$X,Y\sim U(0,1)$ are independent, $W=\max (X,Y)$

Question

$X,Y\sim U(0,1)$ are independent, $W=\max (X,Y)$.

1. How do I find the PDF of $W$?
2. How do I find the expectation of $W$ at two ways: 1. with the PDF of $W$ and without the PDF of $W$.

I'd like to get any idea how to solve it...
Thank you!!!

Answer 1

The pdf of $W$ is the non-negative function $f_W$ satisfying $$ {\rm E}[u(W)]=\int_\mathbb{R}u(w)f_W(w)\,\mathrm dw\tag{1} $$ for any bounded, measurable function $u:\mathbb{R}\to\mathbb{R}$.

So in order to find $f_W$, we let $u$ be such a function. Then $$ \begin{align} {\rm E}[u(W)]&={\rm E}[u(\max(X,Y))]=\int_{\mathbb{R}^2} u(\max(x,y)) f_{X,Y}(x,y)\,\mathrm dx\,\mathrm dy\\ &=\int_{[0,1]^2} u(\max(x,y)) \,\mathrm dx\,\mathrm dy=2\int_{[0,1]^2} u(x)\mathbf{1}_{x>y}\,\mathrm dx\,\mathrm dy. \end{align} $$ The task is now to rewrite this integral into an integral of the form $$ \int_{\mathbb{R}} u(w)g(w)\,\mathrm dw $$ and recognize $g$ as the pdf. To find the expectation of $W$ without having to find the pdf of $W$, you can let $u$ be the identity in $(1)$.

Answer 2

The answer to your first question is to first find the CDF of $W$, and then differentiate the CDF to find the pdf. The CDF of $W$, $F_W(\alpha)$, is, by definition, the probability that $W$ is no larger than $\alpha$, that is, $$F_W(\alpha) = P\{W \leq \alpha\} = P\{\max(X,Y) \leq \alpha\}.$$ But, if the maximum of $X$ and $Y$, the larger of $X$ and $Y$, is no larger than $\alpha$, then both $X$ and $Y$ must be no larger than $\alpha$, no? So we have $$P\{\max(X,Y) \leq \alpha\} = P\{X \leq \alpha, Y \leq \alpha\}.$$ What does independence of $X$ and $Y$ tell you about the right hand side and its relationship to $P\{X \leq \alpha\}$ and $P\{Y \leq \alpha\}$? Does the fact that $X$ and $Y$ are uniformly distributed on $(0,1)$ allow you to write down the value of $P\{X \leq \alpha\}$ and $P\{Y \leq \alpha\}$ without the formality of integrating? If you can solve all these small steps successfully, you will have obtained $F_W(\alpha)$ and then all that remains is to differentiate with respect to $\alpha$ to obtain the pdf.