Finding $P(X+Y<2)$ given joint density of $(X,Y)$

Question

I have a joint density function $f(x,y)=xy, $ for $ 0<x<1, 0<y<2 $. If I am to find $P(X+Y<2)$, would the double integral be $\int_0^2 \int_{0}^{y-2}xy\,dx\,dy$?

Answer 1

Not quite. You have to pay attention to the support.

The bound for $x$ must also not exceed $1$, which $2-y$ will do when $y<1$. So you have to partition the integral.

$$\begin{align}\mathsf P(X+Y<2)&=\iint_{0<x<1,0<y<2,x<2-y} xy~\mathrm d (x,y)\\[2ex] &= \int_0^2\int_0^{\min\{1,2-y\}} xy~\mathrm d x~\mathrm d y\\[2ex]&=\int_0^1\int_0^1 xy~\mathrm d x~\mathrm d y+\int_1^2\int_0^{2-y} xy~\mathrm d x~\mathrm dy\end{align}$$

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However, since $2-x$ cannot exceed $2$ (or go below $0$) when $x\in[0..1]$ reversing the order of integration gives:

$$\begin{align}\mathsf P(X+Y<2)&=\iint_{0<x<1,0<y<2,y<2-x} xy~\mathrm d (x,y)\\[2ex] &= \int_0^1\int_0^{2-x} xy~\mathrm d y~\mathrm d x\end{align}$$

Answer 2

You need to look at the area over which the PDF is being integrated, which looks like this:

You are better off integrating over $x$ as your outer integral, with

$$ \int_{x=0}^1 \int_{y=0}^{2-x} xy \, dy \, dx $$

The reason is that if you integrate over $y$ as your outer integral, your inner limit will require you to use the minimum function over the upper limit, as follows:

$$ \int_{y=0}^2 \int_{x=0}^{\min\{2-y, 1\}} xy \, dx \, dy $$

The usual way to resolve this issue is to break it up into two separate integrals (one with $y$ running from $0$ to $1$, and the other with $y$ running from $1$ to $2$), but you can save yourself this hassle by just reversing the order of integration.