Joint Cumulative Distribution Function

Study guide: Exam P-1

The joint cumulative distribution function (joint CDF) of random variables X and Y gives the probability that X \leq x and Y \leq y simultaneously.

F(x,y) = P(X \leq x,\; Y \leq y)

• It is non-decreasing in each argument
• F(-\infty, y) = F(x, -\infty) = 0 and F(\infty, \infty) = 1
• For discrete variables, F adds up the joint probability function over every cell at or below x and at or below y:

F(x,y) = \sum_{x_i \le x}\ \sum_{y_j \le y} p(x_i, y_j)

• For integer-valued X and Y, a single cell is recovered from F by inclusion–exclusion on the rectangle:

p(x,y) = F(x,y) - F(x-1,y) - F(x,y-1) + F(x-1,y-1)

• For continuous jointly distributed variables, the joint PDF is recovered by:

f(x,y) = \frac{\partial^2 F(x,y)}{\partial x\, \partial y}

• The joint CDF of Independent Random Variables factors as F(x,y) = F_X(x)\cdot F_Y(y)

A Discrete Joint CDF X and Y are the numbers of claims on two policies, with joint probability function

	Y=0	Y=1	Y=2
X=0	0.30	0.15	0.05
X=1	0.20	0.20	0.10

Find F(1,1) and F(0,2), then recover P(X=1, Y=1) from F alone.

Answer Add every cell with x_i \le x and y_j \le y:

\begin{align*}
F(1,1) &= 0.30 + 0.15 + 0.20 + 0.20 = 0.85 \\
F(0,2) &= 0.30 + 0.15 + 0.05 = 0.50
\end{align*}

With F(0,1) = 0.45, F(1,0) = 0.50 and F(0,0) = 0.30, the rectangle rule gives the cell back:

\begin{align*}
P(X=1, Y=1) &= F(1,1) - F(0,1) - F(1,0) + F(0,0) \\
            &= 0.85 - 0.45 - 0.50 + 0.30 \\
            &= 0.20
\end{align*}

which matches the table.

Computing a Joint Probability X and Y are independent, each Uniform on [0,1]. Find P(X \leq 0.4,\; Y \leq 0.6).

Answer Since X and Y are independent with F_X(x) = x and F_Y(y) = y: F(0.4, 0.6) = F_X(0.4) \cdot F_Y(0.6) = 0.4 \times 0.6 = 0.24