STAT 211 Topic 5

Lecture 11 Notes

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Review

Topic 3: Discrete Random Variables

• PMF, CDF
• Expected value and variance

Topic 4: Continuous Random Variables

• PDF, CDF
• Epxected value and variance

Topic 5 Overview

Consider 2 or more random variables.

• Joint PMF or PDF
• Expected value and variance

Central Limit Theorem (CLT)

What is Joint Distribution?

Discrete

Given 2 Bernoulli discrete random variables $X$ and $Y$ (not independent)

Joint Probability Mass Function
$p(x,y)=P(X=x\text{and}Y=y)$

1. $\mathrm{\forall }x\mathrm{\forall }y(p(x,y)\ge 0)$
2. $\sum _x\sum _{y}p(x,y)=1$
3. Marginal PMF of X: $p_X(x)=\sum _{y}p(x,y)$ (also satisfies 1 & 2)
4. Marginal PMF of Y: $p_Y(y)=\sum _{x}p(x,y)$ (also satisfies 1 & 2)

Continuous

Let $X$ and $Y$ be continuous random variables:

$f(x,y)=P[(X,Y)\in A]=\int \underset{A}{\int }f(x,y)dxdy$

$P(a\le X\le b,c\le Y\le d)=\underset{a}{\overset{b}{\int }}\underset{c}{\overset{d}{\int }}f(x,y)dxdy$

Note that whatever comes first in $P(\dots )$ is the outer integral

Marginal Cases:

$\begin{array}{rl}{f}_X(x)& =\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}f(x,y)dy\\ f_Y(y)& =\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}f(x,y)dx\end{array}$

Example

$\begin{array}{rl}\underset{0}{\overset{1}{\int }}\underset{0}{\overset{1}{\int }}\frac{6}{5}(x+y^2)dydx& =\underset{0}{\overset{1}{\int }}{\left[\frac{6}{5}(xy+\frac{y^3}{3})\right]}_0^{1}dx\\ & =\underset{0}{\overset{1}{\int }}\frac{6}{5}(x+\frac{1}{3})dx\\ & =\frac{6}{5}(\frac{x^2}{2}+\frac{x}{3}){|}_0^1\\ & =1\end{array}$

Independence

if discrete

$p(x,y)=p_X(x)\cdot p_Y(y)$

if continuous

$f(x,y)=f_X(x)\cdot f_Y(y)$

You should be able to factor joint PDF into two functions: $x$ alone and $y$ alone

For example, if $X$ and $Y$ are independent, $p(1,1)=p_X(1)\cdot p_Y(1)$.

Conditional Distributions

$P(X>0.5|Y=0.3)=\frac{P(X>0.5,Y=0.3)}{P(Y=0.3)}$

If $X$ and $Y$ are continuous, the result of the above equation will be $\frac{0}{0}$ since $Y$ is set to a certain constant number.

The way to get around it is by looking at the joint PDF and marginal PDF:

$f_{X|Y}(x|y)=\frac{f(x,y)}{f_Y(y)}$

After finding symbolic conditional PDF, integrate over X and plug in Y

Example

Suppose $X$ and $Y$ denote the air pressure in the front tires of a car (left and right) and their joint PDF is given by:

$f(x,y)=\{\begin{array}{ll}K(x^2+y^2)& 20\le x,y\le 30\\ 0& \text{otherwise}\end{array}$

The tires are supposed to be filled to 26 psi.

1. What is $K$?

   do double integral, set equal to 1, and solve for K

2. What is the probability that both tires are underfilled?

   double-integrate from 20 to 26 on both integrals

Lecture 12

Lecture 12 Notes

Thursday, March 3, 2011

(continued from lecture 11)

Homework Example

Given 2 independent continuous rv's $X,Y$ as Uniform distribution between 0 and 3, find $P(|X-Y|<1)$

$P(|X-Y|<1)=\int \underset{\{(x,y):|x-y|<1\}}{\int }f(x,y)dxdy$

We don't know $f(x,y)$, but we do know the marginal PDFS:

$\{\begin{array}{ll}{f}_X(x)=\frac{1}{3}& 0<x<3\\ f_Y(y)=\frac{1}{3}& 0<y<3\end{array}$

By independence, the joint PDF is just the product of the two marginal PDFs:

$f(x,y)=\frac{1}{9}0<x,y<3$

Plot the function as a square with bottom left corner at origin and sides of length 3

Solve function inside probability ($|x-y|<1$) for $y$ and plot bounded area:

$y<x+1,y>x-1$

Find the area of the bounded area using either the graph or the integral above

$=\frac{5}{9}$

Expected Value

For any arbitrary function $h(x)$

$E(h(x,y))=\{\begin{array}{l}\sum _x\sum _{y}h(x,y)\cdot p(x,y)\\ \underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}h(x,y)\cdot f(x,y)dxdy\end{array}$

Easier way:

$E(aX+aY+\dots )=aE(X)+aE(Y)+\dots$

Variance

ONLY IF X AND Y ARE INDEPENDENT!

$V(aX+aY+\dots )=a^{2}V(X)+a^{2}V(Y)+\dots$

(lecture 12)

Random Sample

A collection of independent random variables with same distribution as population(recall from Topic 1) that can be used to estimate a parameter. The numbers we get to estimate parameters are called statistics

Central Limit Theorem

The original rv X may have any distribution.

If we let $E(X)=\mu$ and $V(X)={\sigma }^2$, we know that:

$E(\overline{X})=\mu$
$V(\overline{X})=\frac{{\sigma }^2}{n}$

Suppose we pick a random sample X₁ through X_n from X:

As sample size $n$ gets larger, the average of the sample approaches a normal distribution:

$\overline{X}\sim N(\mu ,\frac{{\sigma }^2}{n})$

Proof

Let X Be a uniform distribution

Take a 15 random samples of size 30. The expected value of the sample average should be the average of the population and have a normal distribution

$\begin{array}{rl}E(\overline{X})& =E(\frac{X_1}{n}+\frac{X_2}{n}+\dots +\frac{X_n}{n})\\ & =\mu \end{array}$

Variance of the sample average should be the variance of the population average divided by the sample size: $\begin{array}{rl}V(\overline{X})& =V(\frac{X_1}{n}+\frac{X_2}{n}+\dots +\frac{X_n}{n})\\ & =\frac{{\sigma }^2}{n}\end{array}$

Example 1

Flipping a fair coin ($p=0.5$ for getting Heads or Tails)

• $\mu =\frac{1}{2}$
• ${\sigma }^2=\frac{1}{4}$

Therefore by CLT, the average of 1000 flips will approach $\overline{X}\sim N(\frac{1}{2},\frac{1/4}{1000})$

Example 2

Suppose the weight of 50 yr-old males has a certain distribution with mean 150 lbs. and std. dev. 32 lbs. What is the approximate probability that the sample mean weight for a random sample of 64 is 160 lbs?

$\overline{X}\sim N(150,\frac{32^2}{64})$

$P(\overline{X}>160)=P(Z>\frac{160-150}{32/8})=P(Z>2.5)=1-\mathrm{\Phi }(2.5)$