# Joint Distributions

## Discrete Definition

Given two discrete random variables, X and Y , p(x, y) = P(X = x, Y = y) is the joint probability mass function for X and Y .

**Important property** X and Y are independent random variables if P(X = x, Y = y) = P(X = x)P(Y = y) for all
possible values of x and y.

\(f(x,y) = P(X=x \, and \, Y=y) = P(X=x,Y=y)\)

Sum of all marginal probabilities is equal to 1. ( P(y=0) + P(y=100) = P(y = 200) = 1 )

Sum of all joint probabilities is equal to 1.

### Marginal Probabilities

### Example

An insurance agency services customers who have both a homeowner’s policy and an automobile policy. For each type of policy, a deductible amount must be specified. For an automobile policy, the choices are $100 or $250 and for the homeowner’s policy, the choices are $0, $100, or $200.

`Recall`

Two events are independent if P(A and B) = P(A)P(B) for all possible values of A and B.

X and y are not independent.

## Continuous Definition

Definition: If X and Y are continuous random variables, then f(x, y) is the joint probability density function for X and Y if \(P(a \leq X \leq b, c \leq Y \leq d)=\int_{a}^{b} \int_{c}^{d} f(x, y) d x d y\) for all possible $a, b, c$, and $d$ Important property: $X$ and $Y$ are independent random variables if $f(x, y)=f(x) f(y)$ for all possible values of $x$ and $y$.

### Example

Example: Suppose a room is lit with two light bulbs. Let \(X_{1}\) be the lifetime of the first bulb and \(X_{2}\) be the lifetime of the second bulb. Suppose \(X_{1} \sim {Exp}\left(\lambda_{1}=1 / 2000\right)\) and \(X_{2} \sim {Exp}\left(\lambda_{2}=1 / 3000\right)\). If we assume the lifetimes of the light bulbs are independent of each other, find the probability that the room is dark after 4000 hours.

\(E\left(X_{1}\right)=\frac{1}{\lambda_{1}}=2000 \mathrm{hrs}, E\left(X_{2}\right)=\frac{1}{\lambda_{2}}=3000 \mathrm{hrs}\).

Light bulbs function independently,so