Probability-generating function: Definition from Answers.com

PGF redirects here, for other uses see PGF (disambiguation).

In probability theory, the probability-generating function of a discrete random variable is a power series representation (the generating function) of the probability mass function of the random variable. Probability-generating functions are often employed for their succinct description of the sequence of probabilities Pr(X = i), and to make available the well-developed theory of power series with non-negative coefficients.

1 Definition
2 Properties
3 Examples
4 Example calculation: two simple univariate probability generating functions
- 4.1 Analysis of the first game
- 4.2 Analysis of the second game
5 Example calculation: use of bivariate generating functions
6 Example calculation: bivariate generating functions and differential equations
7 Related concepts
8 External links

Definition

If X is a discrete random variable taking values on some subset of the non-negative integers, {0,1, ...}, then the probability-generating function of X is defined as:

$G(z) = \textrm{E}(z^X) = \sum_{x=0}^{\infty}p_X(x)z^x,$

where p_X is the probability mass function of X. Note that the equivalent notation G_X is sometimes used to emphasize the dependence on X.

Properties

Power series

Probability-generating functions obey all the rules of power series with non-negative coefficients. In particular, G(1⁻) = 1, since the probabilities must sum to one, and where G(1⁻) = lim_z→1G(z) from below. So the radius of convergence of any probability-generating function must be at least 1, by Abel's theorem for power series with non-negative coefficients.

Probabilities and expectations

The following properties allow the derivation of various basic quantities related to X:

1. The probability mass function of X is recovered by taking derivatives of G

$\quad p(k) = \textrm{Pr}(X = k) = \frac{G^{(k)}(0)}{k!}.$

2. It follows from Property 1 that if we have two random variables X and Y, and G_X = G_Y, then f_X = f_Y. That is, if X and Y have identical probability-generating functions, then they are identically distributed.

3. The normalization of the probability density function can be expressed in terms of the generating function by

$E(1)=G(1^-)=\sum_{i=0}^\infty f(i)=1.$

The expectation of X is given by

$\textrm{E}\left(X\right) = G'(1^-).$

More generally, the kth factorial moment, E(X(X − 1) ... (X − k + 1)), of X is given by

$\textrm{E}\left(\frac{X!}{(X-k)!}\right) = G^{(k)}(1^-), \quad k \geq 0.$

So the variance of X is given by

$\textrm{Var}(X)=G''(1^-) + G'(1^-) - \left [G'(1^-)\right ]^2.$

4.G_X( $e t$ ) = M_X(t) where X is a random variable, G(t) is the probability generating function and M(t) is the moment-generating function.

Functions of independent random variables

Probability-generating functions are particularly useful for dealing with functions of independent random variables. For example:

If X₁, X₂, ..., X_n is a sequence of independent (and not necessarily identically distributed) random variables, and

$S_n = \sum_{i=1}^n a_i X_i,$

where the a_i are constants, then the probability-generating function is given by

$G_{S_n}(z) = E(z^{S_n}) = E(z^{\sum_{i=1}^n a_i X_i,}) = G_{X_1}(z^{a_1})G_{X_2}(z^{a_2})\cdots G_{X_n}(z^{a_n}).$

For example, if

$S_n = \sum_{i=1}^n X_i,$

then the probability-generating function, G_Sn(z), is given by

$G_{S_n}(z) = G_{X_1}(z)G_{X_2}(z)\cdots G_{X_n}(z).$

It also follows that the probability-generating function of the difference of two independent random variables S = X₁ − X₂ is

$G_S(z) = G_{X_1}(z)G_{X_2}(1/z).$

Suppose that N is also an independent, discrete random variable taking values on the non-negative integers, with probability-generating function G_N. If the X₁, X₂, ..., X_N are independent and identically distributed with common probability-generating function G_X, then

$G_{S_N}(z) = G_N(G_X(z)).$

This can be seen as follows:

$G_{S_N}(z) = E(z^{S_N}) = E(z^{\sum_{i=1}^N X_i}) = E\big(E(z^{\sum_{i=1}^N X_i}| N) \big) = E\big( (G_X(z))^N\big) =G_N(G_X(z)).$

This last fact is useful in the study of Galton–Watson processes.

Suppose again that N is also an independent, discrete random variable taking values on the non-negative integers, with probability-generating function G_N. If the X₁, X₂, ..., X_N are independent, but not identically distributed random variables, where $G_{X_i}$ denotes the probability generating function of

X i

, then it holds

$G_{S_N}(z) = \sum_{i \ge 1} f_i \prod_{k=1}^i G_{X_i}(z).$

For identically distributed X_i this simplifies to the identity stated before. The general case is sometimes useful to obtain a decomposition of S_N by means of generating functions.

