Generating functions

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# Generating functions - PowerPoint PPT Presentation

Generating functions. Generating functions. Definition: Let a 0 ,a 1 ,……a n ,….be a sequence of real numbers and let If the series converges in some real interval (-x 0 , x 0 ), |x|≤ x 0 the function A(x) is called a generating function for {a j }.

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### Generating functions

MA 4030-probability generating fnctions

Generating functions
• Definition: Let a0,a1,……an,….be a sequence of real numbers and let
• If the series converges in some real interval (-x0, x0), |x|≤ x0 the function A(x) is called a generating function for {aj}.

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The generating function may be regarded as a transformation which carries the sequence {aj} into A(x).In general x will be a real number. However, it is possible to work with the complex numbers as well.

• If, the sequence {aj}. is bounded , then a comparison with the geometric series shows that A(x) converges at least for |x| ≤1.

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Probability Generating function
• If we have the additional property that:

aj ≥0 and

• then A(x) is called a probability generating function.

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Proposition
• A generating function uniquely determines its sequence.
• This single function A(x) can be used to represent the whole collection of individual items {aj}.
• Uses of probability generating functions

To find the density/mass function

To find the Moments in stochastic models

To Calculate limit distributions

In difference equations or recursions

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Example:
• Let us consider a random variable X,

where the probability, P[X=j]=pj

• Suppose X is an integral valued random variable with values 0,1,2,…….
• Then we can define the tail probabilities as

Pr[X > j] = qj

• The distribution function is thus

Pr[X≤j] = 1- qj

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The probability generating function is

• The generating function,
• is not a p.g.f. since .

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Some useful results:
• 1) 1-P(x) = (1-x) (Q(x))
• 2.) P’(1) = Q(1)
• 3.) P’’(1) = 2Q’(1)
• 4.) V(x)=P’’(1)+P’(1)-[P’(1)]2

=2’Q’(1) +Q(1)-[Q(1)]2

• 5) the rth factorial moment or rth moment about the origin,
• µ’(r) = E [X(X-1) (X-2)……(X-r+1)
• = P(r) (1) = rQ (r-1) (1)

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Convolutions
• Consider two non negative independent integral valued random variables X and Y, having the probability distribution,
• P[X=j] = aj and P[Y=k] = bk
• The probability of the event (X=j,Y=k) is therefore
• Pr[ (X=j,Y=k)]= aj bk

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Suppose we form a new random variable S=X+Y

• Then the event S=r comprises the mutually exclusive events,

(X=0,Y=r),(X=1,Y=r-1),….(X=m,Y=r-m), ………..(X=r-1,Y=1),(X=r,Y=0)

If the distribution of S is given by P[S=r]=cr

Then it follows that :

Cr=a0br+a1br-1+…………..+arb0

This method of compoundingtwo sequences

of numbers is called a convolution {cj}={aj}{bj]

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Generating functions and Convolutions:
• Proposition : The generating function of a convolution ({cj} ) is the product of the generating functions ({aj},{bj}) .

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Define the associated probability generating functions of the sequences defined earlier.

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• Then the corresponding generating function is
• Ctd…

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Extensions to convolution:

• More generally the generating function of {aj},{bj},{cj},…is the product of the generating functions of {aj},{bj},{cj},….
• F(x)= A(x).B(x).C(x)….
• Also let X1,X2,…..,Xn be i.i.d r.v.s
• and Sn=X1+X2+…..+Xn
• Then the g.f.of Sn is [P(x)]n

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Some properties of convolution

1.The convolution of two probability mass functions on the non negative integers is a pr. Mass function.

2.Convolution is a commutative operation

X+Y and Y+X , have the same distribution.

3.It is an associative operation. (order is immaterial) X+(Y+Z) = (X+Y)+ Z , have the same distribution

MA 4030-probability generating fnctions

Examples
• Find the p.g.f , the mean and the variance of :

1.Bernoulli distribution where, p=P[success]=P[X=1], and q=P[X=0]. Where X= number of successes in a trial.

2.Poisson distribution P[X=r]= e-µ µr/r!

eg: X -. number of phone calls in a unit time interval. µ is the average number of phone calls/unit

3.Geometric distribution P[X=j]=pqj X? define

4.Binomial distribution (X_ number of successes in n number of trials.) P[X=r] = nCr pr q n-r ,

where p= P[success] and q= P[failure]

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1. For the Bernoulli r.v.

• P(s)=[q+ps]

the mean =P’[1]=p

the variance =P’’[1]+P’[1]-{P’{1]}2

=0+p-p2=p(1-p)

=qp

4. For the Binomial r.v. ,which is the sum of n independent r.v.s, P(s)=[q+ps]n

Mean=np

Variance =npq

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2. For the Poisson r.v.

• P[x]=
• The mean =P’[1]=
• And the variance=P’’[1]+P’[1]-{P’{1]}2
• =

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3. For the Geometric r.v.

• P[x]=
• The mean =P’[x] | x=1=

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To find the variance,

• The variance =P’’[1]+P’[1]-{P’{1]}2

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Some more examples:

Negative Binomial random variable Y,

• The :

Where the P{success]=p and P[failure]=q

• Number of independent trials for the kth success : Y+k, Or the number of failures before the kth success r.v. Y;
• This distribution is called the negative binomial because he probabilities correspond to the successive terms of the binomial expansion of

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Let Y+k=n and y=n-k

• The m.g.f is M( ,t)={q+pe-)}-k
• And the p.g.f is P(x,t)=={q+px-1)}-k

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Note that when k=1, the probability mass function becomes qyp, which is the Geometric distribution.

• It can be shown that: P(x)=
• Which is the nth power of the p.g.f. of the geometric distribution.
• Then it is clear that negative binomial r.v. is the convolution of n geometric random variables.
• Its mean

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Compound Distributions
• Consider the sum of n independent random variables, where the number of r.v. contributing to the sum is also a r.v.
• Suppose SN=X1+X2+…+XN
• Pr[Xi=j]=fj, Pr[N=n]=gn, Pr[SN=k]=hk,
• with the corresponding p.g.fs.

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Moment generating function
• Moment generating function (m.g.f) of a r.v. Y is defined as M()=E[e Y]
• If Y is a discrete integer valued r.v. with probability P[Y=j]=pj
• Then M()=

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Taylor expansion of of M() generates the

• moments given by
• M() =
• is the rth moment about the origin,
• M’()|  = =. =E(X), M’’ ()|  =0 =(E(X(X-1))

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• And all the other properties as in the discrete case.

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M.g.f of the Binomial distribution
• M()=(q+peθ)n
• That is replace x in p.g.f. P(x) by eθ..
• M’()=n(q+peθ)n-1p
• E[X}=M’()| =0 =n(q+peθ)n-1p| =0
• = n(q+p)n-1p , since eθ=1 when θ=0
• =np
• Similarly it can be shown that V(X)=npq

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