FS1-7 Probability generating functions — printable coverage pack

FS1-7.1 · Definitions, derivations and applications. Use of the probability generating function for the negative binomial, geometric, binomial and Poisson distributions.

For a non-negative integer-valued random variable, the probability generating function is $G_X(t)=E(t^X)=\sum_xP(X=x)t^x$ .
The standard PGFs are $(1-p+pt)^n$ for binomial, $e^{\lambda(t-1)}$ for Poisson, $\dfrac{pt}{1-(1-p)t}$ for geometric, and $\left(\dfrac{pt}{1-(1-p)t}\right)^r$ for negative binomial.
The coefficient of $t^x$ in $G_X(t)$ equals $P(X=x)$ , so a PGF encodes the whole distribution.
A common error is to omit the factor $t$ in the geometric PGF; the Edexcel convention starts at trial $1$ , not at zero failures.

1. Given $X\sim\operatorname{Bin}(4,0.3)$ , write down $G_X(t)$ and use its coefficient of $t^2$ to find $P(X=2)$ .[3 marks]
Answer
- $G_X(t)=(0.7+0.3t)^4$
- $P(X=2)=0.2646$
Method: The binomial PGF is $(1-p+pt)^n$ , giving $(0.7+0.3t)^4$ . Its $t^2$ coefficient is $\binom42(0.3)^2(0.7)^2=0.2646$ .

1. A geometric random variable $X$ has parameter $p=0.4$ and counts the trial of the first success. Derive its PGF and hence find $P(X=5)$ .[5 marks]
Answer
- $G_X(t)=\dfrac{0.4t}{1-0.6t}$
- $P(X=5)=0.05184$
Method: $G_X(t)=\sum_{x=1}^{\infty}0.4(0.6)^{x-1}t^x=0.4t\sum_{j=0}^{\infty}(0.6t)^j=\dfrac{0.4t}{1-0.6t}$ . The coefficient of $t^5$ is $0.4(0.6)^4=0.05184$ .

1. The random variable $X$ counts the trial on which the third success occurs in independent trials with success probability $0.25$ . Derive $G_X(t)$ and use it to find $P(X=6)$ .[7 marks]
Answer
- $G_X(t)=\left(\dfrac{0.25t}{1-0.75t}\right)^3$
- $P(X=6)=0.06592$ to $4$ significant figures
Method: A waiting time to the third success is the sum of three independent geometric waiting times, so its PGF is the cube of the geometric PGF: $G_X(t)=[0.25t/(1-0.75t)]^3$ . Using $(1-z)^{-3}=\sum_{j=0}^{\infty}\binom{j+2}{2}z^j$ , the coefficient of $t^6$ is $\binom52(0.25)^3(0.75)^3=0.06591797\ldots$ .

For a PGF $G_X$ , $G'_X(1)=E(X)$ and $G''_X(1)=E[X(X-1)]$ .
Therefore $\operatorname{Var}(X)=G''_X(1)+G'_X(1)-[G'_X(1)]^2$ .
The check $G_X(1)=1$ confirms that the encoded probabilities sum to one before derivatives are used.
A common error is to treat $G''_X(1)$ as $E(X^2)$ ; the missing $G'_X(1)$ must be added.

1. The PGF of $X$ is $G_X(t)=(0.8+0.2t)^6$ . Use derivatives of the PGF to find $E(X)$ and $\operatorname{Var}(X)$ .[4 marks]
Answer
- $E(X)=1.2$
- $\operatorname{Var}(X)=0.96$
Method: $G'_X(t)=1.2(0.8+0.2t)^5$ , so $G'_X(1)=1.2$ . Also $G''_X(1)=6(5)(0.2)^2=1.2$ . Hence $\operatorname{Var}(X)=1.2+1.2-1.2^2=0.96$ .

1. A random variable has PGF $G_X(t)=\dfrac{2t}{3-t}$ . Use the PGF to find its mean and variance.[5 marks]
Answer
- $E(X)=\frac32$
- $\operatorname{Var}(X)=\frac34$
Method: $G'_X(t)=6/(3-t)^2$ and $G''_X(t)=12/(3-t)^3$ . Thus $G'_X(1)=3/2$ and $G''_X(1)=3/2$ . Therefore $\operatorname{Var}(X)=3/2+3/2-(3/2)^2=3/4$ .

If $X$ and $Y$ are independent, then $G_{X+Y}(t)=G_X(t)G_Y(t)$ .
Multiply and simplify the PGFs before identifying a familiar distribution or extracting coefficients.
For example, multiplying $e^{\lambda(t-1)}$ and $e^{\mu(t-1)}$ gives $e^{(\lambda+\mu)(t-1)}$ , the PGF of $\operatorname{Po}(\lambda+\mu)$ .
A common error is to multiply PGFs without establishing independence; dependence prevents the expectation from factorising.

1. Independent variables have distributions $X\sim\operatorname{Po}(1.4)$ and $Y\sim\operatorname{Po}(2.6)$ . Use PGFs to identify the distribution of $X+Y$ and find $P(X+Y=0)$ .[4 marks]
Answer
- $X+Y\sim\operatorname{Po}(4)$
- $P(X+Y=0)=e^{-4}=0.01832$
Method: $G_X(t)G_Y(t)=e^{1.4(t-1)}e^{2.6(t-1)}=e^{4(t-1)}$ , which is the PGF of $\operatorname{Po}(4)$ . Its constant coefficient is $P(X+Y=0)=e^{-4}=0.0183156\ldots$ .

1. Independent variables satisfy $X\sim\operatorname{Bin}(5,0.4)$ and $Y\sim\operatorname{Bin}(7,0.4)$ . Use PGFs to identify $S=X+Y$ and calculate $P(S=3)$ .[5 marks]
Answer
- $S\sim\operatorname{Bin}(12,0.4)$
- $P(S=3)=0.1419$ to $4$ significant figures
Method: $G_S(t)=(0.6+0.4t)^5(0.6+0.4t)^7=(0.6+0.4t)^{12}$ , so $S\sim\operatorname{Bin}(12,0.4)$ . The coefficient of $t^3$ is $\binom{12}{3}(0.4)^3(0.6)^9=0.14189396\ldots$ .

1. Independent geometric variables $X$ and $Y$ each have parameter $0.4$ and count trials to first success. Use PGFs to identify the distribution of $S=X+Y$ and find $P(S>5)$ .[7 marks]
Answer
- $S$ is negative binomial with $r=2$ , $p=0.4$
- $P(S>5)=0.33696$
Method: $G_S(t)=\left(\dfrac{0.4t}{1-0.6t}\right)^2$ , the PGF of the trial of the second success. Hence $P(S>5)=1-\sum_{s=2}^{5}\binom{s-1}{1}(0.4)^2(0.6)^{s-2}=0.33696$ .