Let $X=\{X_1, X_2,...X_n\}$ where $X_1, X_2,...X_n$ are i.i.d $\mathcal{N}(\mu,\sigma)$-distributed.

Given $q\in(0,1)$ and $n\in\mathbb{N}$, when I simulate the probability $$p=P(q < \int_{\min (X)}^{\max (X)} \frac{e^{-\frac{(x-\mu )^2}{2 \sigma ^2}}}{\sqrt{2 \pi } \sigma } \, dx)$$

I always get the same result no matter the values I assign to $\mu$ and $\sigma$, so I think the confidence of the region over $\mu$ and $\sigma$ defined by the inequality is independent from the true parameter. But how to eliminate $\mu$ and $\sigma$ in the expression for $p$ so as to compute the significance level without simulation?

Edit: It holds that $$p = n (q-1) q^{n-1}-q^n+1$$


1 Answer 1


Define $Y_i = \frac{X_i - \mu}{\sigma}$ which are i.i.d $\mathcal{N}(0,1)$, and let $y = \frac{x - \mu}{\sigma}$. Then, noting that $dy = \frac{dx}{\sigma}$ gives us that the above integral is equivalent to $$ \int_{\min Y}^{\max Y} \frac{e^{-\frac{y^2}{2}}}{\sqrt{2\pi}}dy $$

Hence, there is no dependency on $\sigma$ nor $\mu$. Intuitively speaking, the $\min{X}$ and $\max{X}$ encapsulate the same probability, regardless of moving the average or spreading the function out more (as they move along and spread out the same way).


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