5.2.2 Joint Cumulative Distribution Function (CDF)
The joint cumulative function of two random variables $X$ and $Y$ is defined as
\begin{align}%\label{}
\nonumber F_{XY}(x,y)=P(X \leq x, Y \leq y).
\end{align}
The joint CDF satisfies the following properties:
- $F_X(x)=F_{XY}(x, \infty)$, for any $x$ (marginal CDF of $X$);
- $F_Y(y)=F_{XY}(\infty,y)$, for any $y$ (marginal CDF of $Y$);
- $F_{XY}(\infty, \infty)=1$;
- $F_{XY}(-\infty, y)=F_{XY}(x,-\infty)=0$;
- $P(x_1<X \leq x_2, \hspace{5pt} y_1<Y \leq y_2)=$ $\hspace{60pt} F_{XY}(x_2,y_2)-F_{XY}(x_1,y_2)-F_{XY}(x_2,y_1)+F_{XY}(x_1,y_1)$;
- if $X$ and $Y$ are independent, then $F_{XY}(x,y)=F_X(x)F_Y(y)$.
Example
Let $X$ and $Y$ be two independent $Uniform(0,1)$ random variables. Find $F_{XY}(x,y)$.
- Solution
-
Since $X,Y \sim Uniform(0,1)$, we have
\begin{equation}
\nonumber F_X(x) = \left\{
\begin{array}{l l}
0 & \quad \textrm{for } x < 0 \\
x & \quad \textrm{for } 0 \leq x \leq 1 \\
1 & \quad \textrm{for } x > 1
\end{array} \right.
\end{equation}
\begin{equation}
\nonumber F_Y(y) = \left\{
\begin{array}{l l}
0 & \quad \textrm{for } y < 0 \\
y & \quad \textrm{for } 0 \leq y \leq 1 \\
1 & \quad \textrm{for } y > 1
\end{array} \right.
\end{equation}
Since $X$ and $Y$ are independent, we obtain
\begin{equation}
\nonumber F_{XY}(x,y)=F_X(x)F_Y(y) = \left\{
\begin{array}{l l}
0 & \quad \textrm{for } y < 0 \textrm{ or } x<0 \\
& \\
xy & \quad \textrm{for } 0 \leq x \leq 1, 0 \leq y \leq 1 \\
& \\
y & \quad \textrm{for } x>1, 0 \leq y \leq 1 \\
& \\
x & \quad \textrm{for } y>1, 0 \leq x \leq 1 \\
& \\
1 & \quad \textrm{for } x>1, y > 1
\end{array} \right.
\end{equation}
Figure 5.7 shows the values of $F_{XY}(x,y)$ in the $x-y$ plane. Note that $F_{XY}(x,y)$ is a continuous function in both arguments. This is always true for jointly continuous random variables. This fact sometimes simplifies finding $F_{XY}(x,y)$. The next example (Example 5.19) shows how we can use this fact.
Figure 5.7: The joint CDF of two independent $Uniform(0,1)$ random variables $X$ and $Y$.
Remember that, for a single random variable, we have the following relationship between the PDF and CDF: \begin{align}\label{} \nonumber F_X(x) &=\int_{-\infty}^{x} f_X(u)du, \\ \nonumber f_X(x) &=\frac{dF_X(x)}{dx}. \end{align} Similar formulas hold for jointly continuous random variables. In particular, we have the following:\begin{align}\label{} \nonumber F_{XY}(x,y) &=\int_{-\infty}^{y}\int_{-\infty}^{x} f_{XY}(u,v)dudv, \\ \nonumber \\ \nonumber f_{XY}(x,y) &=\frac{\partial^2}{\partial x \partial y} F_{XY}(x,y) \end{align}
-
Since $X,Y \sim Uniform(0,1)$, we have
\begin{equation}
\nonumber F_X(x) = \left\{
\begin{array}{l l}
0 & \quad \textrm{for } x < 0 \\
x & \quad \textrm{for } 0 \leq x \leq 1 \\
1 & \quad \textrm{for } x > 1
\end{array} \right.
