math f113 -chapter-3.pdf

BITS Pilani Pilani Campus

BITS Pilani presentation

Dr RAKHEE Department of Mathematics


MATH F111 & AAOC C111 Probability and Statistics


Chapter 3

Discrete Distribution

BITS Pilani, Pilani Campus

Random variables are variables whose values are determined by a chance. (This can be thought as a sample space of a random experiment whose outcomes are real numbers).

Random variables


• The outcomes of random experiment may be numerical or non-numerical (descriptive).

For example, when we throw a die, we get the outcomes as 1, 2, 3, 4, 5, 6 which is a numerical value, whereas when we toss a coin we get either a head or a tail. This is a non-numerical values. Instead of dealing the non-numerical values, we can assign some numerical value to them, say 1 for head and 0 for tail.

Random variables


• Random variable is a real valued function which maps the numerical or non-numerical sample space (domain) of the random experiment to a real values (co domain or range)

• It should be mapped such that an outcome of an event should correspond to only one real value.

Random Variable


• Random variable is a real valued function which is also a single valued function and not a multi-valued.

• That means it can be one-to- one or many-to-one but never be one-to-many mapping.

Random Variable


Example Suppose that we toss three coins and consider the sample space associated with the experiment

S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT} X: number of tails obtained in the toss of three coins Hence, X(HHH) = 0, X(TTT) = 3, X(THT) = 2, X(HTT) = 2, X(TTH) = 2, X(THH) = 1, X(HHT) = 1, X(HTH) = 1


Random Variable

Definition: Let E be a random experiment and S a sample space associated with it. A function X assigning to every element s ∈ S, a real number X(s) is called random variable. Though, X is a function yet we call it a random variable.

s X(s) X S


• Generally random variables are denoted by capital letters X, Y, Z etc or X1, X2 etc. whereas their possible values are denoted by the corresponding lower case letters x, y, z or x1, x2 etc. respectively.


Random variable


Let E be the experiment of rolling two fair dice Let X be the random variable that is defined as the sum of numbers shown then X takes values 2, 3, 4,…, 10, 11, 12

Examples

P[X=2]= P[(1,1)] = 1/36 P[X=3]= P[(2,1),(1,2)] = 2/36 P[X=4]= P[(2,2),(3,1),(1,3)] = 3/36 P[X=5]= P[(2,3),(3,2),(1,4),(4,1)] = 4/36 ………… P[X=11]= P[6,5),(5,6)] = 2/36 P[X=12]= P[6,6)] = 1/36


Let Z = the time of peak demand for electricity at a power plant. Time is measured continuously, and Z can conceivably assume any value in the interval [0,24), 0: means mid night one day 24: means 12 mid night next day. In this case, the set of real numbers is neither finite nor countably infinite and hence Z is not discrete random varable.

Example


The random variable denoting the life time of a car, when the car’s lifetime is assumed to take on any value in some interval [a, c]. So this is not discrete r.v.

Example


T : the turnaround time for a computer job

-- not discrete random variable M : the number of meteorites hitting a

satellite per day. -- discrete random variable

Section 3.1 (page no 81)


The uncertain behavior of the random variable is predicted by:

(i) Probability density function f(x) (ii) Cumulative distribution

function F(x)


Definition : The density function of a discrete random variable X is defined by f(x) = P(X=x) for all real x.

Discrete Probability Density


From the density, one can evaluate the probability of any subset A of real numbers (i.e. event): ∑

∈

=XAxxfAP of valuea is

)()(

Conversely if we are given probabilities of all events of a discrete random variable, we get a Density function.


∑ =

≥

x

xf

xf

all

and x allfor

1)(

0)(

The necessary and sufficient condition for a function f to be a discrete density function :


The cumulative distribution function F of a discrete random variable X, is defined by

for any real number x, here f denote the density of X.

∑≤

=≤=xk

f(k)x)P(XF(x)


The density and cumulative distribution function determine each other. If random variable takes integer values then f(n) = F(n)-F(n-1).


F is CDF (cumulative distribution function) of a discrete random variable X then

P(a < X ≤ c ) = P( X ≤ c) - P(X ≤ a) = F(c) – F(a) as set of all x such that X ≤ a is subset of set of all x such that X ≤ c.