Examples

The probability-generating function of a constant random variable, i.e. one with Pr(X = c) = 1, is

$G(z) = \left(z^c\right).$

The probability-generating function of a binomial random variable, the number of successes in n trials, with probability p of success in each trial, is

$G(z) = \left[(1-p) + pz\right]^n.$

Note that this is the n-fold product of the probability-generating function of a Bernoulli random variable with parameter p.

The probability-generating function of a negative binomial random variable, the number of failures occurring before the rth success with probability of success in each trial p, is

$G(z) = \left(\frac{p}{1 - (1-p)z}\right)^r.$

Note that this is the r-fold product of the probability generating function of a geometric random variable.

The probability-generating function of a Poisson random variable with rate parameter λ is

$G(z) = \textrm{e}^{\lambda(z - 1)}.\;\,$

Example calculation: two simple univariate probability generating functions

This example illustrates basic computations with basic PGFs. We analyse the expected gain of the player in two probabilistic games. The first game works like this: the player rolls a die n times, obtaining a sequence of values such as e.g. 126346656665. For each consecutive sequence of sixes of length k, she wins $6 k + 2$ Euros. We ask about the expected amount won after n rolls of the die. For example, for the sequence given, we would win $63 + 6 4 + 6 5 = 9288$ Euros. In the second game, the player starts with a capital of zero Euros and starts flipping coins. If the coin comes up heads, his capital increases by one Euro, if it comes up tails, his capital is reduced by half. We ask about the expected amount won after n coin flips. We will use one to denote heads and zero to denote tails. For example, the sequence $110110$ gives 3/2.

Analysis of the first game

The random variable that we will use is $G n$ , the number of sixes at the end of the sequence. Let $p_n(u) = \sum_{k=0}^n p_{n, k} u^k = E\left[u^{G_n}\right]$ be the PGF where the term $p n, k$ represents the probability that there is a sequence of k sixes at the end of the total sequence of length n.

We immediately have $p 0 (u) = 1$ and for $n\ge 1$ ,

$p_{n+1}(u) = \frac{1}{6} u p_n(u) + \frac{5}{6} p_n(1).$

Given the fact that $p n (u)$ is a PGF, one has $p n (1) = 1$ and one finds

$p_{n+1}(u) = \frac{1}{6} u p_n(u) + \frac{5}{6}.$

This lets us calculate the closed form of $p n (u)$ , although we will not need it in the calculation that follows. It is:

$p_n(u) = \left( 1/6\,u \right) ^{n}-5\, \left( -6+u \right) ^{-1}+5\,{\frac { \left( 1/6\,u \right) ^{n}}{-6+u}}.$

Using the definition of the expected gain, one has

$E_n = 6^2 \left( \frac{5}{6} \sum_{m=1}^{n-1} (p_m(6) - p_m(0) ) + p_n(6) - p_n(0) \right).$

This formula represents two cases. First, $\frac{5}{6} 6^2 ( p_m(6) - p_m(0))$ represents the expected gain in going from roll m to the next. On average, one receives a quantity of $62 (p m (6) - p m (0))$ Euros with a probability of $\frac{5}{6}$ (end of a run of sixes in a row). Second, if the last roll was a 6, then one receives, on average, the sum $62 (p n (6) - p n (0))$ with probability one. To find $E n$ , we only need to sum these contributions.

The recurrence for $p n (u)$ implies immediately that $p_n(0) = \frac{5}{6}$ when $n\ge 1$ . What's more, $p_{n+1}(6) = p_n(6) + \frac{5}{6}$ , which gives $p_n(6) = 1 + n \frac{5}{6}.$ Using these values, one has

$E_n = 6^2 \left( \frac{5}{6} \sum_{m=1}^{n-1} \left(1 + m \frac{5}{6} - \frac{5}{6} \right) + 1 + n \frac{5}{6} - \frac{5}{6} \right).$

We only need some additional simplifications to conclude.

