# Thread: cdf of a transformation of two random variables

1. ## cdf of a transformation of two random variables

X and Y have the following joint pdf.

$f(x,y)=\frac{1}{4}$ for $-1, zero elsewhere.

Find the cdf of $Z=X+Y$

2. Originally Posted by akolman
X and Y have the following joint pdf.

$f(x,y)=\frac{1}{4}$ for $-1, zero elsewhere.

Find the cdf of $Z=X+Y$

$F(z) = \Pr(Z < z) = \Pr(X + Y < z) = \Pr(Y < -X + z)$.

A diagram is essential to see why there are two cases to consider and where the integral terminals come from in each case.

Case 1: $z \leq 0$ (that is, $-2 < z \leq 0$)

$F(z) = \int_{x = -1}^{x = z + 1} \int_{y = -1}^{y = -x + z} \frac{1}{4} \, dy \, dx = \, ....$

Case 2: $z > 0$ (that is, $0 < z \leq 2$)

$F(z) = \int_{x = -1}^{x = z - 1} \int_{y = -1}^{y = 1} \frac{1}{4} \, dy \, dx + \int_{x = z-1}^{x = 1} \int_{y = -1}^{y = -x + z} \frac{1}{4} \, dy \, dx = \, ....$

As I've said, a diagram is essential to see where these results have come from. The calculations are left for you to do.