Cumulative distribution function of a discrete random variable is a step function, its values change at points where density is positive. Note : F(x) is non-decreasing and ,

1)(lim =∞→

xFx

0)(lim =∞−→

xFx


Exercise : Given that f(x)= k/(2x), x=0, 1, 2, 3 and 4 for a density function of a random variable taking only these values, find k.

Exercise : Given that f(x) = k /(2x) x=0, 1, 2, 3,- - - for a density function of a random variable taking only these values (a)Find k. (b) Find P( 3 < X < 100). (c) The cumulative distribution function of

X.


Tabular way of defining density (pdf): Can tabulate values of density at points where it is nonzero. Tabular way of defining cumulative distribution function (cdf): Can tabulate values of F(x) where steps change.


The density for X, the number of holes that can be drilled per bit while drilling into limestone is given by the following table : x 1 2 3 4 5 6 7 8 f(x) 0.02 0.03 0.05 0.2 0.4 0.2 0.07 ?

(i) Find f(8), (ii) Find the table for F(x). (iii) Use F to find the probability that a randomly selected bit can be used to drill between three and five holes inclusive.

Exercise 8


x -1 0 1/3 1/2 2/3 1 3

F (x) 0.1 0.3 0.35 0.4 0.5 0.8 1.0

(i)Find probability density function f(x) for all x (ii) Find P(2 < X ≤ 3) & P(2 ≤ X < 3) (iii)Find F(-2) & F(4) (iv)Find P(X < 3) & P(X > 0)

Example

If CDF F(x) for a r.v. is given as


x -1 0 1/3 1/2 2/3 1 3

F(x) 0.1 0.3 0.35 0.4 0.5 0.8 1.0

(i) x -1 0 1/3 1/2 2/3 1 3

f(x) 0.1 0.2 0.05 0.05 0.1 0.3 0.2

f(x) = 0 at all other real number x


x -1 0 1/3 1/2 2/3 1 3

F(x) 0.1 0.3 0.35 0.4 0.5 0.8 1.0

(iii) Find F(-2) & F(4) F(-2)=0 & F(4)=1 (iv) Find P(X<3) & P(X>0) P(X< 3)= 0.8 P(X>0) = 1-P(X≤0)= 1-F(0)= 1-0.3


It is known that the probability of being able to log on to a computer from a remote terminal at any given time is 0.7. Let X denote the number of attempts that must be made to gain access to the computer. (a)Find the first 4 terms of the density table. (b)Find a closed form expression for f(x). (c)Find a closed form expression for F(x). (d)Use F to find the probability that at most 4 attempts are required to access the computer.

Exercise 10 :


The density function of a random variable completely describe the behavior of the random variable. Random variables can also be characterized by the knowledge of numerical values of three parameters, Mean(µ), Variance (σ2) and Standared deviation (σ).

Expectation


Consider the roll of a single fair die, and X denote the number that is obtain. The possible values for X are 1, 2, 3, 4, 5, 6, and since the die is fair, the probability associated with each value is 1/6. So the density function for X is given by

Example

6,5,4,3,2,1,61)( == xxf


When we repeat the rolling over and over and recording the values of X in each roll, We ask what is the theoretical average value of the rolls as the number of rolls approaches infinity. Since the density is symmetric and is known, this average can be found intuitively. As, P[X = 1] = P[X = 6] = 1/6, the average value is (1 + 6)/2 = 3.5


Similarly, P[X = 2] = P[X = 5] = 1/6, the average value is (2 + 5)/2 = 3.5 In long run, 3.5 dictates the average value. So we write it as E[X] = 3.5


Definition : Let X be a discrete random variable and H(X) be a function of X. Then the expected value of H(X), denoted by E(H(X)), is defined by

∑=

Xofvalueanyx

xfxHXHE

)()())((

Where f(x) is density of X provided

∑x

xfxH finite is )(|)(|

Expectations


1) E[H(X)] can be interpreted as the weighted average value of H(X).

2) If ∑all x|H(x)|f(x) diverges then E[H(X)] does not exist irrespective of convergence of ∑all xH(x)f(x), see Ex. 22.

3) E[X] measures average value of X and is called the mean of X and denoted by µX or µ

4) Distribution is scattered around µ. Thus it indicates location of center of values of X and hence called a location parameter.

