STAT 211 Topic 3

Lecture 5

Lecture 5 Notes

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Random Variables

a function that maps each element in the sample space to a numeric value

Note: Random variables always denoted by uppercase letters

When flipping a coin: if we get heads, we assign 1, if we get tails, we assign 0.

Random variable X so that X(H) = 1 and X(T) = 0

Discrete Variables

Finite number of possible values

EX: gender (male or female); rolling a die (1-6)

Continuous Variables

Possibly infinite number of possible outcomes or along an interval

A random variable that is either 0 or 1 is a Bernoulli random variable

Probability Distribution

Probability Mass Function (PMF)

The PMF is the probability that a trial will result in a certain outcome.

Probability distribution function of a discrete random variable X is a little p_X.

$p(x)=P(X=x)$

Notice that $p(x)\ge 0$ because we can't have negative probability

$\sum p(x)=1$

Example

Suppose six lots of components are ready to be shipped by a certain supplier. The number of defective components in each lot is as follows:

Lot	1	2	3	4	5	6
Defective Items	0	1	3	0	1	2

Let X be the number of defective items in each lot:

• $P(X=0)=2/6$
• $P(X=1)=2/6$
• $P(X=2)=1/6$
• $P(X=3)=1/6$

Cumulative Distribution Function (CDF)

The CDF represents the probability that an observed result will be at most $x$ (ex. rolling a die will be less than 3)

CDF of discrete random variable X with PMF $p(x)$:

$F(x)=P(X\le x)=\sum _{y:y\le x}p(y)$

Example

Suppose a discrete random variable X gives the face value we get after rolling a die such that x = 1, 2, …, 6

	1	2	3	4	5	6
$p(x)$	1/6	1/6	1/6	1/6	1/6	1/6

• $F(x<1)=0$
• $F(1\le x<2)=1/6$
• $F(2\le x<3)=1/3$
• $F(3\le x<4)=1/2$
• $F(4\le x<5)=2/3$
• $F(5\le x<6)=5/6$
• $F(x\ge 6)=1$

Complex probability: P(2 ≤ x ≤ 4)

$p(2)+p(3)+p(4)=F(4)-F(1)=1/2$

In General: P(a ≤ x ≤ b)

$P(a\le X\le b)=F(b)-F(a-1)$

Unfair Coin

Probability of landing on heads and tails is not 50%.

We can "guess" based on many, many, many coin flips, then divide the results by the total trials.

Example

Suppose 3 independent electronic components connected in serial. Let $X_i$ be 1 if the $i$^th component works, and 0 if it fails. Suppose that each component works with probability $p$

$X_i$ is a discrete variable

PMF

• $p_{X_i}(0)=1-p$
• $p_{X_i}(1)=p$

Let $X$ be the # of components that work: $X$ can be 0..3

$x$	0	1	2	3
$p_X(x)$	$(1-p{)}^3$	$3(1-p{)}^{2}p$	$3p^2(1-p)$	$p^3$

CDF

$x$	0	1	2	3
$F_X(x)$	...

On average, what is the number of components that work?

Expected Value

Expected value (mean value) of a random variable$X$ is the sum of all possibilities for X multiplied by their PMF; denoted by:

$E(X)=\sum x\cdot p_X(x)$

From previous example:

$x$	0	1	2	3
$p_X(x)$	$(1-p{)}^3$	$3(1-p{)}^{2}p$	$3p^2(1-p)$	$p^3$

$E(X)=0\times (1-p{)}^3+1\times 3p(1-p{)}^2+2\times 3p^2(1-p)+3\times p^3=3p$

Variance of Random Variable

Variance of a random variable $X$ when given PMF:

$\begin{array}{rl}V(X)& =\sum (x-E(X){)}^2\cdot p_X(x)\\ & =E(X^2)-E(X{)}^2\end{array}$

Given our previous example $E(X)=3p$, we can find the variance:

...

Lecture 6

Lecture 6 Notes

Thursday, February 3, 2011

Review Question

Suppose a local TV station sells 15, 30, and 60 second advertising spots. If X is the length of a randomly selected commercial, we are given the PMF:

$x$	15	30	60
$p_X(x)$	0.1	0.3	0.6

What is the average length of all commercials on the station? (Expected value)

$E(X)=\sum x\cdot p_X(x)=1.5+9+36=46.5\text{seconds}$

What is the variance of the length among all commercials on the station?

