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).


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.