# Change of variable for normal ditribution

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?

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.