Notes :


Variability is not being measured by the mean. Parameter must reflects consistency or the lack of it. The measure a large (small) positive values if the random variable fluctuates in the sense that it often assumes values far (closer) from its mean.

Variance and Standard Deviation


Definition : If a discrete random variable X has mean µ, its variance Var(X) or σ2 is defined by Var(X) = E[(X-µ)2]. The standard deviation σ is the nonnegative square root of Var(X).

Variance and Standard deviation


Notes : 1) Note that Var(X) is always nonnegative, if it

exists. 2) Variance measures the dispersion or variability

of X. It is large if values of X away from µ have large probability, i.e. values of X are more likely to be spread. This indicates inconsistency or instability of random variable.


Theorem : If X is a random variable and c is a real number then :

E[c] = c and E[cX] = cE[X]. Proof : E[c] = ∑c f(x) = c ∑f(x) = c(1) = c. E[cX] = ∑c xf (x) = c ∑xf (x)= cE[X]. Ex.: Prove for reals a, b, E[aX + b] = aE[X] + b.

Properties of Mean


Theorem : Var [X] = E[X2] – (E[X])2. Theorem : For a real number c, Var [c] = 0 and Var [cX] = c2Var[X].

Properties of Variance


Exercise 15 : The density for X, the number of holes that can be drilled per bit while drilling into limestone is given by the following table :

x 1 2 3 4 5 6 7 8

f(x) 0.02 0.03 0.05 0.2 0.4 0.2 0.07 0.03

Find E[X], E[X2], Var[X], σX. Find the unit of σX.


Example : Let X be random variable with density function

=

=−

therwisex

xf

Solx

o,0....3,2,1),3/2()3/1(

)(

.1

E(X).o,0

....3,2,1),3/2()3/1()(

1

Findtherwise

xxf

x

=

=−


Section 3.3, page 84, 17

The probability of being able to log on to a computer from a remote terminal at any given time is 0.7. Let X denote the number of attempts that must be made to gain access to the computer. Find E[X]. Can you express E[X] in terms of p?


Consider the function f defined by a) Verify that this is the density function for a

discrete r.v. b) Let

show that Σ g(x) f(x) < ∞ c)

Section 3.3, page 84, 22

,...3,2,1,221)( || ±±±== − xxf x

.1||2

2)1()(||

1||

−

−= −

xxg

xx


c) Show that Σ |g(x)| f(x) does not converges.


Ordinary Moments : For any positive integer k, the kth ordinary moment of a discrete random variable X with density f(x) is defined to be E[Xk].

Thus for k = 1 we get mean. Using 1st and 2nd ordinary moment, we can evaluate variance. There is a tool, moment generating function (m.g.f) which helps to evaluate all ordinary moments in one go.


Definition: Let X be any random variable with density f. The m.g.f. for X is denoted by mX(t) and is given by

provided the expectation is finite for all

real numbers t in some open interval (-h, h).

Moment generating function

][)( tXX eEtm =


Theorem 3.4.2: If mX(t) is the m.g.f. for a random variable X, then

][0

)( kXEt

kdt

tmkd X =

=


...!/...!2/1: 22 +++++= nXtXttXe ntXProof

][)( Hence tXX eEtm =

...!/][...!2/][][1)(

...]!/...!2/1[)(22

22

+++++=

+++++=

nXEtXEtXtEtmnXtXttXEtm

nnX

nnX

.result get the to0put Now

...!/][...][][)( times, atingDifferenti

1

=

++++= −+

t

kXEtXtEXEdt

tmdk

nknkkkX

k


Consider the random variable X whose density is given by: a) Verify that this function is a density for a

discrete random variable.

Section 3.4 page no. 87, 31

5,4,3,5

)3()(2

=−

= xxxf


c) Find moment generating function for X. b) Find E[X] directly. That is evaluate Σall x xf(x).

e) Find E[X2] directly. f) Use m.g.f. to find E[X2] g) Find σ2 and σ.

d) Use m.g.f. to find E[X]


A trial which has exactly 2 possible outcomes, success s and failure f, is called Bernoulli trial. For any random experiment, if we are only interested in occurrence or not of a particular event, we can treat it as Bernoulli trial. Thus if we toss a dice but are interested in whether top face has even number or not, we can treat it as a Bernoulli trial.

