# PDF of $\min$ and $\max$ of $n$ iid random variables

The Problem: Suppose that $$X_1,\dots,X_n$$ are independent random variables with the same absolutely continuous distribution. Let $$f$$ denote their common marginal PDF. Set $$Y=\min(X_1,\dots,X_n)$$ and $$Z=\max(X_1,\dots,X_n)$$. Show that $$Y$$ and $$Z$$ are both absolutely continuous, and find their marginal PDFs.

My Thoughts: We begin by finding the CDF of $$Y$$. For $$t\in\mathbb R$$ we have $$$$\begin{split} F_Y(t)&=P(Y\leq t)=P(\min(X_1,\dots,X_n)\leq t)=1-P(X_1>t,\dots,X_n>t)\\ &=1-P(X_1>t)\cdots P(X_n>t)\\ &=1-[1-F(t)]^n, \end{split}$$$$ where in the fourth step we used the independence of the random variables $$X_1,\dots,X_n$$, and in the last step we used the fact the latter random variables have the same distribution which we call $$F$$. By the absolute continuity of the random variables $$X_1,\dots,X_n$$, we may use the chain rule to differentiate the expression for $$F_Y$$ to obtain $$f_Y(t)=nf(t)[1-F(t)]^{n-1}=nf(t)\left[1-\int_{-\infty}^tf(s)\,ds\right]^{n-1}.$$ It follows that $$Y$$ is an absolutely continuous random variable.
Next, we find the CDF of $$Z$$. For $$t\in\mathbb R$$ we have $$\begin{equation*}\begin{split} F_Z(t)&=P(Z\leq t)=P(\max(X_1,\dots,X_n)\leq t)=P(X_1\leq t,\dots,X_n\leq t)\\ &=P(X_1\leq t)\cdots P(X_n\leq t)\\ &=F(t)^n, \end{split}\end{equation*}$$ where in th fourth step we used the independence of the random variables $$X_1,\dots,X_n$$. Since the latter mentioned random variables are absolutely continuous, we may use the chain rule to differentiate the expression for $$F_Z$$ to obtain $$f_Z(t)=nF(t)^{n-1}f(t)=nf(t)\left[\int_{-\infty}^tf(s)\,ds\right]^{n-1}.$$ It follows that $$Z$$ is an absolutely continuous random variable.

Could anyone please provide feedback on the correctness of my proof above?

The derivative $$F_X'$$ of the distribution function of a random variable $$X$$, if it exists, is always measurable and non-negative, but its integral need not be $$1$$. So just proving that the derivative exists is not enough. In both cases you can see that the derivative you obtained actually integrates to $$1$$ and the formula $$F_X(x)=\int_{-\infty} ^{x} F_X'(t)dt$$ holds. Hint for computing the integral: putting $$h(t)=\int_{-\infty}^{t} f(s)ds$$ helps in both cases.