$E_n = 6^2 \left( \frac{25}{72} n^2 + \frac{5}{8} n + \frac{1}{36} \right) = \frac{25}{2} n^2 + \frac{45}{2} n + 1.$

The first few values are $36,96,181,291,426.$

Analysis of the second game

Let $p_n(u) = \sum_{k=0}^n p_{n, k} u^k$ be the PGF $E\left[u^{G_n}\right]$ where the term $p n, k$ represents the probability of winning k Euros with n coin flips. Thus one has $p 0 (u) = 1$ and

$p_{n+1}(u) = \frac{1}{2} \, u \, p_n(u) + \frac{1}{2} \, p_n\left( u^\frac{1}{2} \right).$

We don't need an explicit form of $p n (u)$ , given that the value we are looking for is

$E_n = \left. \frac{d}{du} p_n(u) \right|_{u=1} .$

We therefore continue with the derivative of the recurrence:

$p'_{n+1}(u) = \frac{1}{2} \, p_n(u) + \frac{1}{2} \, u \, p'_n(u) + \frac{1}{2} \, p'_n\left( u^\frac{1}{2} \right) \, \frac{1}{2} \left( u^{-\frac{1}{2}} \right) .$

From the definition of $p n (u)$ one finds $p n (1) = 1,$ , which gives

$p'_{n+1}(1) = \frac{1}{2} + \frac{1}{2} \, p'_n(1) + \frac{1}{2} \, p'_n(1) \, \frac{1}{2} .$

We conclude that the recurrence for $E n$ is $\left(E_0 = 0\right)$ :

$E_{n+1} = \frac{3}{4} E_n + \frac{1}{2} \quad \mbox{or} \quad E_n = 2 - 2 \left( \frac{3}{4} \right)^n .$

The first few values are $1/2,{\frac {7}{8}},{\frac {37}{32}},{\frac {175}{128}},{\frac {781}{512}},{\frac {3367}{2048}},{ \frac {14197}{8192}},{\frac {58975}{32768}},{\frac {242461}{131072}},{\frac {989527}{524288}}.$

It is important to note that we might have calculated this recurrence directly, starting from

$E_{n+1} = \frac{1}{2} ( E_n + 1) + \frac{1}{2} (E_n/2) = \frac{3}{4} E_n + \frac{1}{2}$

What we cannot calculate without the PGF is the variance, because only the PGF provides all factorial moments. This calculation is very simple and quite similar to the calculation of the expectation.

We wish to find $p'' n (1)$ and indeed, the cases $p n'(1)$ and $p'' n (1)$ are alike.

We calculate the derivative of the last recurrence from above (the one for $p n + 1'(u)$ ) and find

$p''_{n+1}(u) = \frac{1}{2} p'_n(u) + \frac{1}{2} p'_n(u) + \frac{1}{2} u p''_n(u) + \frac{1}{4} p''_n\left( u^{\frac{1}{2}} \right) \frac{1}{2} u^{-\frac{1}{2}} u^{-\frac{1}{2}} - \frac{1}{8} p'_n\left( u^{\frac{1}{2}} \right) u^{-\frac{3}{2}} ,$

which immediately yields $\left( \mbox{with} \quad F_n = p''_n(1) = E\left[G_n*(G_n-1)\right]\right)$ :

$F_{n+1} = E_n + \frac{1}{2} F_n + \frac{1}{8} F_n - \frac{1}{8} E_n \quad \mbox{or} \quad F_{n+1} = \frac{5}{8} F_n + \frac{7}{8} \left( 2 - 2 \left( \frac{3}{4} \right)^n \right).$

This recurrence is easy to solve, be it manually or with a CAS. With MAPLE, one finally finds

$F_n = \frac{14}{3}+{\frac {28}{3}}\, \left( \frac{5}{8} \right) ^{n}-14\, \left( \frac{3}{4} \right) ^{n}.$

It is an open problem to prove that the number of different winnings (possible values for the capital) after n coin flips is $f n + 3 - 1,$ with $f n$ the Fibonacci numbers.

These are the first few sets of possible values:

{0},{0,1},{0,1,2,1 / 2},{0,1,2,3,1 / 2,1 / 4,3 / 2},

$\{0, 1, 2, 3, 4, 1/2, 1/4, 1/8, 3/2, 3/4, 5/2, 5/4\}, \ldots$

The sequence is 1, 2, 4, 7, 12 -- this is indeed $f n + 3 - 1.$

There is a large amount of additional material at les-mathematiques.net, where these two problems originated (consult the external links for more information).

Example calculation: use of bivariate generating functions

The following example illustrates a very common technique the manipulation of PGFs: the use of bivariate super generating functions to compute the ordinary generating function (OGF) of the PGFs of a sequence of random variables.

