FS1 7.1 - Generating functions of standard distributions | Edexcel A-Level Further Mathematics

A probability generating function stores an entire discrete distribution in one expression. You will construct and check these functions, derive the four standard forms, recover exact probabilities and transform the values of a random variable (FS1 7.1).

Store the value in the power and its probability in the coefficient

The expression $(1+2t+t^3)/4$ contains a complete probability distribution. Could evaluating it at just $t=-1$ tell you how likely an even outcome is?

Start by reading its pieces: the constant stores probability $1/4$ at value $0$ , the coefficient of $t$ stores probability $1/2$ at value $1$ , and the coefficient of $t^3$ stores probability $1/4$ at value $3$ . The missing $t^2$ means probability zero at value $2$ .

You already know

For a discrete variable, $\mathrm E(g(X))=\sum_x g(x)\mathrm P(X=x)$ . Core Pure Maclaurin work gives the coefficient of $t^r$ as $f^{(r)}(0)/r!$ , where the expansion applies. Use the standard exponential and binomial series with their conditions.

Here the function applied to the random value is $t^X$ . The letter $t$ is an input, not another random variable.

Probability generating function

For a non-negative integer-valued random variable $X$ , its probability generating function, or PGF, is

G_X(t)=\mathrm E(t^X)=\sum_{x=0}^{\infty}\mathrm P(X=x)t^x

For finite support the sum is a polynomial. For infinite support it is a power series, used where it converges. In particular it converges for $|t|\le1$ . The definition $G_X(t)=\mathrm E(t^X)$ is in the formula booklet.

A PGF is a weighted list in algebraic form, rather than a density or a cumulative probability. Its value at an arbitrary input is usually neither a point probability nor a cumulative probability.

Reading and checking a PGF

For a non-negative integer-valued variable,

G_X(1)=1,\qquad G_X(0)=\mathrm P(X=0)

The coefficient of $t^r$ is $\mathrm P(X=r)$ . Equivalently, using the Maclaurin coefficient rule,

\mathrm P(X=r)=\frac{G_X^{(r)}(0)}{r!},\qquad r=0,1,2,\ldots

Every coefficient must be non-negative. Normalisation alone is not enough to establish a valid PGF.

Not in the formula booklet: learn it.

Substituting $1$ adds all the coefficients, whereas substituting $0$ isolates the constant term. Substituting $-1$ adds the even-power probabilities and subtracts the odd-power probabilities. Thus $G_X(-1)$ measures the difference between even and odd probabilities: a surprisingly compact way to ask about parity. We will finish the opening calculation after working with the encoding.

Key idea

The exponent is the possible value; its coefficient is the probability. Check both non-negative coefficients and total probability $G_X(1)=1$ .

Use both checks rather than accepting any expression that equals one at $t=1$ .

Normalising a PGF and reading exact probabilities

Find an unknown normalising constant using $G_X(1)=1$ , then read the required coefficients.

For a finite distribution, form one term $\mathrm P(X=x)t^x$ for each possible value. Include a missing power with coefficient zero when adding a range of probabilities. For example, $\mathrm P(X\le2)$ adds the coefficients at powers $0,1,2$ ; it does not mean substitute $t=2$ .

Worked example 1: an unknown scale factor

Question. A non-negative integer-valued variable has PGF $G_X(t)=k(1+2t+t^3)$ . Find $k$ and $\mathrm P(X\le1)$ .

Plan. Normalise at $1$ , then add the constant and linear coefficients.

\begin{aligned} G_X(1)&=k(1+2+1)=1 &&\text{total probability}\\ k&=\frac14 &&\text{normalise}\\ G_X(t)&=\frac14+\frac12t+\frac14t^3 &&\text{display the probabilities}\\ \mathrm P(X\le1)&=\frac14+\frac12=\frac34 &&\text{values zero or one} \end{aligned}

Why this step? Setting $t=0$ would give only $\mathrm P(X=0)=k$ , which need not equal $1$ .

Answer. $k=1/4$ and $\mathrm P(X\le1)=3/4$ .

Check. All three probabilities are non-negative and add to $1$ ; the probability at value $2$ is zero.

Worked example 2 changes only the unnormalised weight at value $1$ , from $2$ to $3$ . The whole expression must be normalised again.

Worked example 2: changing one weight

Question. A variable has PGF $G_X(t)=k(1+3t+t^3)$ . Find $k$ and $\mathrm P(X\le1)$ .

Plan. Recalculate the total coefficient weight before reading probabilities.

\begin{aligned} G_X(1)&=k(1+3+1)=1 &&\text{new total weight}\\ k&=\frac15 &&\text{new normalising constant}\\ \mathrm P(X\le1)&=\frac15+\frac35=\frac45 &&\text{constant plus linear term} \end{aligned}

Why this step? Retaining $k=1/4$ would make the probabilities add to $5/4$ . Changing one weight changes the scale of every probability.

Answer. $k=1/5$ and $\mathrm P(X\le1)=4/5$ .

Check. The remaining probability is $\mathrm P(X=3)=1/5$ , complementing $4/5$ .

You can reverse the process: a table with values $0,1,3$ and probabilities $1/5,3/5,1/5$ generates precisely this polynomial.

Watch out

Wrong: $\mathrm P(X\le1)=G_X(1)$ . That value is always $1$ . Right: add the coefficients belonging to outcomes at most $1$ ; keep the argument of a function separate from the outcome threshold.

