F X y Xy is Continuous

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5.2.1 Joint Probability Density Function (PDF)

Here, we will define jointly continuous random variables. Basically, two random variables are jointly continuous if they have a joint probability density function as defined below.

Definition
Two random variables $X$ and $Y$ are jointly continuous if there exists a nonnegative function $f_{XY}:\mathbb{R}^2 \rightarrow \mathbb{R}$, such that, for any set $A\in \mathbb{R}^2$, we have \begin{align}\label{eq:double-int} P\big((X,Y) \in A\big) =\iint \limits_A f_{XY}(x,y)dxdy \hspace{30pt} (5.15) \end{align} The function $f_{XY}(x,y)$ is called the joint probability density function (PDF) of $X$ and $Y$.

In the above definition, the domain of $f_{XY}(x,y)$ is the entire $\mathbb{R}^2$. We may define the range of $(X,Y)$ as \begin{align}%\label{} \nonumber R_{XY}=\{(x,y) | f_{X,Y}(x,y)>0\}. \end{align} The above double integral (Equation 5.15) exists for all sets $A$ of practical interest. If we choose $A=\mathbb{R}^2$, then the probability of $(X,Y) \in A$ must be one, so we must have

\begin{align}%\label{} \nonumber \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} f_{XY}(x,y)dxdy=1 \end{align}

The intuition behind the joint density $f_{XY}(x,y)$ is similar to that of the PDF of a single random variable. In particular, remember that for a random variable $X$ and small positive $\delta$, we have \begin{equation}%\label{} \nonumber P(x<X \leq x+\delta) \approx f_X(x) \delta. \end{equation} Similarly, for small positive $\delta_x$ and $\delta_y$, we can write \begin{equation}%\label{} \nonumber P(x<X \leq x+\delta_x, y \leq Y \leq y+\delta_y ) \approx f_{XY}(x,y) \delta_x \delta_y. \end{equation}

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Example
Let $X$ and $Y$ be two jointly continuous random variables with joint PDF \begin{equation} \nonumber f_{XY}(x,y) = \left\{ \begin{array}{l l} x+cy^2 & \quad 0 \leq x \leq 1,0 \leq y \leq 1 \\ & \quad \\ 0 & \quad \text{otherwise} \end{array} \right. \end{equation}

1. Find the constant $c$.
2. Find $P(0 \leq X \leq \frac{1}{2}, 0 \leq Y \leq \frac{1}{2})$.

• Solution
  • 1. To find $c$, we use \begin{align}%\label{} \nonumber \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} f_{XY}(x,y)dxdy=1. \end{align} Thus, we have \begin{align}%\label{} \nonumber 1&=\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} f_{XY}(x,y)dxdy\\ \nonumber &=\int_{0}^{1} \int_{0}^{1} x+cy^2 \hspace{5pt} dxdy\\ \nonumber &=\int_{0}^{1} \bigg[ \frac{1}{2} x^2+cy^2x \bigg]_{x=0}^{x=1} \hspace{5pt} dy\\ \nonumber &=\int_{0}^{1} \frac{1}{2}+cy^2 \hspace{5pt} dy\\ \nonumber &=\bigg[ \frac{1}{2}y+\frac{1}{3}cy^3 \bigg]_{y=0}^{y=1}\\ \nonumber &=\frac{1}{2}+\frac{1}{3}c. \end{align} Therefore, we obtain $c=\frac{3}{2}$.
    2. To find $P(0 \leq X \leq \frac{1}{2}, 0 \leq Y \leq \frac{1}{2})$, we can write \begin{align}%\label{} \nonumber P\big((X,Y) \in A\big) =\iint_{A} f_{XY}(x,y)dxdy, \hspace{10pt} \textrm{for }A=\{(x,y)| 0 \leq x,y \leq 1\}. \end{align} Thus, \begin{align}%\label{} \nonumber P (0 \leq X \leq \frac{1}{2}, 0 \leq Y \leq \frac{1}{2})&= \int_{0}^{\frac{1}{2}}\int_{0}^{\frac{1}{2}} \left(x+\frac{3}{2}y^2\right) dxdy \\ \nonumber &= \int_{0}^{\frac{1}{2}} \bigg[\frac{1}{2}x^2+\frac{3}{2}y^2x\bigg]_{0}^{\frac{1}{2}}dy\\ \nonumber &=\int_{0}^{\frac{1}{2}} \left(\frac{1}{8}+\frac{3}{4}y^2\right) dy\\ \nonumber &=\frac{3}{32}. \end{align}

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We can find marginal PDFs of $X$ and $Y$ from their joint PDF. This is exactly analogous to what we saw in the discrete case. In particular, by integrating over all $y$'s, we obtain $f_X(x)$. We have

Marginal PDFs

\begin{align}%\label{} \nonumber f_X(x)=\int_{-\infty}^{\infty} f_{XY}(x,y)dy, \hspace{10pt} \textrm{ for all }x,\\ \nonumber f_Y(y)=\int_{-\infty}^{\infty} f_{XY}(x,y)dx, \hspace{10pt} \textrm{ for all }y. \end{align}

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Example
In Example 5.15 find the marginal PDFs $f_X(x)$ and $f_Y(y)$.

