Let $(X,Y)$ be a uniformly chosen point of a region $A \subset \mathbb{R}^2$

Given we have the following joint pdf:

$$ f(x,y) = \begin{cases} \dfrac{a}{ \text{area of}~ A} & (x,y)\in A \\ 0 & \text{else} \end{cases}$$

And Let $$A = {(x,y),|y| \leq x \cdot e^{-x}}$$

the provided solutions state that A is 2 and it computes

$$EX = \int_{0}^{\infty} \int_{-xe^{-x}}^{e^{-x}} \frac{x}{2} dydx = \int_{0}^{\infty} x^2 e^{-x} = 2$$

Not that doesn't make sense to me:

1) how is the area of $A = 2$ , when the graph of $|y| = x e^{-x}$ looks like this:

enter image description here


1 Answer 1


The area of $A$ is given by $$ \int_0^\infty xe^{-x}-(-xe^{-x})\, dx=2\int_0^\infty xe^{-x}\, dx=2. $$ To see that $\int_0^\infty xe^{-x}\, dx=1$ use integration by parts. Indeed, $$ \int_0^\infty xe^{-x}\, dx=-xe^{-x}]_{0}^\infty+\int_0^\infty e^{-x}\, dx=0+1=1. $$


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