Suppose you sample a system that can assume two states, X and Y, X with probability p and Y with probability 1 − p, e.g. a coin being flipped, obtaining the sequence of samples

$S_1, \, S_2, \, S_3, \,\ldots \,S_n,$

where the system was sampled n times and has no memory.

Define the random variable $M n$ to be the number of changes from one sample to the next in a sequence of n samples, i.e. how often $S m$ was different from $S m - 1$ . For example, the sequence

$X \,X \,X \,Y \,X \,X$

has two changes, as does

$Y \,X \,X \,X \,X \,X \,Y \,Y \,Y \,Y.$

We want to calculate the PGF of $M n$ , which we will do by using bivariate generating functions.

We introduce the bivariate GF $G (z, u)$ given by

$G(z, u) = \sum_{n\ge 1} E\left[u^{M_n}\right] z^n,$

i.e. $G (z, u)$ is the ordinary generating function of the PGFs of the $M n .$ This step is completely general and indeed the core of the method.

Now let $x n, k$ be the probability of having k changes in a sequence of n samples, where the last sample was an X. Similarly, let $y n, k$ be the probability of having k changes in a sequence of n samples, where the last sample was a Y, and put

$X(z, u) = \sum_{n\ge 1, k\ge 0} x_{n, k} \, u^k z^n \quad \mbox{and} \quad Y(z, u) = \sum_{n\ge 1, k\ge 0} y_{n, k} \, u^k z^n$

so that

$G(z, u) = X(z, u) + Y(z, u).\,$

Now we clearly have

$x_{n, 0} = p^n \quad \mbox{and} \quad y_{n, 0} = (1-p)^n,$

because having zero changes means getting a sequence of all Xs or Ys.

For $k\ge 1$ we find

$x_{n, k} = p \, y_{n-1, k-1} + p \, x_{n-1, k} \quad \mbox{and} \quad y_{n, k} = (1-p) \, x_{n-1, k-1} + (1-p) \, y_{n-1, k},$

because e.g. to have k changes in a sequence of length n that ends in X, we either append an X to a sequence having k − 1 changes and ending in Y, or append an X to a sequence having k changes and ending in X.

Summing these equations over n and k and writing X for X(z, u) and Y for Y(z, u), we obtain

$X - \frac{p z}{1 - p z} = p u z Y + p z \left( X - \frac{p z}{1 - p z} \right)$

and

$Y - \frac{(1-p) z}{1 - (1-p) z} = (1-p) u z X + (1-p) z \left( Y - \frac{(1-p) z}{1 - (1-p) z} \right).$

The solution of this system is

$X = -{\frac { \left( -pz+puz+z-uz-1 \right) pz}{-z+1+p{z}^{2}-{p}^{2}{z}^{2}-{u}^{2}{z}^{2}p+{p}^{2}{u}^{2}{z}^{2}}}$

and

$Y = -{\frac {z \left( -puz-1+p+pz-{p}^{2}z+{p}^{2}uz \right) }{-z+1+p{z}^{2}-{p}^{2}{z}^{2}-{u}^{2}{z}^{2}p+{p}^{2}{u}^{2}{z}^{2}}}.$

We may now use the general identity

$\sum_{n\ge 1} E\left[M_n (M_n-1) \ldots (M_n-r)\right] z^n = \left( \left(\frac{d}{du}\right)^{r+1} G(z, u) \right)_{u=1}$

to calculate the factorial moments of $M n .$ E.g. the OGF of the expectations is given by

$\sum_{n\ge 1} E[M_n] z^n = \left( \frac{d}{du} (X + Y) \right)_{u=1} = -2\,{\frac { \left( -1+p \right) {z}^{2}p}{ \left( -1+z \right) ^{2}}},$

from which we find (extracting coefficients) that

$E[M_n] = 2 \, p (1-p) \,(n-1).$

An extensive discussion of this problem, as well as solutions by other methods, may be found on Les-Mathematiques.net (external links).

Example calculation: bivariate generating functions and differential equations

Consider the following balls and urns problem: suppose we have an urn containing n distinguishable balls, i.e. bearing labels from 1 to n. We pick one of the balls at random and remove it from the urn. We also remove all balls whose labels are larger than the one we picked from the urn. E.g. if we picked ball number one, the urn is emptied after one operation. We repeat until the urn is empty. E.g. for an urn containing ten balls, the sequence of picks 6-3-1 would empty the urn in three operations. We introduce the random variable $X n$ , which gives the number of picks needed to empty the urn. Our goal is to compute all of its moments, and we will do so using exactly the same bivariate generating function as in the previous example, namely the OGF of the PGFs:

