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
$p(x,y)=P(X=x\ {\mbox{and}}\ Y=y)$

1. $\forall x\forall y(p(x,y)\geq 0)$
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:

$f(x,y)=P[(X,Y)\in A]=\int \int _{A}f(x,y)\ dx\ dy$

$P(a\leq X\leq b,c\leq Y\leq d)=\int _{a}^{b}\int _{c}^{d}f(x,y)\ dx\ dy$

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

Marginal Cases:

${\begin{aligned}f_{X}(x)&=\int _{-\infty }^{\infty }f(x,y)\ dy\\f_{Y}(y)&=\int _{-\infty }^{\infty }f(x,y)\ dx\end{aligned}}$

Example

${\begin{aligned}\int _{0}^{1}\int _{0}^{1}{\frac {6}{5}}\left(x+y^{2}\right)\ dy\ dx&=\int _{0}^{1}\left[{\frac {6}{5}}\left(xy+{\frac {y^{3}}{3}}\right)\right]_{0}^{1}\ dx\\&=\int _{0}^{1}{\frac {6}{5}}\left(x+{\frac {1}{3}}\right)\ dx\\&={\frac {6}{5}}\left({\frac {x^{2}}{2}}+{\frac {x}{3}}\right){\bigg |}_{0}^{1}\\&=1\end{aligned}}$

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 ${\tfrac {0}{0}}$ 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:

$f(x,y)={\begin{cases}K\left(x^{2}+y^{2}\right)&20\leq x,y\leq 30\\0&{\mbox{otherwise}}\end{cases}}$

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 $P(|X-Y|<1)$

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

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

${\begin{cases}f_{X}(x)={\frac {1}{3}}&0<x<3\\f_{Y}(y)={\frac {1}{3}}&0<y<3\end{cases}}$

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

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

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 $h(x)$

$E(h(x,y))={\begin{cases}\sum _{x}\sum _{y}h(x,y)\cdot p(x,y)\\\int _{-\infty }^{\infty }\int _{-\infty }^{\infty }h(x,y)\cdot f(x,y)\ dx\ dy\end{cases}}$

Easier way:

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

Variance

ONLY IF X AND Y ARE INDEPENDENT!

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

(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:

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

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:

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

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

${\begin{aligned}E({\bar {X}})&=E\left({\frac {X_{1}}{n}}+{\frac {X_{2}}{n}}+\ldots +{\frac {X_{n}}{n}}\right)\\&=\mu \end{aligned}}$

Variance of the sample average should be the variance of the population average divided by the sample size: ${\begin{aligned}V({\bar {X}})&=V\left({\frac {X_{1}}{n}}+{\frac {X_{2}}{n}}+\ldots +{\frac {X_{n}}{n}}\right)\\&={\frac {\sigma ^{2}}{n}}\end{aligned}}$

Example 1

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

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

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

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?

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

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