Change of variable for normal ditribution

Question

Given $X_1, X_2,\ldots,X_n$ are independent random variable such that $\hat{\mu} = \frac{\sum_i^n{X_1 + X_2 + \cdots + X_n}}{n}$. The pdf of normal distribution is $f(x) = \frac{1}{2\pi} e^{\frac{-(x-\mu)^2}{2\sigma^2}}$. Also, let $\Phi^{-1}(t) $ be in inverse function of cdf of normal distribution. Define another random variable $I$ such that $I = \hat{\mu} - \frac{\sigma}{\sqrt{n}} \Phi^{-1}(q)$ where $q \in (0, 1).$

Q: I need to show that $P(I < \mu) = q.$ For this question, I think change of variable technique is the only way to prove the statement. Is there any other method to get around messy calculation?

Answer 1

Since $X_i\sim\mathcal N(\mu,\sigma^2)$ for all $i$ and $X_1,X_2,\ldots,X_n$ are idependent, the sum $X_1+\ldots+X_n\sim \mathcal N(n\mu, n\sigma^2)$ also has normal distribution. Then $$ \hat\mu=\frac{X_1+\ldots+X_n}{n} \sim \mathcal N\left(\mu, \frac{\sigma^2}{n}\right) $$ and $$ \sqrt{n}\frac{\hat\mu-\mu}{\sigma} \sim \mathcal N(0,1). $$ Therefore $$ \mathbb P(I < \mu) =\mathbb P(\hat{\mu} - \frac{\sigma}{\sqrt{n}} \Phi^{-1}(q) < \mu)=\mathbb P\left(\sqrt{n}\frac{\hat\mu-\mu}{\sigma} < \Phi^{-1}(q)\right) = \ldots $$ You only need to finish the calculations.