$P(z, u) = \sum_{n\ge 1} E\left[u^{X_n}\right] z^n.$

We let $p n, k$ be the probability of emptying an urn containing n balls with k operations, so that

$P(z, u) = \sum_{n\ge 1, k\ge 1} p_{n, k} u^k z^n.$

We find that

$p_{n, 1} = \frac{1}{n} \quad \mbox{and} \quad p_{n, k} = \frac{1}{n} \sum_{r=1}^{n-(k-1)} p_{n-r, k-1}, \quad k\ge 2$

because to empty the urn with one operation, we must pick the ball labelled 1. The remaining probabilities are computed recursively, e.g. we pick the ball with the largest label with probability $1 / n$ , leaving $n - 1$ balls (this is $r = 1$ ). We pick the ball with the next-to-largest label with probability $1 / n$ , leaving $n - 2$ balls (this is $r = 2$ ), etc. The upper bound for r is $n - (k - 1)$ , because we must have $n-r\ge k-1$ (we cannot e.g. empty an urn containing six balls using seven operations).

Next we set $p n, k = 0$ for $n < k$ , so that we may replace the recursion by

$p_{n, k} = \frac{1}{n} \sum_{r=1}^{n-1} p_{n-r, k-1}, \quad k\ge 2.$

Using the coefficient-extraction operator for formal power series, we thus have

$[z^{n-1} u^k] \frac{d}{dz} P(z, u) = [z^{n-1} u^{k-1}] \frac{1}{1-z} P(z, u) = [z^{n-1} u^k] \frac{u}{1-z} P(z, u), \quad k\ge 2.$

We note furthermore that

$[z^{n-1} u] \frac{d}{dz} P(z, u) = [z^{n-1}] \frac{d}{dz} \sum_{n\ge 1} \frac{1}{n} z^n = [z^{n-1}] \frac{d}{dz} \log \frac{1}{1-z} = [z^{n-1}] \frac{1}{1-z} = 1$

and that

$[z^{n-1} u] \frac{u}{1-z} P(z, u) = 0,$

where the first equation results from our "boundary condition" that the probability of emptying an urn with one operation is $1 / n .$

Summing over $k\ge 1$ (there are two contributions to both sides of the equation, one for $k\ge 2$ and another one for $k = 1$ ), we obtain

$\frac{d}{dz} P(z, u) = \frac{u}{1-z} + \frac{u}{1-z} P(z, u),$

e.g. through

$\sum_{n\ge 1} z^{n-1} u [z^{n-1} u] \frac{d}{dz} P(z, u) = \sum_{n\ge 1} z^{n-1} u = \frac{u}{1-z}.$

The solution to the differential equation is

$P(z, u) = -1 + C(u) \left( \frac{1}{1-z} \right)^u$ ,

with $C (u)$ a formal power series in u. We note

$[z^0] P(z, u) = 0 = -1 + C(u)\,$

which follows from the formal series

$\left( \frac{1}{1-z} \right)^u = \sum_{k\ge 0} \frac{u^k}{k!} \left( \log \frac{1}{1-z} \right)^k.$

Hence $C (u)$ is constant and equal to one, and we finally have

$P(z, u) = -1 + \left( \frac{1}{1-z} \right)^u,$

which is, incidentally, the generating function of the Stirling numbers of the first kind.

The moments now follow trivially from the formula given in the first example. E.g. for the expectation, we find

$\sum_{n\ge 1} E[X_n] z^n = \left. \frac{d}{du} P(z, u) \right|_{u=1} = \left. \left( \frac{1}{1-z} \right)^u \log \frac{1}{1-z} \right|_{u=1} = \frac{1}{1-z} \log \frac{1}{1-z}$

which gives

$E[X_n] = [z^n] \frac{1}{1-z} \log \frac{1}{1-z} = H_n,$

the nth harmonic number, and we need about $\log n\,$ operations to empty the urn.

An extensive discussion of this problem, as well as solutions by other methods, may be found on Les-Mathematiques.net (external links).

Related concepts

The probability-generating function is occasionally called the z-transform of the probability mass function. It is an example of a generating function of a sequence (see formal power series).

Other generating functions of random variables include the moment-generating function and the characteristic function.

External links

v • d • e Theory of probability distributions

probability mass function (pmf) · probability density function (pdf) · cumulative distribution function (cdf) · quantile function

raw moment · central moment · mean · variance · standard deviation · skewness · kurtosis

moment-generating function (mgf) · characteristic function · probability-generating function (pgf) · cumulant

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