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Probability distribution

Dr. Deshi YeCollege of Computer Science, Zhejiang University

yedeshi@zju.edu.cn

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Outline

Random variable

The Binomial distribution

The Hypergeometric Distribution

The Mean and the Variance of the a Probability distribution.

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Random variables We concern with one number or a few

number that associated with the outcomes of experiments.

EX. Inspection: number of defectives road test: average speed and average

fuel consumption. In the example of tossing dice, we are

interested in sum 7 and not concerned whether it is (1,6) or (2, 5) or (3, 4) or (4, 3) or (5, 2) or (6, 1).

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Definition

A random variable: is any function that assigns a numerical value to each possible outcome of experiments.

Discrete random variable: only a finite or a countable infinity of values.

Otherwise, continuous random variables.

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Probability distribution The probability distribution of the random variable: is

the probabilities that a random variable will take on any one value within its range.

The probability distribution of a discrete random variable X is a list of possible values of X together with their probabilities

The probability distribution always satisfies the conditions

][)( xXPxf

( ) 0, ( ) 1x

f x and f x

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Checking probability distribution

x 0 1 2 3

Prob. .26 .5 .22 .02

Another

1) f(x) = (x-2)/2, for x=1,2, 3, 4

2) H(x) =x2/25, x=0,1,2,3,4

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Probability histogram & bar chart

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Bar chart

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Histogram

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Histogram

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Cumulative distribution

F(x): value of a random variable is less than or equal to x.

xt

tfxXPxF )()()(

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EX.x 0 1 2 3

Prob. .26 .76 .98 1.0

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Binomial Distribution

Foul Shot: 1. Min Yao (Hou) .862. 2. O’Neal Shaquille .422

The Question is: what is the probability of them in two foul shots that they get 2 points, respectively?

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Binomial distribution

Study the phenomenon that the probability of success in repeat trials.

Prob. of getting x “success” in n trials, otherwords, x “success” and n – x

failures in n attempt.

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Bernoulli trials 1. There are only two possible outcomes for

each trial.

2. The probability of success is the same for each trial.

3. The outcomes from different trials are independent .

4’. There are a fixed number n of Bernoulli trials conducted.

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Let X be the random variable that equals the number of success in n trials. p and 1- p are the probability of “success” and “failure”, the probability of getting x success and n-x failure is

xnx pp )1(

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Def. of Binomial Dist.

#number of ways in which we can select the x trials on which there is to be a success is

Hence the probability distribution of Binomial is

x

n

xnx ppx

npnxb

)1(),;( nx ,,2,1,0

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Expansions

n

x

n

x

xnxn pnxbppx

npp

00

),;()1()1(

x

n Binomial coefficient

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Table 1

Table 1: Cumulative Binomial distribution

x

k

pnkbpnxB0

),;(),;(

),;1(),;(),;( pnxBpnxBpnxb

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EX Solve

Foul shot example:Here n=2, x=2, and p=0.862 for Yao, and p=0.422 for

Shaq.

0 1 2

Yao .02 .24 .74

Shaq. .33 .49 .18

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Bar Chats

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Minitab for Binomial

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Skewed distributionPositively skewed

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Hypergeometric Distr.

Sampling with replacement Sampling without replacement

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Hyergeometric distr. Sampling without replacement

The number of defectives in a sample of n units drawn without replacement from a lot containing N units, of which are defectives.

Here: population is N, and are total defectives

Sampling n units, what is probability of x defectives are found?

a

a

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formulations

Hypergeometric distr.

n

N

xn

aN

x

a

Nanxh ),,;(

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Discussion

Is Hypergeometric distribution a Bernolli trial?

Answer: NO! The first drawing will yield a defect

unit is a/N, but the second is (a - 1)/(N-1) or a/(N-1).

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EX

A shipment of 20 digital voice recorders contains 5 that are defective. If 10 of them are randomly chosen for inspection, what is the probability that 2 of the 10 will be defective?

Solution: a=5, n=10, N=20, and x=2348.0)20,5,10;2( h

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Expectation

Expectation: If the probability of obtaining the amounts

then the mathematical expectation is

kk pandppareaoraa ,,,,,, 2121

nn papapaE 2211

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Motivations

The expected value of x is a weighted average of possible values that X can take on, each value being weighted by the probability that X assumes it.

Frequency interpretation

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4.4 The mean and the variance

Mean and variance: two distributions differ in their location or variation

The mean of a probability distribution is simply the mathematical expectation of a random variable having that distribution.

Mean of discrete probability distribution

xall

xfx )(

)(XEAlternatively,

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EX

The mean of a probability distribution measures its center in the sense of an average.

EX: Find the mean number of heads in three tosses.

Solution: The probabilities for 0, 1, 2, or 3 heads are 1/8, 3/8, 3/8, and 1/8

2

3

8

13

8

32

8

31

8

10

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Mean of Binomial distribution

Contrast: please calculate the following

?)5.0,4;(4

0

x

xxb

16

0

( ;16,0.2) ?x

xb x

16

0

( ;16,0.8) ?x

xb x

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Mean of b()

Mean of binomial distribution:

pn

Proof.

np

pmybnp

pnxbnp

ppxnx

nx

m

y

n

x

n

x

xnx

0

1

0

),;(

);1;1(

)1()!(!

!

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Mean of Hypergeometric Distr.

N

an

Proof.

n

N

xn

aN

x

a

Nanxh ),,;(

n

x

Nanxxh0

),,;(

Similar proof or using the following hints:

k

sm

rk

s

r

mk

r 0

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EX

5 of 20 digital voice records were defectives ,find the mean of the prob. Distribution of the number of defectives in a sample of 10 randomly chose for inspection.

Solution: n=10, a= 5, N=20. Hence

5.2N

an

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Expectation of a function of random variable

Let X denote a random variable that takes on any of the values -1,0,1 respective probabilities P{x=-1}=0.2, P{x=0}=0.5, P{x=1}=0.3

Compute E[X2] Answer = 0.5

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Proposition

If X is a discrete random variable that takes on one of the value of xi, with respective to probability p(xi), then for any real-valued function g,

( ( )) ( ) ( )i ii

E g x g x p x

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Variance of probability

Variance of a probability distribution f(x), or that of the random variable X which has that probability distribution, as

)()(22 xfx

xall

We could also denote it as )(XD

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Standard deviation

Standard deviation of probability distribution

xall

xfx )()( 2

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Relation between Mean and Variance

22

22

2

))(()(

]2[

])[()(

XEXE

xxE

xEXD

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Ex.

Find the variance of the number of heads in four tosses.

Solution: 2

14

116

1)24(

16

4)23(

16

6)22(

16

4)21(

16

1)20(

22

2222

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Variance of binomial distr.

Variance of binomial distribution:

)1(2 ppn

Proof. Detailed proof after the section of disjoint probability distribution

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Some properties of Mean C is a constant, then E(C) = C. X is a random variable and C is a constant E(CX) = CE(X) X and Y are two random variables, then E(X+Y) = E(X)+E(Y)

If X and Y are independent random variables E(XY) = E(X)E(Y)

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Variance of hypergeometric distr.

)1

)(1(2

N

nN

N

a

N

an

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K-th moment

K-th moment about the origin

K-th moment about the mean

xall

kk xfx )(

( ) ( )kk

all x

x f x

Case study

Occupancy Problem

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Homework

problems