# Build an approximated confidence interval for $\sigma$ based on its maximum likelihood estimator

Let $$X_{1},X_{2},\ldots,X_{n}$$ be a random sample whose distribution is given by $$\mathcal{N}(0,\sigma^{2})$$. Build an approximated confidence interval for $$\sigma$$ based on its maximum likelihood estimator.

MY ATTEMPT

To begin with, let us find the maximum likelihood estimador of $$\sigma$$: \begin{align*} & L(\textbf{x}|\sigma) = \left(\frac{1}{\sigma\sqrt{2\pi}}\right)^{n}\exp\left(-\frac{1}{2\sigma^{2}}\sum_{k=1}^{n}x^{2}_{k}\right) \Rightarrow\\ & \ln L(\textbf{x}|\sigma) = -n\ln(\sigma) - \frac{n}{2}\ln(2\pi) - \frac{1}{2\sigma^{2}}\sum_{k=1}^{n}x^{2}_{k}\Rightarrow\\ & \frac{\partial\ln L(\textbf{x}|\sigma)}{\partial\sigma} = -\frac{n}{\sigma} + \frac{1}{\sigma^{3}}\sum_{k=1}^{n}x^{2}_{k} = 0 \Rightarrow \hat{\sigma} = \sqrt{\frac{1}{n}\displaystyle\sum_{k=1}^{n}x^{2}_{k}} \end{align*}

However, I do not know how to proceed from here. Can someone help me out? Thanks in advance!

You are just a few steps away; I will show most of them. The MLE of $$\sigma^2$$ is $$V = \hat \sigma^2 = \frac 1 n \sum_i X_i^2.$$ So $$\frac{nV}{\sigma^2} \sim \mathsf{Chisq}(n).$$ Choosing $$L$$ and $$U$$ to cut 2.5% from the lower and upper tails, respectively, of $$\mathsf{Chisq}(n),$$ we have $$P(L < nV/\sigma^2 < U) = P(nV/U < \sigma^2 < nV/L)\\ = P\left(\sqrt{nV/U} < \sigma < \sqrt{nV/L}\right) = 0.95.$$ Thus, a 95% confidence interval for $$\sigma$$ is of the form $$\left(\sqrt{nV/U},\; \sqrt{nV/L}\right).$$
Example: Consider a sample of size $$n = 25$$ from $$\mathsf{Norm}(\mu = 0, \sigma=10).$$ The R code below simulates such a sample, and finds the 95% confidence interval $$(9.198,\, 16.190)$$ for $$\sigma.$$
set.seed(412);  n = 25;  sg = 10

Because we sampled from a population with $$\sigma = 10,$$ we know this is one of the 95% of such intervals that include $$\sigma.$$ [If you use the same seed, you will get the same 25 observations, and so the same CI; choose a different seed or omit the set.seed statement for a fresh dataset.]