STAT 211 Topic 3

Lecture 5

Lecture 5 Notes

« previous | Tuesday, February 1, 2010 | next »

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 $p(x)\geq 0$ 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 $p(x)$:

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

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

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

Complex probability: P(2 ≤ x ≤ 4)

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

In General: P(a ≤ x ≤ b)

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

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:

${\begin{aligned}V(X)&=\sum {(x-E(X))^{2}\cdot p_{X}(x)}\\&=E(X^{2})-E(X)^{2}\end{aligned}}$

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)

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

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

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

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: ${\binom {n}{k}}={}_{n}C_{k}={\frac {n!}{(n-k)!k!}}$

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

$P(X=x)={\binom {n}{x}}p^{x}(1-p)^{n-x}$

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

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?

   ${\binom {4}{1}}p(1-p)^{3}$

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)

   ${\begin{aligned}P(X\geq 3)&=P(x=3)+P(x=4)+P(x=5)\\&=\ldots \end{aligned}}$

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)

   ${\begin{aligned}P(X\leq 3)&=P(x=0)+P(x=1)+P(x=2)+P(x=3)\\&=\sum _{x=0}^{3}{\binom {7}{x}}\,0.4^{x}\,0.6^{7-x}\end{aligned}}$

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)

$p_{X}(x)=p(x;\lambda )={\frac {e^{-\lambda }\lambda ^{x}}{x!}}\quad x=0,1,2,\ldots$

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)