• Solution
  • For $0 \leq x \leq 1$, we have \begin{align}%\label{} \nonumber f_X(x)&=\int_{-\infty}^{\infty} f_{XY}(x,y)dy \\ \nonumber &=\int_{0}^{1}\left(x+\frac{3}{2}y^2\right)dy\\ \nonumber &=\bigg[xy+\frac{1}{2}y^3 \bigg]_{0}^{1}\\ \nonumber &=x+\frac{1}{2}. \end{align} Thus, \begin{equation} \nonumber f_{X}(x) = \left\{ \begin{array}{l l} x+\frac{1}{2} & \quad 0 \leq x \leq 1 \\ & \quad \\ 0 & \quad \text{otherwise} \end{array} \right. \end{equation} Similarly, for $0 \leq y \leq 1$, we have \begin{align}%\label{} \nonumber f_Y(y)&=\int_{-\infty}^{\infty} f_{XY}(x,y)dx \\ \nonumber &=\int_{0}^{1}\left(x+\frac{3}{2}y^2\right)dx\\ \nonumber &=\bigg[\frac{1}{2}x^2+\frac{3}{2}y^2x \bigg]_{0}^{1}\\ \nonumber &=\frac{3}{2}y^2+\frac{1}{2}. \end{align} Thus, \begin{equation} \nonumber f_{Y}(y) = \left\{ \begin{array}{l l} \frac{3}{2}y^2+\frac{1}{2} & \quad 0 \leq y \leq 1 \\ & \quad \\ 0 & \quad \text{otherwise} \end{array} \right. \end{equation}

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Example
Let $X$ and $Y$ be two jointly continuous random variables with joint PDF \begin{equation} \nonumber f_{XY}(x,y) = \left\{ \begin{array}{l l} cx^2y & \quad 0 \leq y \leq x \leq 1 \\ & \quad \\ 0 & \quad \text{otherwise} \end{array} \right. \end{equation}

1. Find $R_{XY}$ and show it in the $x-y$ plane.
2. Find the constant $c$.
3. Find marginal PDFs, $f_X(x)$ and $f_Y(y)$.
4. Find $P(Y\leq \frac{X}{2})$.
5. Find $P(Y\leq \frac{X}{4}|Y\leq \frac{X}{2})$.

• Solution
  • 1. From the joint PDF, we find that \begin{align}%\label{} \nonumber R_{XY}=\{(x,y)\in \mathbb{R}^2| 0 \leq y \leq x \leq 1 \}. \end{align} Figure 5.6 shows $R_{XY}$ in the $x-y$ plane.
       

       Figure 5.6: Figure shows $R_{XY}$ as well as integration region for finding $P(Y\leq \frac{X}{2})$.

    2. To find the constant $c$, we can write \begin{align}%\label{} \nonumber 1&=\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} f_{XY}(x,y)dxdy\\ \nonumber &=\int_{0}^{1} \int_{0}^{x} cx^2y \hspace{5pt} dydx\\ \nonumber &=\int_{0}^{1} \frac{c}{2} x^4 dx\\ \nonumber &=\frac{c}{10}. \end{align} Thus, $c=10$.
    3. To find the marginal PDFs, first note that $R_X=R_Y=[0,1]$. For $0 \leq x \leq 1$, we can write \begin{align}%\label{} \nonumber f_X(x)&=\int_{-\infty}^{\infty} f_{XY}(x,y)dy\\ \nonumber &=\int_{0}^{x}10x^2ydy\\ \nonumber &=5x^4. \end{align} Thus, \begin{equation} \nonumber f_X(x) = \left\{ \begin{array}{l l} 5x^4 & \quad 0 \leq x \leq 1\\ & \quad \\ 0 & \quad \text{otherwise} \end{array} \right. \end{equation} For $0 \leq y \leq 1$, we can write \begin{align}%\label{} \nonumber f_Y(y)&=\int_{-\infty}^{\infty} f_{XY}(x,y)dx\\ \nonumber &=\int_{y}^{1}10x^2ydx\\ \nonumber &=\frac{10}{3}y(1-y^3). \end{align} Thus, \begin{equation} \nonumber f_Y(y) = \left\{ \begin{array}{l l} \frac{10}{3}y(1-y^3) & \quad 0 \leq y \leq 1\\ & \quad \\ 0 & \quad \text{otherwise} \end{array} \right. \end{equation}
    4. To find $P(Y\leq \frac{X}{2})$, we need to integrate $f_{XY}(x,y)$ over region $A$ shown in Figure 5.6. In particular, we have \begin{align}%\label{} \nonumber P\left(Y\leq \frac{X}{2}\right)&=\int_{-\infty}^{\infty} \int_{0}^{\frac{x}{2}} f_{XY}(x,y)dydx\\ \nonumber &=\int_{0}^{1} \int_{0}^{\frac{x}{2}} 10x^2y \hspace{5pt} dydx\\ \nonumber &=\int_{0}^{1} \frac{5}{4} x^4 dx\\ \nonumber &=\frac{1}{4}. \end{align}
    5. To find $P(Y\leq \frac{X}{4}|Y\leq \frac{X}{2})$, we have \begin{align}%\label{} \nonumber P\left(Y\leq \frac{X}{4}|Y\leq \frac{X}{2}\right)&=\frac{P\left(Y\leq \frac{X}{4},Y\leq \frac{X}{2}\right)}{P\left(Y\leq \frac{X}{2}\right)}\\ \nonumber &=4P\left(Y\leq \frac{X}{4}\right)\\ \nonumber &=4\int_{0}^{1} \int_{0}^{\frac{x}{4}} 10x^2y \hspace{5pt} dydx\\ \nonumber &=4\int_{0}^{1} \frac{5}{16} x^4 dx\\ \nonumber &=\frac{1}{4}. \end{align}

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