5.2.2 Joint Cumulative Distribution Function (CDF)

We have already seen the joint CDF for discrete random variables. The joint CDF has the same definition for continuous random variables. It also satisfies the same properties.
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:
  1. $F_X(x)=F_{XY}(x, \infty)$, for any $x$ (marginal CDF of $X$);
  2. $F_Y(y)=F_{XY}(\infty,y)$, for any $y$ (marginal CDF of $Y$);
  3. $F_{XY}(\infty, \infty)=1$;
  4. $F_{XY}(-\infty, y)=F_{XY}(x,-\infty)=0$;
  5. $P(x_1<X \leq x_2, \hspace{5pt} y_1<Y \leq y_2)=$ $\hspace{60pt} F_{XY}(x_2,y_2)-F_{XY}(x_1,y_2)-F_{XY}(x_2,y_1)+F_{XY}(x_1,y_1)$;
  6. 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}


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}





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