# Proof for the p.d.f of minimum and maximum of a sample

The following is a question from a past paper for one of my university statistical inference modules, and I know how to use the formula for each the max/min, but

Assume that the sample $X_1, X_2, . . . , X_n$ comes from a continuous distribution with cumulative distribution function $F(x)$ and probability density function $f(x)$. Show that the probability density functions of the maximum $(z)$ and minimum $(w)$ of the sample are respectively given by:

$$g(z) = nf(z)[F(z)]^{n−1}$$

and

$$h(w) = nf(w)[1 − F(w)]^{n-1}.$$

If someone could provide a proof for one, or both, that would be much appreciated.

Thank you

• Please add the [self-study] tag & read its wiki. Then tell us what you understand thus far, what you've tried & where you're stuck. We'll provide hints to help you get unstuck. – gung - Reinstate Monica Jun 2 '15 at 17:30
• Hint: differentiate the CDFs. One will be slightly easier to do than the other, but once you have the PDF for one extreme, it is simple to obtain the PDF for the other extreme: just negate the variable. – whuber Jun 2 '15 at 17:37

I'll do the maximum one. Let $Z = \max\{X_1, \ldots X_n\}$. First we find the cdf of $Z$, $F_Z(z)$. $P(Z \le z) = P(X_1 \le z \cap \cdots \cap X_n \le z) = \prod_{i=1}^nP(X_i \le z) = [F_X(z)]^n$. Where the second equality follows from independence.