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Suppose $X$ is exponential with mean $1$ and $Y$ uniform on $[0,1]$ and they are independent. Calculate $\mathbb{E}[\max (X,Y)]$

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As they are independent, we can easily calculate the joint density $f_{X,Y}(x,y) = 1 \cdot e^{-x} = e^{-x} $. Therefore,

$$ \mathbb{E}[ \max(X,Y) ] = \int \int \max(x,y) e^{-x} dx dy $$

We have to consider cases now:

$$ \max(x,y) e^{-x} = \begin{cases} x e^{-x}, & y<x \\ y e^{-x}, & x<y \end{cases} $$

Thus,

if $y<x$, we obtain

$$ \int\limits_0^1 \int\limits_y^{\infty} x e^{-x} dx dy = \frac{1}{e}-2$$

Now, if $x<y$, then

$$ \int_0^1 \int_0^y y e^{-x} dxdy = \frac{2}{e} - \frac{1}{2}$$

Thus the required expectation is

$$ \boxed{ \frac{3}{e} - \frac{5}{2} }$$

Is this a correct solution?

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Your solution cannot be correct because $e > 2$ implies $6 - 5e < 0$, thus $3/e - 5/2 < 0$, and the expectation cannot be negative since neither $X$ nor $Y$ are ever negative.

Looking more closely at your computation, it is the first iterated integral that is incorrect, again simply by checking the sign, since $1/e < 1$ hence $1/e - 2 < 0$. We have instead $$\begin{align*} \int_{y=0}^1 \int_{x=y}^\infty x e^{-x} \, dx \, dy &= \int_{y=0}^1 \left[ -(1+x) e^{-x} \right]_{x=y}^\infty \, dy \\ &= \int_{y=0}^1 (1+y)e^{-y} \, dy \\ &= \left[ -(2+y)e^{-y} \right]_{y=0}^1 \\ &= -\frac{3}{e} + 2. \end{align*}$$ The second integral is correct, hence the result should be $$\operatorname{E}[\max(X,Y)] = \frac{3}{2} - \frac{1}{e},$$ and this value is positive.


The following Mathematica code estimates this expectation from $10^7$ simulations:

Mean[Max[#, RandomReal[]] & /@ RandomVariate[ExponentialDistribution[1], 10^7]]
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