\end{equation}
\begin{equation}
\nonumber F_Y(y) = \left\{
\begin{array}{l l}
0 & \quad \textrm{for } y < 0 \\
y & \quad \textrm{for } 0 \leq y \leq 1 \\
1 & \quad \textrm{for } y > 1
\end{array} \right.
\end{equation}
Since $X$ and $Y$ are independent, we obtain
\begin{equation}
\nonumber F_{XY}(x,y)=F_X(x)F_Y(y) = \left\{
\begin{array}{l l}
0 & \quad \textrm{for } y < 0 \textrm{ or } x<0 \\
& \\
xy & \quad \textrm{for } 0 \leq x \leq 1, 0 \leq y \leq 1 \\
& \\
y & \quad \textrm{for } x>1, 0 \leq y \leq 1 \\
& \\
x & \quad \textrm{for } y>1, 0 \leq x \leq 1 \\
& \\
1 & \quad \textrm{for } x>1, y > 1
\end{array} \right.
\end{equation}
Figure 5.7 shows the values of $F_{XY}(x,y)$ in the $x-y$ plane. Note that $F_{XY}(x,y)$ is a continuous function in both arguments. This is always true for jointly continuous random variables. This fact sometimes simplifies finding $F_{XY}(x,y)$. The next example (Example 5.19) shows how we can use this fact.
Example
Find the joint CDF for $X$ and $Y$ in Example 5.15
- Solution
- In Example 5.15, we found \begin{equation} \nonumber f_{XY}(x,y) = \left\{ \begin{array}{l l} x+\frac{3}{2}y^2 & \quad 0 \leq x,y \leq 1 \\ & \quad \\ 0 & \quad \text{otherwise} \end{array} \right. \end{equation} First, note that since $R_{XY}=\{(x,y)|0 \leq x,y \leq 1\}$, we find that \begin{align}\label{} \nonumber &F_{XY}(x,y)=0, \textrm{ for }x<0 \textrm{ or } y<0,\\ \nonumber &F_{XY}(x,y)=1, \textrm{ for }x \geq 1 \textrm{ and } y \geq 1. \end{align} To find the joint CDF for $x>0$ and $y>0$, we need to integrate the joint PDF: \begin{align}\label{} \nonumber F_{XY}(x,y) &=\int_{-\infty}^{y}\int_{-\infty}^{x} f_{XY}(u,v)dudv \\ \nonumber &=\int_{0}^{y}\int_{0}^{x} f_{XY}(u,v)dudv \\ \nonumber &=\int_{0}^{\min(y,1)}\int_{0}^{\min (x,1)} \left(u+\frac{3}{2}v^2\right) dudv. \end{align} For $0 \leq x,y \leq 1$, we obtain \begin{align}\label{} \nonumber F_{XY}(x,y) &=\int_{0}^{y}\int_{0}^{x} \left(u+\frac{3}{2}v^2\right) dudv\\ \nonumber &=\int_{0}^{y} \bigg[\frac{1}{2}u^2+\frac{3}{2}v^2u \bigg]_{0}^{x} dv\\ \nonumber &=\int_{0}^{y} \left(\frac{1}{2}x^2+\frac{3}{2}xv^2\right) dv\\ \nonumber &=\frac{1}{2}x^2y+\frac{1}{2}xy^3. \end{align} For $0 \leq x \leq 1$ and $y \geq 1$, we use the fact that $F_{XY}$ is continuous to obtain \begin{align}\label{} \nonumber F_{XY}(x,y) &=F_{XY}(x,1)\\ \nonumber &=\frac{1}{2}x^2+\frac{1}{2}x. \end{align} Similarly, for $0 \leq y \leq 1$ and $x \geq 1$, we obtain \begin{align}\label{} \nonumber F_{XY}(x,y) &=F_{XY}(1,y)\\ \nonumber &=\frac{1}{2}y+\frac{1}{2}y^3. \end{align}