Bernoulli trials


If we perform a series of identical and independent trials, X = number of trials required to get the first success is a discrete random variable, which is known as geometric random variable. It’s probability distribution is called Geometric distribution.

Geometric distribution


Sample space of this experiment is S = {s, fs, ffs, fffs, …}.

1,2,...ifor )1()( 1 =−== − ppiXP i

Probability of success on any trial = p is same.


In fact the function f is called the density of a geometric distribution with parameter p for 0 < p < 1, if

variable)random discrete a ofdensity a isit (Verify otherwise. ;0

,..3,2,1 ;1)1()(

=−−= xpxpxf

We write q = 1- p.


Then c.d.f. of geometric distribution is F(x) = 1 - q[x] for any real x > 0 and F(x) = 0 otherwise.


.1 where

;lnfor ;1

)(

pq

qtqe

petm t

t

X

−=

−<−

=

The m.g.f. of geometric random variable with parameter p, 0 < p < 1, is

Theorem 3.4.1


The density of a geometric distribution By def. The series on the right is a geometric series with first term qet, common ratio qet.

Proof:

=−

=otherwise. ;0

,..3,2,1 ;1)( xpxqxf

∑∑∞

=

−∞

=

===1

1

1)()(][)(

x

xt

x

txtXX qepqxfeeEtm


.1 where;1

)( pqqe

petm t

t

X −=−

=

Provided | r | < | qet | < 1. Since the exponential function is nonnegative and 0 < q < 1, this restriction implies that (qet) < 1, implies that et < (1/q)

qtqtq

et ln)ln1(ln,1ln)ln( −<⇒−<⇒

<


.][Var and 1][

Then .parameter with variablerandom geometric a be Let

2pqX

pXE

pX

==

Theorem 3.4.3


Proof

.1 where;1

)( pqqe

petm t

t

X −=−

=

( )

( ) pqptm

dtdXE

qepetm

dtd

tX

t

t

X

11

)(][

1)(

20

2

=−

==

−=

=


Now take the second derivative at t = 0

( )

( ) 2302

22

32

2

)1(1

)1()(][

1)1()(

pq

qqptm

dtdXE

qeqepetm

dtd

tX

t

tt

X

+=

−+

==

−

+=

=


Thus,

22 )1(][1][

pqXEand

pXE +

==

222

1)1(][pq

ppqXVar =−

+=


Expectation of Geometric r.v. by Definition

( )...4321

)(][

32

1

1

1

1

1

++++==

==

∑

∑∑∞

=

−

∞

=

−∞

=

qqqpxqp

xpqxxfXE

x

x

x

x

x


ppp

qpXE 11

)1(1][ 22 =

=

−

=∴

Sum of AGP is

...32

...432132

32

+++=

++++=

qqqqSqqqS

2

32

)1(1

...1)1(

qS

qqqSq

−=

++++=−


( )

233

32

1

12

1

12

1

22

11)1(

1

...16941

)(][

pq

pqp

qqp

qqqpqxp

pqxxfxXE

x

x

x

x

x

+=

+=

−+

=

++++==

==

∑

∑∑∞

=

−

∞

=

−∞

=


=

−

+=

−=

2

2

2

2

11

][][][

pq

ppq

XEXEXVar


The zinc phosphate coating on the threads of steel tubes used in oil and gas wells is critical to their performance. To monitor the coating process, an uncoated metal sample with known outside area is weighed and treated along with the lot of tubing. This sample is then stripped and reweighed. From this it is possible to determine whether or not the proper amount of coating was applied to the tubing.

Exercise 25


Assume that the probability that a given lot is unacceptable is 0.05. Let X denote the number of runs conducted to produce an unacceptable lot. Assume that the runs are independent in the sense that the outcome of one run has no effect on that of any other. Verify X is geometric. What is success? p =? What is density, E[X], E[X2], σ2? m.g.f.? Find the probability that the number of runs required to produce an unacceptable lot is at least 3.


In a Video game the player attempts to capture a treasure lying behind one of five doors. The location of treasure varies randomly in such a way that at any given time it is just as likely to be behind one door as any other. When the player knocks on a given door, the treasure is his if it lies behind that door.

