Expected Value of max of two uniform, independent and non-identical random variables

Question

I have two random variables, $X \sim \mathcal{U}(1-a,1)$ and $Y \sim \mathcal{U}(1-b,1)$, and want to calculate $E[max(X,Y)]$.

I tried

$\int_{-\infty}^{\infty}$$\int_{-\infty}^{\infty}max(X,Y)F_XF_Ydxdy$

$=\int_{(1-a)}^{1}$$\int_{(1-b)}^{1}max(X,Y)\frac{1}{ab}dxdy$

$=\int_{(1-a)}^{1}$$\int_{(1-b)}^{x}\frac{x}{ab}dydx$ + $\int_{(1-b)}^{1}$$\int_{(1-a)}^{y}\frac{y}{ab}dxdy$

$=\int_{(1-a)}^{1}$$(\frac{x^2}{ab}-\frac{x(1-b)}{ab})dx$ + $\int_{(1-b)}^{1}$$(\frac{y^2}{ab}-\frac{y(1-a)}{ab})dy$

But this seems to lead nowhere with the final expression being long and, at least when I simulated for test values, incorrect. I believe I messed up the integrals.. Thanks in advance!

Answer 1

Suppose $(1-a)<(1-b)<1$. Then $$\iint_{\Bbb R^2}\max(x,y)F_{X,Y}(x,y)~\mathsf d(x,y) = {\int_{(1-a)}^{(1-b)} \int_{(1-b)}^ 1 \dfrac{y}{ab}~\mathsf d y~\mathsf d x+\int_{(1-b)}^1\left(\int_{(1-b)}^x \dfrac x{ab}\mathsf d y+\int_x^1\dfrac y{ab}\mathsf d y\right)\mathsf d x}$$

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The case for $(1-b)<(1-a)<1$ will be similar.