Expectation of maximum of two independent random variable with known distribution [closed]

Assume $X$ and $Y$ are two random variables such that $X\sim \textrm{Unif}(0,1)$ and $Y=e^{-t}\times a$ where $t\sim \mathrm{Exp}(\lambda)$ and $a\sim \textrm{Unif}(0,1)$. What is $\mathbb{E}[\max(X,Y)]$?

Update: $X$, $t$ and $a$ are all independent random variables. By $Exp(\lambda)$, I meant the exponential distribution with mean $\frac{1}{\lambda}$.

closed as off-topic by Did, Johanna, heropup, Daniel W. Farlow, dustinMar 8 '15 at 2:07

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• You haven't told us the joint distribution of $t$ and $a$. Are they independent? – Michael Hardy Mar 7 '15 at 23:18
• "$\mathrm{Exp}(\lambda)$" sometimes means exponentially distributed with expected value $\lambda$ and sometimes means exponentially distributed with expected value $1/\lambda$. Which is intended here? ${}\qquad{}$ – Michael Hardy Mar 7 '15 at 23:20
• Thanks for your comment. $t$, $X$ and $a$ are independent random variables. Also $Exp(\lambda)$ is exponential distribution with mean $1/\lambda$. – user54626 Mar 8 '15 at 0:01
• What did you try? There are some classical venues here... – Did Mar 8 '15 at 0:20
• It is standard convention to use CAPITAL letters for random variables, and lower case for parameters. In your case, $t$ and $a$ are actually random variables, and so should be notated with capitals to avoid unnecessary confusion. – wolfies Mar 8 '15 at 3:19

A quick start, based on what seems obvious and a bit of simulation, but without doing all the steps:

If T ~ EXP(1), then exp(-T) ~ BETA(1, 1) = UNIF(0, 1), Max has support (0,1) and E(Max) = .5555.

If T ~ EXP(rate=2), then Y ~ BETA(2, 1) and E(Max) = .58.

In R (and probably about the same in Matlab):

m = 10^6; x = runif(m); a = runif(m); lam = 1

t = rexp(m, lam); y = exp(-t)*a; w = pmax(x,y)

hist(exp(-t), prob=T); mean(w)

 0.5554506

I'm new to this, if partial answers and simulated clues for questions with "classical elements" are not appreciated here, then tell me so, and I will desist.