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.

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

Find the cdf of $\displaystyle Z=X+Y$

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

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

Find the cdf of $\displaystyle Z=X+Y$

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

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

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

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

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

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