Discrete Random Variables

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.

Suppose an individual, let’s say Bob, is selected at random

from the agency’s files. Let $X$ be the deductible amount on the auto policy and let Y be the deductible amount on the homeowner’s policy.

We want to understand the relationship between $X$ and $Y$.

Suppose the joint probability table is given by the insurance company as follows:

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\[P(X=100, Y=0)=0.2\\ P(X=250,Y=0)=0.05\\ \implies P(Y=0)=0.25\]

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 $Y$.

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Recall: Two events, $A$ and $B$, are independent if $P(A\cap B)=P(A)P(B)$.

In insurance example: are $X$ and $Y$ independent?

\[\begin{align} &P(X=100,Y=100)=0.1\\ &P(X=100)P(Y=100)=(0.5)(0.25)=1.25\\ &\implies X\text{ and }Y\text{ are not indept.} \end{align}\]

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 $y$.

Continuous Random Variables

Definition: if $X$ and $Y$ are continuous random variables, then $f(x,y)$ is the joint probability density function for $X$ and $Y$ if

\[P(a\le X\le b,c\le Y\le d)=\int_{a}^{b}\int_{c}^{d}f(x,y)dxdy\]

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: 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 $X_1\sim\text{Exp}(\lambda_1=1/2000)$ and, $X_2\sim\text{Exp}(\lambda_2=1/3000)$. 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.

\[E(X_1)=\frac{1}{\lambda_1}=2000\text{ hours}\\ E(X_2)=\frac{1}{\lambda_2}=3000\text{ hours}\\\]

Light bulbs function independently, so

\[\begin{align} P(X_1\le4000,X_2\le4000)&=P(X_1\le4000)P(X_2\le4000)\\ &=\bigg(\int_0^{4000}\lambda_1e^{-\lambda_1x_1}dx_1\bigg)\bigg(\int_0^{4000}\lambda_2e^{-\lambda_2x_2}dx_2\bigg)\\ &=\bigg(-e^{-\lambda_1x_1}\bigg)\Bigg\rvert_{0}^{4000}\bigg(-e^{-\lambda_2x_2}\bigg)\Bigg\rvert_{0}^{4000}\\ &=\bigg(1-e^{-4000/2000}\bigg)\bigg(1-e^{-4000/3000}\bigg)\\ &=(1-e^{-2})(1-e^{-4/3})\\ &\approx0.6368 \end{align}\]