17: Continuous Joint Distributions

Joint Probability Density Function

Joint Probability Density Function

The joint probability density function $f(x,y)$ to handle simultaneous calculations of random variables $X$ and $Y$ can be expressed as

$$P(a_1<X<a_2,b_1<Y<b_2)={\displaystyle {\int }_{a_1}^{a_2}{\int }_{b_1}^{b_2}f(x,y)dydx}$$

Properties

• Each probability is between zero and one inclusively $$0\le f(x,y)\le 1\text{ for all }\$x\$,\$y\$$$
• All probabilities add up to 100 percent $${\displaystyle {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}}{\displaystyle {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}}f(x,y)dxdy=1$$

Setting

For the examples in this lecture session, we will model the queues at In-n-Out with random variables

In-n-Out

• $X$: wait time to order food
• $Y$: wait time to receive food

and a function of the form

$$f(x,y)=kxy{e}^{-x}{e}^{-y/5},$$

$$x>0,y>0$$

Normalization

Find the value of $k$ so that $f$ is a probability density function.

$$f(x,y)=kxy{e}^{-x}{e}^{-y/5},x>0,y>0$$

Joint Probability

$$f(x,y)={\displaystyle \frac{1}{25}xy{e}^{-x}{e}^{-y/5},x>0,y>0}$$

Compute the probability that you will take {between 1 and 2 minutes to order} and wait {between 3 and 4 minutes to receive} your food.

Joint Cumulative Distribution Function

Joint Cumulative Distribution Function

In general, we handle the probability calculations with the joint cumulative distribution function $F(a,b)$

% joint cumulative distribution function $$F(a,b)=P(X\le a,Y\le b)={\displaystyle {\int }_{-\mathrm{\infty }}^a}{\displaystyle {\int }_{-\mathrm{\infty }}^b}f(x,y)dydx$$

Properties

We can verify that the joint CDF starts at zero

$$F(0,0)=0$$

and that the joint CDF collects all probabilities

$${\displaystyle \underset{a\to \mathrm{\infty },b\to \mathrm{\infty }}{lim}F(a,b)=1}$$

If need be, we can recover the joint PDF from the joint CDF as the mixed second-order partial derivatives

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

What is the joint CDF for the In-n-Out setting?

Marginal Probabilities

Marginal Cumulative Distribution Functions

The marginal cumulative distribution functions can be computed as

% marginal CDF $$\begin{array}{rcl}{F}_X(a)& =& {\displaystyle \underset{b\to \mathrm{\infty }}{lim}F(a,b)}\\ F_Y(b)& =& {\displaystyle \underset{a\to \mathrm{\infty }}{lim}F(a,b)}\end{array}$$

Properties

Intuition: the marginal CDF is seeking to analyze the probabilities in just one variable regardless of the other variables, so ``eliminate’’ the other varibles by taking their limits to infinity.

We can verify that the marginal CDFs start at zero

$$F_X(0)=0\text{ and }{F}_Y(0)=0$$

and that the marginal CDFs collect all probabilities

$${\displaystyle \underset{a\to \mathrm{\infty }}{lim}{F}_X(a)=1\text{ and }{\displaystyle \underset{b\to \mathrm{\infty }}{lim}{F}_Y(b)=1}}$$

What are the marginal CDFs for the In-n-Out setting?

Marginal Probabilities

Marginal Cumulative Distribution Functions

The marginal probability density functions can be computed as

$$\begin{array}{rcl}{f}_X(x)& =& {\displaystyle {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}}f(x,y)dy\\ f_Y(y)& =& {\displaystyle {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}}f(x,y)dx\end{array}$$

Intuition

Intuition: the marginal PDF is seeking to analyze the probabilities in just one variable regardless of the other variables, so ``integrate out’’ the other variables.

Alternatively,

$$f_X(x)={\displaystyle \frac{d}{dx}{F}_X(x)}\text{ and }{f}_Y(y)={\displaystyle \frac{d}{dy}{F}_Y(y)}$$

What are the marginal PDFs for the In-n-Out setting?

Marginal Expectation

In-N-Out

• What is the expected wait time to order food?
• What is the expected wait time to receive food?

Independence

Independence

Recall that two events $A$ and $B$ are independent if

$$P(AB)=P(A)\cdot P(B)$$

Here, the variables $X$ and $Y$ in the In-n-Out example were independent, which can be easily verified by noting that the integrals were separable.

$$\begin{array}{rcl}f(x,y)& =& f_X(x)\cdot f_Y(y)\\ {\displaystyle \frac{1}{25}}xy{e}^{-x}{e}^{-y/5}& =& x{e}^{-x}\cdot {\displaystyle \frac{y}{25}{e}^{-y/5}}\end{array}$$

and

$$\begin{array}{rcl}F(a,b)& =& {\displaystyle {\int }_{-\mathrm{\infty }}^a}{\displaystyle {\int }_{-\mathrm{\infty }}^b}f(x,y)dydx\\ & =& {\displaystyle \frac{1}{25}}{\displaystyle {\int }_0^a}{\displaystyle {\int }_0^b}xy{e}^{-x}{e}^{-y/5}dydx\\ & =& {\displaystyle \frac{1}{25}}\left({\displaystyle {\int }_0^{a}x{e}^{-x}dx}\right)\left({\displaystyle {\int }_0^{b}y{e}^{-y/5}dy}\right)\end{array}$$

Dependence

Upcoming lectures: dependent variables

Looking Ahead

• due Fri., Mar. 10:

  • WHW7
  • LHW6
  • Internet Connection (survey)

• Exam 2 will be on Mon., Apr. 10

• no lecture on Mar. 10, Mar. 24

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