Compute probability from joint density function

Question

Let $X$ and $Y$ be two random variables with joint density function $f(x,y) = x+y, \ 0 \leq x, y \leq 1,$ zero elsewhere. Find $\mathbb{P}(X<2Y).$

As of now what I think I have an idea of how to do it but I'm not certain. I need to do a double integral of $(x+y) dx \ dy$ with the inner integral (the one with dx) spanning from 0 to 2y and the outer integral spanning from 0 to 1? Am I right to do it this way?

Answer 1

The idea you have is correct - you can get $\mathbb{P}(X < 2Y)$ by integrating $f(x,y)$ over the region where $X<2Y$ is true.

First draw the unit square. Over it draw the line segment where $x=2y.$ It connects the points $(0,0)$ and $(1, 1/2).$ The portion of the square above this line is where $ X<2Y.$ In this region, $x$ varies between $0$ and $1$ and for a fixed $x,$ $y$ varies from $x/2$ to $1.$ So the integral you want is

$$ \mathbb{P}(X<2Y) = \int^1_0 \int^{1}_{x/2} f(x,y) dy dx $$

Answer 2

The PDF of $Z:=X/Y$, which has support $[0,\,\infty)$, is$$\int_0^1yf(zy,\,y)dy=\int_0^{\min\{1/z,\,1\}}y^2(z+1)dy=\frac13(z+1)[\min\{1/z,\,1\}]^3.$$So$$\begin{align}P(Z<2)&=\int_0^2\frac13(z+1)[\min\{1/z,\,1\}]^3dz\\&=\int_0^1\frac13(z+1)dz+\int_1^2\frac13(z^{-2}+z^{-3})dz\\&=\frac12+\frac13[-z^{-1}-z^{-2}/2]_1^2\\&=\frac{19}{24}.\end{align}$$Python agrees.