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

density functionextreme valuerandom variableself-study

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:

$$g(z) = nf(z)[F(z)]^{nāˆ’1}$$

and

$$h(w) = nf(w)[1 āˆ’ F(w)]^{n-1}.$$

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

Thank you

Best Answer

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.

Take the derivative, and you get what you want. The trick for finding the density of the minimum is similar.

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