Example


Otherwise he must return to his original starting point and approach the doors through a dangerous maze again. If the treasure is captured, the game ends. Let X be the number of trials needed to capture the treasure. Find the average number of trials needed to capture the treasure. Find P(X ≤ 3) and P(X > 3).


Let an experiment consist of fixed number ‘n’ of Bernoulli trials. Assume all trials are identical and independent. Thus p = probability of success is same for each trial. X = number of successes in these n trials. What is P(X = x)?

Binomial Distribution


Consider a case in which n = 3 then it’s the sample space is

S = {fff, sff, fsf, ffs, ssf, sfs, fss, sss} Since trials are independent, the probability assigned to each sample point is found by multiplying. For instance the probability assigned to the sample points are as follows:

(1-p)(1-p)(1-p) = (1-p)3 and p(1-p)(1-p) = p(1-p)2

Example


The r. v. X assumes the value 0 only if the experiment result in the outcome fff. That is, P[X = 0] = (1- p)3

However, X assumes the value 1 if the any one of the outcome is success (sff, fsf, ffs), then P[X = 1] = 3(1- p)2

Similarly, P[X = 2] = 3(1- p)2 and P[X = 3] = p3


It is evident that x = 0, 1, 2, 3

P[X = x] = c(x) px(1-p)3-x

where

=

xxc

3)(

xx ppx

xf −−

= 3)1(

3)(


A discrete random variable X has binomial distribution with parameters n and p, n is a positive integer and 0 < p < 1, if its density function is

theorem).binomial use density, isit Verify (otherwise. 0

,...,2,1,0;)1()(

=−−

=

nxxnpxpxn

xf


Theorem: Let X be a binomial random variable with parameters n and p. Then

.1 with )()(

is of m.g.f. The )1

pqntpeqtm

X

X −=+=

.][ and ][)2 npqXVarnpXE ==


.1 where)(

)()1(

)1( 1)

:

0

0

pqpeq

pepxn

eppxn

]E[e(t)m

nt

xtxnn

x

txxnxn

x

tXX

−=+=

−

=

−

==

−

=

−

=

∑

∑

Proof


.)(

)()(E[X] Thus

.)()( )2

0

1

0

nppqnp

peqnpedt

tdmpeqtm

t

ntt

t

X

ntX

=+=

+==

+=

=

−

=


.)1(][][][

Thus.)1()()1(

])()()1([

])([)(][ Also

2222222

220

1222

0

1

02

22

npqpnppnnpnppnXEXEXVar

nppnnnppqpnn

peqnpepeqepnn

dtpeqnped

dttmdXE

t

nttntt

t

ntt

t

X

=−=−−+=−=

+−=++−=

+++−=

+==

=

−−

=

−

=


Expectation of Bionimal r.v. by Definition

xnn

x

x

xnn

x

x

xnxn

x

qpxnxx

nx

qpxnx

nx

ppxn

xE[X]

−

=

−

=

−

=

∑

∑

∑

−−=

−=

−

=

0

0

0

)!()!1(!)!(!

!

)1(


Since the term x = 0 is zero

xnn

x

xqpxnx

nE[X] −

=∑ −−

=0 )!()!1(

!

)1()1(1

1

1

))!1()1(()!1()!1( −−−

−

=

−∑ −−−−−

= xnn

x

x qppxnx

nnE[X]

Let s = x - 1 and x assumes value 1 to n, therefore s assumes value 0 to n - 1

snn

s

sqppsns

nnE[X] −−−

=∑ −−

−= 1

1

0 )!1(!)!1(


npppnpqpnp

qps

nnp

qppsns

nnE[X]

nn

n

s

sns

snn

s

s

=−+=+=

−=

−−−

=

−−

−

=

−−

−−−

=

∑

∑

11

1

0

)1(

11

0

)1()(

1)!1(!

)!1(


Expectation of X2 by Definition

xnn

x

x

xnn

x

x

xnn

x

x

xnxn

x

2

qpxnx

nx

qpxnx

nxx

qpxnx

nxxx

ppxn

x]E[X

−

=

−

=

−

=

−

=

∑

∑

∑

∑

−+

−−=

−+−=

−

=

0

0

0

0

2

)!(!!