$V(X)=\sum (x-E(X){)}^2\cdot p_X(x)=99.225+81.675+109.35=290.25{\text{seconds}}^2$

Functions of Random Variables

Linear conversion ratios for any function $h$:

$E(h(x))=\sum h(x)\cdot p_X(x)$

$V(h(X))=\sum (h(x)-E(h(X)){)}^2\cdot p_X(x)$

Special (linear) Cases:

$E(aX+b)=aE(X)+b$

$V(aX+b)=a^{2}V(X)$

Binomial Distribution

A Binomial Experiment stisfies the following condisitons:

1. $n$ = number of independent trials / components
2. results in success or failure ($X$ = # of successes)
3. $p$ = probability of success (consistent from trial to trial)

Note: $(\genfrac{}{}{0ex}{}{n}{k})={}_n{C}_k=\frac{n!}{(n-k)!k!}$

Written as: $X\sim Bin(n,p)$, and if this is the case, the probability of getting $x$ successes is:

$P(X=x)=(\genfrac{}{}{0ex}{}{n}{x})p^x(1-p{)}^{n-x}$

$x=0,1,\dots ,n$

Expected Value of Binomial Distribution

$E(X)=pn$

Lecture 7

Lecture 7 Notes

Tuesday, February 8, 2011

Binomial Experiments (cont'd)

Example 1

The probability of getting heads by tossing a coin is $p$. Suppose we toss this coin 4 times:

1. How many ways can we get only one success (heads)?

   SFFF, FSFF, FFSF, FFFS

2. What is the probability of getting 1 heads in 4 flips?

   $(\genfrac{}{}{0ex}{}{4}{1})p(1-p{)}^3$

Example 2

I asked 36 students in my past classes how many times they go out in a week. From the data, the fraction of people who go out at least 4 times per week is 5/36. Suppose I randomly select 5 students among those students:

1. What is the chance that at least 3 of the 5 students go out at least 4 times a week?

   X = # of students that go out ≥ 4 times/wk among 5 selected students

   X ~ Bin(n = 5, p = 5/36)

   $\begin{array}{rl}P(X\ge 3)& =P(x=3)+P(x=4)+P(x=5)\\ & =\dots \end{array}$

Example 3

Consider a system with 7 components serially connected. Suppose each component works with probability $p=0.4$ and work independently:

1. What is the probability that at most 3 components work?

   X ~ Bin(n = 7, p = 0.4)

   $\begin{array}{rl}P(X\le 3)& =P(x=0)+P(x=1)+P(x=2)+P(x=3)\\ & =\sum _{x=0}^3(\genfrac{}{}{0ex}{}{7}{x})0.4^{x}0.6^{7-x}\end{array}$

Poisson Distribution

Amount of events happening in a fixed unit parameter (time, volume, area, count, etc.)

• Even though the number of possible outcomes goes to infinity, it is still considered a discrete random variable
• We need to be given an average rate, well call it λ
• Written as X ~ Poisson(λ)

$p_X(x)=p(x;\lambda )=\frac{e^{-\lambda }{\lambda }^x}{x!}x=0,1,2,\dots$

Expected value = $E(X)=\lambda$ Variance = $V(X)=\lambda$

Example 1

It has been observed that the average number of traffic accidents on the Hollywood Freeway between 7 and 8 AM on Tuesday mornings is 1 per hour. What is the chance that there will be 2 accidents on the Freeway, on some specified Tuesday morning (per hour)?

Let X = # of accidents on freeway in one hour on Tuesday morning: X ~ Poisson(λ = 1)

$P(X=2)=p_X(2)=\frac{e^{-1}{1}^2}{2!}=\frac{1}{2e}$

Example 2

Suppose bacteria distribution in river is 1 per 20 cc. If we draw 10 cc in a test tube:

1. what is the chance that the sample contains 2 bacteria?

   X = # of bacteria in test tube.

   $\frac{1}{20}=\frac{\lambda }{10}$

   $P(X=2)=p_X(2)=\frac{e^{-\frac{1}{2}}{\left(\frac{1}{2}\right)}^2}{2!}$

1. what is the chance that the sample contains more than 2 bacteria?

   X ≥ 2 ⇒ P(X ≥ 2) = 1 - P(X ≤ 1)