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Home / How to Combine Continuous Probability Density Functions

How to Combine Continuous Probability Density Functions


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}

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}

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}

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}

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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Source: https://www.probabilitycourse.com/chapter5/5_2_1_joint_pdf.php