)!(!!)1(

)!(!!])1([

)1(


Since the term x = 0 and x = 1 are zero,

npqpxnx

n]E[X xnn

x

x2 +−−

= −

=∑

0 )!()!2(!

npqppxnx

nnnE[X] xnn

x

x +−−−−

−−= −−−

−

=

−∑ )2()2(1

2

22

))!2()2(()!2()!2)(1(

Let s = x - 2 and x assumes value 2 to n, therefore s assumes value 0 to n - 2

npqppsns

nnn]E[X snn

s

s2 +−−−−

= −−−

=∑ 2

2

0

2

)!2(!)!2)(1(

BITS Pilani, Pilani Campus nppnn

nppppnnnpqppnn

npqps

npnn

qppsns

nnn]E[X

n

n

n

s

sns

snn

s

s2

+−=

+−+−=

++−=

+

−−=

−−−−

=

−

−

−

=

−−

−−−

=

∑

∑

2

22

22

1

0

)2(2

22

0

2

)1()1()1(

)()1(

2)1(

)!2(!)!2)(1(


Var(X) by definition

npqpnpnpnpnpnppnn

XEXEXVar

=−=−=

−+−=

−=

)1()()1(

][][][

2

22

22


It is difficult to write explicit formula. So values are given in Table I App. A, p. 687-691.

c.d.f. of binomial distribution


From c.d.f., we can find density f(x)= F(x) - F(x-1) if x = 0, 1, 2,…, n.

P(a ≤ X ≤ b) = F(b) - F(a-1) for integers a, b.


Let X represent the number of signals properly identified in a 30 minute time period in which 10 signals are received. Assuming that any signal is identified with probability p=1/2 and identification of signals is independent of each other. (i) Find the probability that at most seven signals

are identified correctly. (ii) Find the probability that at most 7 and at least

2 signals are identified correctly.

Example 3.5.3


i) *----*-----*-----*----*----*----*----*------------- 0 1 2 3 4 5 6 7 8 9 10

Here, P[X ≤ 7] = 0.9453 includes the Probability associated with 0 and 1

n = 10, p = 0.5 and look at the table (in row 8, column labeled 0.5 for F, we will see the value is 0.9453).


(ii) ----------*---*--*--*--*--*----------------- 0 1 2 3 4 5 6 7 8 9 10

Thus, P[2 ≤ X ≤ 7] = P[X ≤ 7] – P[X < 2] = P[X ≤ 7] – P[X ≤ 1] = F(7) – F(1)


P[2 ≤ X ≤ 7] = 0.945 – 0.011 = 0.934


Assume that an experiment is conducted and that the outcome is considered to be either a success or a failure. Let p denote the probability of success. Define X to be 1 if the experiment is a success and 0 if it is a failure. X is said to have a point binomial distribution {Bernoulli distribution) with parameter p. i) Argue that X is a binomial random variable with n = 1.

Section 3.4, page no 90, 45


iii) Find the moment generating function for X. iv) Find mean and variance of X. v) In DNA replication error can occur that are

chemically induced. Some of these errors are “silent” in that they do not lead to an observable mutation. Growing bacteria are exposed to a chemical that has probability 0.14 of inducing an observable error. Let X be 1 if an observable mutation results and let X be 0 otherwise. Find E[X].

ii) Find the density of X.


If we choose randomly with replacement a sample of n objects from N objects of which r are favorable and X = number of favorable objects in the sample chosen then X has binomial distribution with parameters n and p = r/N.

Sampling with replacement


From a usual pack of 52 cards, 10 cards are picked randomly with replacement. Find the probability that they will contain at least 4 and at most 7 spades. Identify Bernoulli trials and success and random variable X together with its distribution.

n = 10, p = 13/52 = 0.25. Required probability = F(7) - F(3)

Example


p n x .01 .05 .10 .15 .20 .25 10 0 0.90438 0.59874 0.34868 0.19687 0.10737 0.05631 1 0.99573 0.91386 0.73610 0.54430 0.37581 0.24403 2 0.99989 0.98850 0.92981 0.82020 0.67780 0.52559 3 1.00000 0.99897 0.98720 0.95003 0.87913 0.77588 4 1.00000 0.99994 0.99837 0.99013 0.96721 0.92187 5 1.00000 1.00000 0.99985 0.99862 0.99363 0.98027 6 1.00000 1.00000 0.99999 0.99987 0.99914 0.99649 7 1.00000 1.00000 1.00000 0.99999 0.99992 0.99958 8 1.00000 1.00000 1.00000 1.00000 1.00000 0.99997


