# 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.

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. P(x=100,y=100) = .1 P(x=100) p(y=200) = (.5)(.25) =.125 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

\begin{align}\begin{aligned}P\left(X_{1} \leq 4000, X_{2} \leq 4000\right)=P\left(X_{1} \leq 4000\right) P\left(X_{2} \leq 4000\right)\\=\left(\int_{0}^{4000} \lambda_{1} e^{-\lambda_{1} x_{1}} d x_{1}\right)\left(\int_{0}^{4000} \lambda_{2} e^{-\lambda_{2} x_{2}} d x_{2}\right)\\=\left.\left.\left(-e^{-\lambda_{1} x_{1}}\right)\right|_{0} ^{4000}\left(-e^{-\lambda_{2} x_{2}}\right)\right|_{0} ^{4000}\\\begin{split}\begin{aligned}=\left(1-e^{-4000 / 2000}\right)\left(1-e^{-4000 / 3000}\right) &=\left(1-e^{-2}\right)\left(1-e^{-4 / 3}\right) \\ & \simeq .6368 \end{aligned}\end{split}\end{aligned}\end{align}