What is the expected minimum of two iid random variables from this distribution?

Question

Two random variables are drawn independently from the distribution $F(x) = 1-\frac{1}{x}$ for $x \geq 1$. How would I find the expected value of the minimum of these two random variables? I'm thinking of using the PDF but I believe there is a trick that uses just the CDF.

Answer 1

Let $X,Y$ be two iid random variables distributed according to $F$.

$\mathbb P(\min(X,Y)\le x)=1-\mathbb P(\min(X,Y)>x)=1-\mathbb P(X>x,Y>x)=1-\mathbb P(X>x)\mathbb P(Y>x)=1-(1-F(x))^2$

So if $G$ denotes the CDF of $\min(X,Y)$, then

$G(x)=0$ if $x<1$ and $G(x)=1-\frac{1}{x^2}$ if $x\ge1$. So for all $x\in\mathbb R$, $G(x)=F(x^2)$.

Since $\min(X,Y)$ is a nonnegative random variable, we have $$ \mathbb E[\min(X,Y)]=\int_{\mathbb R_+}(1-G(x))\,dx=\int_0^11\,dx+\int_1^{+\infty}\frac{1}{x^2}\,dx=2. $$

You can also deduce from $G$ the density function of $\min(X,Y)$ which is $G'(x)=\frac{2}{x^3}1_{\{x\ge1\}}$ and we find again $$ \mathbb E[\min(X,Y)]=\int_1^{+\infty}x\frac{2}{x^3}\,dx=2. $$