Required probability = F(7) - F(3) = 0.99958 - 0.77588 (By tables) = 0.22370


If we are choosing without replacement a sample of size n from N objects of which r are favorable, and X = number of favorable objects in the sample, then

Hypergeometric distribution

otherwise. 0 and r)min(n,xr)]-(N-n max[0, if

;][

≤≤

−−

==

nN

xnrN

xr

xXP


N objects

n objects

‘r’ have trait (success)

(N - r) do not have trait (failure)


• The experiment consists of drawing a random sample of size n without replacement and without regard to order from a collection of N objects.

• Of the N objects, r have a trait of interest to us; the other (N – r) do not have the trait.

• The random variable X is the number of objects in the sample with the trait.

Properties:


Definition: A random variable X with integer values has a hypergeometric distribution with parameters N, n, r if its density is

r)min(n,xr)]-(N-n max[0, if

≤≤

−−

= ;)(

nN

xnrN

xr

xf


Theorem : If X is a hypergeometric random variable with parameters N, n, r then E[X] = n(r / N)

−−

−

=

1)(

NnN

NrN

NrnXVar


Suppose that X is hypergeometric with N = 20, r = 17, n = 5. What are the possible values for X? What is E[X] and Var (X)? Sol:

Section 3.7 page no 91, 54

r)min(n,xr)]-(N-n max[0, if

≤≤

−−

= ;)(

nN

xnrN

xr

xf


max[0, 5 – (20-17)] ≤ x ≤ min(5, 17) max[0, 2] ≤ x ≤ min(5, 17) i.e. X = 2, 3, 4 and 5. E[X] = 5(17/20) = 4.25 Var(X) = 5(17/20)(3/20)(15/19) = 0.5033


When the sample size n is small compared to population size N, then the composition of the sampled group does not change much from trial to trial if sampling is without replacement, This we can use binomial distribution with parameters are n and p = r/N. This is done if n/N ≤ 0.05.

Hypergeometric binomial


During a course of an hour, 1000 bottles of beer are filled by a particular machine. Each hour a sample of 20 bottles is randomly selected and number of ounces of beer per bottle is checked. Let X denote the number of bottles selected that are underfilled. Suppose during a particular hour, 100 underfilled bottles are produced. Find the probability that at least 3 underfilled bottles will be among those sampled.

Example 3.7.3


X = denote the number of bottles selected that are underfilled

N = 1000, n = 20, r = 100 Required probability = P[X ≥3] = 1- P[X =0] – P[X=1] – P[X=2]

Solution (using hypergeometric)


Required probability = P[X ≥3] = 1 – P[X=0] – P[X=1] – P[X=2]

3224.020

100018900

2100

201000

19900

1100

201000

20900

0100

1

=

−

−

−=


n = 20, p = 100/1000 = 0.1 (n/N = 20/1000 = 0.02 < 0.05) P[X ≥ 3] = 1 - F(2) = 1 - 0.6769 = 0.3231.

Using Binomial Approximation with


Sometimes population size is large but not known. Proportion of favorable population is given. Then we can use binomial distribution for both sampling with or without replacement where p is the proportion of favorable population.


Ex : A vegetable vendor has a large pile of tomatoes of which 30% are green. A buyer randomly puts 10 tomatoes in his basket. What is the probability that more than 5 of them are green?


This distribution is named on the French mathematician Simeon Denis Poisson. Let k > 0 be a constant and, for any real number x,

Poisson Distribution

=

−

=

otherwise

for

0

,...2,1,0;!)( x

x

xkkexf


f(x) is nonnegative. A random variable X with this density f is said to have a Poisson distribution with parameter k.

Verify f(x) is probability density function.

+++−=

−= ∑∑

∞

=

∞

=

...!2!1

1!

)(2

00

kkkex

xkkexfxx

1=−= keke


The m.g.f. of a Poisson random variable X with parameter k > 0 is

)1()( −=teketmX

E[X] = k and Var[X] = k.

Theorem


Proof:

( )

)1(.....!2)(1

!

