joint exponential distribution range problem

Question

So the question asks:

Let $X, Y$ be random variables, with the following joint probability density function:

$$f_{X,Y}(x,y) = \left\{ \begin{array}{ 1 l } kye^{-y} & \mbox{if $0≤|x|≤y$}\\ 0 & \mbox{ otherwise} \end{array} \right.$$

Find the value of the constant $k$.

So so far, I got:

$$ 1 = \int_{-y}^y \int_{ 0 }^{∞} kye^{-y} \,dy \, dx $$

while I kind of have a problem calculating the equation, and I ended up getting that $ 1 = 2ky $, which leads $k=0.5/y$ which is clearly wrong.

So do I have a wrong range of the $dy\, dx$?

Answer 1

Draw a picture, Our joint density function lives above the "curve" $y=|x|$.

Let $C$ be the first quadrant region that is above $y=|x|$. By symmetry we want:

$$2 \iint_C kye^{-y}=1$$

Express as an iterated integral. We want:

$$2\int_{x=0}^\infty \left(\int_{y=x}^\infty kye^{-y}\,dy\right)\,dx=1\tag{1}$$

The integrations are fairly straightforward. Note that integration by parts shows that an antiderivative of $ye^{-y}$ is $-ye^{-y}-e^{-y}$.

So the inner integral on the left side of (1) is $2k(xe^{-x}+e^{-x})$.

Answer 2

Your order of integration is incorrect, which is why your computation doesn't work. The outermost integral cannot be a function of any other variables of integration. If you write $$\int_{x=-y}^y \int_{y=0}^\infty f(x,y) \, dy \, dx,$$ then what you have essentially done is integrated a function $f$ with respect to $y$, but then the interval of integration with respect to $x$, itself being a function of $y$, makes no sense.

Instead, your order of integration for the region under consideration should be $$\int_{y=0}^\infty \int_{x=-y}^y f(x,y) \, dx \, dy.$$ For a given value of $y$, the inner integral with respect to $x$ is evaluated on the interval $x \in [-y,y]$, which now makes sense: we are integrating $f$ along horizontal "strips" in the $xy$-plane.