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Generating Function

oyth94

Member
Jun 2, 2013
33
Suppose a gambler starts with one dollar and plays a game in which he or she wins one dollar with probability p and loses one dollar with probability 1 - p. Let fn be the probability that he or she fi rst becomes broke at time n for n = 0, 1, 2... Find a generating function for these probabilities.

So I think this is a binomial distribution is it? because it is giving me the fn = probability when first become broke.
since it is asking to find a generating function is use the
mx(s) = rx(es).
so since (i think) it is a binomial dist
then i let X ~ Binomial(n, theta)
and we know that the rx(t) = (t x theta + 1 - theta)n
so mx(s) = rx(es) = (estheta + 1 - theta)n
am i on the right track? i think i am not.. please help?
 

chisigma

Well-known member
Feb 13, 2012
1,704
Re: generating function

Suppose a gambler starts with one dollar and plays a game in which he or she wins one dollar with probability p and loses one dollar with probability 1 - p. Let fn be the probability that he or she first becomes broke at time n for n = 0, 1, 2... Find a generating function for these probabilities.

So I think this is a binomial distribution is it? because it is giving me the fn = probability when first become broke.
since it is asking to find a generating function is use the
mx(s) = rx(es).
so since (i think) it is a binomial dist
then i let X ~ Binomial(n, theta)
and we know that the rx(t) = (t x theta + 1 - theta)n
so mx(s) = rx(es) = (estheta + 1 - theta)n
am i on the right track? i think i am not.. please help?
The first step is the computation of $p_{n}$, i.e. the probability that he/she first becomes broke at the n-th step. It is not too difficult to realize that $p_{n}=0$ for n even and for n odd is...


$\displaystyle p_{2 n+ 1} = (1 - p)\ h_{n}\ [p\ (1-p)]^{n}\ (1)$


... where $h_{n}$ obeys to the recursive relation...


$\displaystyle h_{n+1} = h_{n} + n,\ h_{1}=1\ (2)$


... so that is...

$\displaystyle p_{n}= (1-p)\ \frac{n^{2} - n + 2}{2}\ [p\ (1-p)]^{n}\ (3)$

The (3) can now be used to valuate the generating function...


Kind regards


$\chi$ $\sigma$
 
Last edited:

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,721
Re: generating function

Suppose a gambler starts with one dollar and plays a game in which he or she wins one dollar with probability p and loses one dollar with probability 1 - p. Let fn be the probability that he or she first becomes broke at time n for n = 0, 1, 2... Find a generating function for these probabilities.

So I think this is a binomial distribution is it? because it is giving me the fn = probability when first become broke.
since it is asking to find a generating function is use the
mx(s) = rx(es).
so since (i think) it is a binomial dist
then i let X ~ Binomial(n, theta)
and we know that the rx(t) = (t x theta + 1 - theta)n
so mx(s) = rx(es) = (estheta + 1 - theta)n
am i on the right track? i think i am not.. please help?
This is a version of the gambler's ruin problem. As chisigma points out, the probability $f_n$ of becoming broke at the $n$th step is $0$ if $n$ is even. In the case where it is odd, $f_{2n+1} = C_np^n(1-p)^{n+1}$, where $C_n$ is the $n$th Catalan number. Using the first of those two links, you can check that the generating function for these probabilities can be expressed in the form $$\frac{1-\sqrt{1-4p(1-p)x^2}}{2px}.$$
 

oyth94

Member
Jun 2, 2013
33
Is there any other way to solve it besides using Catalan number and recursive relation and gamblers ruin?? Like using expected value or different distributions of some sort?