!][)(

2

0

0

−=

+++=

=

==

−

∞

=

−

∞

=

−

∑

∑

tekekekee

xkee

xkeeeEtm

ttk

x

xtk

x

xktxtX

X


[ ]

[ ]kXEXEXVarHence

kkkeekee

tmdtdXE

kkeetmdtdXE

ttektek

tX

ttek

tX

tt

t

=−=

+=+=

=

==

=

=−−

=

=−

=

22

20

2)1()1(

02

22

0)1(

0

])[(][)(,

)()(

)((][

)()((][


Expectation of Poisson r.v. by Definition

kkee

kkkeykkeXE

yxLetxkke

xke

xkexXE

kk

k

y

yk

x

xk

x

xk

x

xk

==

+++==

=−−

=−

==

−

−∞

=

−

∞

=

−−

∞

=

−∞

=

−

∑

∑∑∑

.....!2

1!

][

1)!1()!1(!

][

2

0

0

1

00


∑∑

∑

∑ ∑

∑∑

∞

=

−−

∞

=

−

∞

=

−

∞

=

∞

=

−−

∞

=

−∞

=

−

+−

=+−

=

+−−

−=

+−=

+−==

0

22

0

0

0 0

00

22

)!2()!2(

)!2)(1()1(

!!)1(

!})1({

!][

x

xk

x

xk

x

xkx x

xkxkx

xk

x

xk

kxkkek

xke

kxxx

kexx

xkex

xkexx

xkexxx

xkexXE


kkkkXEXEXVar=−+=

−=22

22 ][][][kkkeke

kkkke

kykkeXE

yxLet

kk

k

y

yk

+=+=

+

+++=

+=

=−

−

−

∞

=

− ∑

22

2

0

22

.....!2

1

!][

2


Poisson Process : A process occurring discretely over a continuous interval of time or length or space is called a Poisson Process. Let λ = average number of successes occurring in a unit interval.


Assumptions of Poisson process : (i) Probability of success in a very small

interval of time ∆t is λ∆t (ii) Probability of more than one success

in such a very small interval of time is negligible.

(iii) Probability success in such a small interval does not depend on what happened prior to that time.


Then λ = average number of successes occurring in a unit of (time or space or length ) Let X = number of times the discrete event occurs in a given interval of length s units

Then X has Poisson distribution with parameter k = λs.

Rajiv


Let X = number of times the discrete event occurs in a given interval of length s in a Poisson process. Then X has Poisson distribution with parameter k = λs. Thus density of X is :

==

−

otherwise

0,1,2,...xfor

0!

)()( x

sexf

xs λλ


Provided by Table on p.692. Values of k=λs, the parameter of Poisson distribution corresponds to columns, values t of random variable correspond to rows and value of cdf F(t) are entries inside table.

c.d.f. of Poisson distribution


• The expected value of X is λs. • The average number of occurrence of the

event of interest in an interval of ‘s’ units = λs.

• Thus the average number of occurrences of the event in 1 unit of time, length, area or space is λs/s = λ


• Determine the basic unit of measurement being used.

• Determine the average number of occurrences of the event per unit. This number is denoted by λ.

• The random variable X, the number of occurrences of the event in the interval of size s follows a Poisson distribution with parameter k = λs.

Steps in solving Poisson Problem


If a binomial random variable X has parameter p very small and n large so that np = k is moderate then X can be approximated by a Poisson random variable Y with parameter k.

Poisson approximation to Binomial


Geophysicists determine the age of a zircon by counting the number of uranium fission tracks on a polished surface. A particular zircon is of such an age that the average number of tracks per square centimeter is five. What is the probability that a 2 centimeter-square sample of this zircon will reveal at most 3 tracks, thus leading to an underestimation of the age of the Material?

Exercise 63


A large microprocessor chip contains multiple copies of circuits . If a circuit fails, the chip knows how to select the proper logic to repair itself. Average number of defects per chip is 300. Find the probability that 10 or fewer defects will be found in a randomly selected region that comprises 5% of the total surface area?

Example:


California is hit by approximately 500 Earthquakes that are large enough to be felt every year. However those of destructive magnitude occur on an average once a year. Find the probability that at least one earthquake of this magnitude occurs during a 6 month period. Would it be unusual to have 3 or more earthquakes of destructive magnitude on a 6 month Period? Explain.

Ex.64 :


Thank You

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