Increase the same unnormalised weight once more.

Deriving the PGFs of count distributions

Insert the probability formula into the PGF definition, then recognise a binomial or exponential expansion.

For $X\sim\mathrm B(n,p)$ , let $q=1-p$ . The binomial theorem gives

\begin{aligned} G_X(t)&=\sum_{x=0}^{n}\binom nx p^xq^{n-x}t^x &&\text{definition and binomial probabilities}\\ &=\sum_{x=0}^{n}\binom nx(pt)^xq^{n-x} &&\text{combine powers}\\ &=(q+pt)^n &&\text{binomial theorem} \end{aligned}

For $Y\sim\operatorname{Po}(\lambda)$ , the exponential series gives

\begin{aligned} G_Y(t)&=\sum_{y=0}^{\infty}\mathrm e^{-\lambda}\frac{\lambda^y}{y!}t^y &&\text{definition and Poisson probabilities}\\ &=\mathrm e^{-\lambda}\sum_{y=0}^{\infty}\frac{(\lambda t)^y}{y!} &&\text{extract the constant}\\ &=\mathrm e^{-\lambda}\mathrm e^{\lambda t} =\mathrm e^{\lambda(t-1)} &&\text{exponential series} \end{aligned}

Both derivations start with the definition; merely copying the final supplied expression would not show a derivation.

Binomial and Poisson PGFs

X\sim\mathrm B(n,p)\quad\Longrightarrow\quad G_X(t)=(1-p+pt)^n

Here $n$ is a non-negative integer, $0\le p\le1$ , and the polynomial is valid for all real $t$ .

Y\sim\operatorname{Po}(\lambda)\quad\Longrightarrow\quad G_Y(t)=\mathrm e^{\lambda(t-1)}

Here $\lambda>0$ , and the exponential series converges for all real $t$ .

In the formula booklet.

The factor $\mathrm e^{-\lambda}$ must remain when extracting Poisson coefficients. For example, $\operatorname{Po}(2)$ gives $G(t)=\mathrm e^{-2}(1+2t+2t^2+\cdots)$ , so the probability at value $2$ is $2\mathrm e^{-2}$ , rather than $2$ .

Worked example 3: adding the first three coefficients

Question. Let $X\sim\mathrm B(4,2/5)$ . Find its PGF and use it to find $\mathrm P(X\le2)$ exactly.

Plan. Substitute into the binomial PGF and retain its constant, linear and quadratic terms.

\begin{aligned} G_X(t)&=\left(\frac35+\frac25t\right)^4 &&\text{standard binomial PGF}\\ &=\frac{(3+2t)^4}{625} &&\text{preserve the outside factor}\\ &=\frac{81+216t+216t^2+\cdots}{625} &&\text{first three coefficients}\\ \mathrm P(X\le2)&=\frac{81+216+216}{625} &&\text{add outcomes zero to two}\\ &=\frac{513}{625} &&\text{exact probability} \end{aligned}

Why this step? The quadratic coefficient includes $\binom42=6$ . A Maclaurin alternative must divide $G_X''(0)$ by $2!$ , giving the same coefficient.

Answer. $G_X(t)=(3+2t)^4/625$ , and $\mathrm P(X\le2)=513/625$ .

Check. Direct binomial probabilities at $0,1,2$ give the same three coefficients, and $G_X(1)=5^4/625=1$ .

Worked example 4 changes only the trial count, from $4$ to $5$ . The success probability and event remain fixed.

Worked example 4: one more trial

Question. Let $X\sim\mathrm B(5,2/5)$ . Find its PGF and use it to find $\mathrm P(X\le2)$ exactly.

Plan. Increase the PGF's power and recalculate the first three coefficients.

\begin{aligned} G_X(t)&=\frac{(3+2t)^5}{3125} &&\text{five trials}\\ &=\frac{243+810t+1080t^2+\cdots}{3125} &&\text{binomial coefficients}\\ \mathrm P(X\le2)&=\frac{243+810+1080}{3125} &&\text{same event}\\ &=\frac{2133}{3125} &&\text{exact sum} \end{aligned}

Why this step? The quadratic numerator is $\binom52 3^3 2^2=1080$ ; changing the trial count changes the coefficient as well as the denominator.

Answer. $G_X(t)=(3+2t)^5/3125$ , and $\mathrm P(X\le2)=2133/3125$ .

Check. The probability is smaller than $513/625$ , as an extra independent trial can add a success but cannot remove an existing one.

Expanding a rational or exponential PGF uses the same coefficient rule. If using derivatives, evaluate at zero for probabilities; the next lesson uses derivatives at one for moments.

Watch out

Wrong: $\mathrm P(X=2)=G_X''(0)$ , or evaluate that derivative at $1$ . Right: divide $G_X''(0)$ by $2!$ . Keep exact fractions when the question requests an exact probability.

Continue this lesson.

You’re previewing NovaLearn without an account. Create a free account to read the rest.

Create free account

Last Accessed

Core Pure Mathematics

Further Pure Mathematics 1

Further Pure Mathematics 2

Further Statistics 1

Further Statistics 2

Further Mechanics 1

Further Mechanics 2

Decision Mathematics 1

Decision Mathematics 2

FS1 7.1 - Generating functions of standard distributions

Store the value in the power and its probability in the coefficient

Normalising a PGF and reading exact probabilities

Deriving the PGFs of count distributions

Continue this lesson.

Learn faster with Nova