Multiple Random Variable for a uniform distribution

Question

A random point $(X,Y)$ is distributed uniformly on the square with vertices $(1,1)$, $(1,-1)$, $(-1,1)$, and $(-1,-1)$. That is, the joint pdf is $f(x,y)=\frac{1}{4}$ on the square. Determine the probability of $P(|X+Y|<2)$.

My attempt: $$\int_{-1}^1\int_{-1}^{2-x} \frac{1}{4}\,dy\,dx$$ Did I set up the double integral right?

Answer 1

Comment: @herbsteinberg is correct that the problem as stated is pointless, because the answer is obviously $1.$ The integral would be $\int_{-1}^1 \int_{-1}^1 \frac 1 4\, dy\,dx.$

The plot below shows the region of the square corresponding to $P(|X+Y|<1).$ Because the joint distribution of $(X,Y)$ on the square is uniform it seems clear that $P(|X+Y|<1) = 3/4.$

If you want to use integral calculus, it is probably best to break the integral into two parts, perhaps to the left and right of the vertical green line.

Note: I used simulation (in R statistical software) as an easy way to make the plot. But the same simulation also provides a way to approximate the result $P(|X+Y|<1) = 3/4,$ correct to two places. [A larger number of points, such as m = 10^6, would give a much closer approximation, but an ugly plot.]

set.seed(1112);  m = 50000
x = runif(m, -1,1);  y = runif(m, -1,1)
plot(x,y, pch=".")
cond = (abs(x+y)<1)
points(x[cond],y[cond], pch=".", col="blue")
abline(h=0, col="green2", lwd=2);  abline(v=0, col="green2", lwd=2)
mean(cond)
[1] 0.7519  # aprx P(|X + Y| < 1) = 3/4

Answer 2

Note thet: $$P(|X+Y|<2)=P(-2<X+Y<2)=P(Y>-X-2 \ \text{or} \ Y<-X+2)=\\ P(Y>-X-2\ge -1 \ \text{or} \ Y<-X+2\le 1)=\int_{-1}^1 \int_{-1}^1 \frac14dydx=1.$$ See the graph:

If the question is to find $P(|X+Y|<1)$, then: $$P(|X+Y|<1)=P(-1<X+Y<1)=P(-X-1<Y \ \text{or} \ Y<-X+1)=\\ \int_{-1}^0 \int_{-x-1}^1 \frac14dydx+\int_0^1 \int_{-1}^{-x+1}dydx=\frac38+\frac38=\frac34.$$

See the graph: