.1$ probability to one of the sides, instead of splitting equally?

$$
\Pr(a<\chi^2_k<b) = 0.9 \tag 1
$$
One way to choose $a$ and $b$ is to choose them so that the values of the chi-square density function at those two points are equal to each other, and at the same time so that line $(1)$ above is true.
That does give you a shorter confidence interval, but it is numerically somewhat complicated to implement, and probably the purpose in the course you're taking is just to show that it is possible to get a confidence interval for $\sigma^2$ by "inverting" the "pivotal quantity" $(n-1)S^2/\sigma^2.$
