Find $E[\max (R_1, R_2)]$ when $R_1$ and $R_2$ are independent and uniformly distributed in $[-1,1]$

Question

Find $E[\max (R_1, R_2)]$ when $R_1$ and $R_2$ are independent and uniformly distributed in $[-1,1]$.

So first I was thinking something along the lines of

$$P(R_1 = n, R_2 \leq R_1)$$ would be the the expected value of getting $n$, but this is the continuous case so maybe $$ P(R_1 \leq n, R_2 \leq R_1) $$ or something? I'm suppose to be using one as a function of the other, but can't think of how to create $R_2$ such that when $R_1 < R_2$, it picks $R_2$. Any help would be great...

Answer 1

For every i.i.d. nonnegative random variables $X$ and $X_i$, $$E(\max(X_1,X_2))=\int_0^\infty P(\max(X_1,X_2)\gt x)\,\mathrm dx=\int_0^\infty\left(1-P(X\lt x)^2\right)\,\mathrm dx.$$ For $X=R+1$ and $X_i=R_i+1$, this yields $$E(\max(R_1,R_2))+1=\int_0^2\left(1-P(R\leqslant x-1)^2\right)\,\mathrm dx=\int_0^2\left(1-\left(\frac{x}2\right)^2\right)\,\mathrm dx=\ldots$$ Likewise, for every $n\geqslant1$, $$E(\max(R_1,R_2,\ldots,R_n))=\frac{n-1}{n+1}.$$

Answer 2

In the square $[{-1},1]^2$ of area $4$ the points $(x,y)$ with $$t\leq \max\{x,y\}<t+dt\qquad(-1\leq t< 1)$$ form an L-shaped domain of area $2(t+1)\>dt$, corresponding to a probability of ${1\over2}(t+1)\>dt$. It follows that $$E\bigl[\max\{X,Y\}\bigr]=\int_{-1}^1 t\ \ {1\over2}(t+1)\ dt={1\over3}\ .$$