Probability Generating Function

Suppose the discrete random variable \(X\) has a PGF given by

$$G_X(t)=\frac{1}{8}(1+t)^3.$$

Then,

$$G_X(1)=\frac{1}{8}{2}^3=1.$$

Let \(X\) have a PGF given by

$$G_X(t)=\frac{1}{8}(1+t)^3.$$

Then

\[\begin{align} G_X'(t)&=\frac{3}{8}(1+t)^2 \\ G_X'(t)&=\frac{3}{8}(2)^2=\frac{3}{2} .\end{align}\]

Therefore

\[ \mathbb{E}(X)=\frac{3}{2}.\]

Let \(X\) have a PGF given by

$$G_X(t)=\frac{1}{8}(1+t)^3.$$

Then

\[\begin{align} G_X'(1)&=\frac{3}{2} \\ G_X''(t)&=\frac{3}{4}(1+x) \\ G_X''(1)&=\frac{3}{2} \end{align}\] Therefore

\[ \begin{align}\text{Var}(X)&=G_X''(1)+G_X'(1)-(G_X'(1))^2 \\ &=\frac{3}{2}+\frac{3}{2}-\left(\frac{3}{2}\right)^2\\ &=\frac{3}{4} .\end{align}\]

Suppose a discrete random variable \(X\) has a PGF given by

$$G_X(t)=\frac{t^2}{(2-t)^5}$$

and a discrete random variable \(Y\) has a PGF given by

$$G_Y(t)=\frac{t}{(4-3t)^2}.$$

Given that \(X\) and \(Y\), find the PGF of \(Z=X+Y\):

Solution:

Using property 8,

\[\begin{align} G_Z(t)&=G_X(t) \cdot G_Y(t) \\ &=\frac{t^2}{(2-t)^5} \cdot\frac{t}{(4-3t)^2} = \frac{t^3}{(2-t)^5(4-3t)^2}. \end{align}\]

The probability generating function of a discrete random variable \(X\) is given by

$$G_X(t)=z(1+2t+2t^2)^2.$$

a) Find the value of \(z\).

b) Give the probability distribution of \(X\).

Solution:

a) Using property 2 above, you know that for any PGF,

\[\begin{align} G_X(1) &=\sum_{x} 1^x\mathbb{P}(X=x) \\ &=\sum_{x}\mathbb{P}(X=x) \\ &=1, \end{align}\]

so

\[\begin{align} G_X(1)&=z(1+2(1)+2(1)^2)^2 \\ 1&=z(1+2+2)^2 \\ z&=\frac{1}{25}. \end{align}\]

b) You have \(G_X(t)=\frac{1}{25}(1+2t+2t^2)^2.\)

Expanding brackets gives

\[\begin{align} G_X(t)&=\frac{1}{25}(1+2t+2t^2)(1+2t+2t^2) \\ &=\frac{1}{25}(1+2t+2t^2+2t+4t^2+4t^3+2t^2+4t^3+4t^4) \\ &=\frac{1}{25}(1+4t+8t^2+8t^3+4t^4) \\ &=\frac{1}{25}+\frac{4t}{25}+\frac{8t^2}{25}+\frac{8t^3}{25}+\frac{4t^4}{25} \\ &=\frac{1t^0}{25}+\frac{4t^1}{25}+\frac{8t^2}{25}+\frac{8t^3}{25}+\frac{4t^4}{25}. \end{align}\]

Now you have a function where you can read off the values of the values of \(x\) with the corresponding probabilities of \(x\) using the fact that the coefficients of \(t^x\) are the probabilities \(\mathbb{P}(X = x)\). Therefore, the probability distribution of X is:

Suppose a random variable \(X\) has a PGF given by

$$G_X(t)=\frac{1}{10}(4t+3t^2+2t^3+t^4).$$

Find the variance of \(X\).

Solution:

Using property 7 above, you have

\[\begin{align} \text{Var}(X) &=\mathbb{E}\left(X^2\right)-(\mathbb{E}(X))^2 \\ &=G_X''(1)+G_X'(1)-(G_X'(1))^2, \end{align}\]

so

\[\begin{align} G_X'(t)&=\frac{\mathrm{d} }{\mathrm{d} t} G_X(t) \\ &= \frac{1}{10}\left(4+6t+6t^2+4t^3\right) \\ G_X'(1)&=2 \\ G_X''(t)&=\frac{\mathrm{d^2} }{\mathrm{d} x^2} G_X(t) \\ &= \frac{3\left(2t^2+2t+1\right)}{5} \\ G_X''(1)&=3. \end{align}\]

Therefore

\[\begin{align} \text{Var}(X)&=G_X''(1)+G_X'(1)-(G_X'(1))^2 \\ &=3+2-2^2\\ &=1. \end{align}\]

The number of website visitors is given by a rate of \(4\) per hour. Given that the random variable \(X\) is the number of visitors to the website that come in a random hour, and that the visits are independent and random, show from first principles, that the probability generating function for \(X\) is

$$G_X(t)=e^{4(t-1)}.$$

Solution:

From the event description, you can see that the random variable has the property that \(X\sim Poi(4)\) since each visit independent of each another, and occur in a fixed time period (an hour) at a constant average rate \(4.\)

Therefore,

\[\mathbb{P}(X=x)=\frac{e^{-4}4^x}{x!},\]

so

\[\begin{align} G_X(t)&=\mathbb{E}(t^X)=\sum_{x} t^x\mathbb{P}(X=x) \\ &=\sum_{x=0}^{\infty} t^x\frac{e^{-4}4^x}{x!} \\ &=e^{-4}\sum_{x=0}^{\infty}\frac{t^x 4^x}{x!} \\ &=e^{-4} \sum_{x=0}^{\infty}\frac{(4t)^x}{x!} \\ &=e^{-4}e^{4t}\\ &=e^{-4+4t} \\ &=e^{4(t-1)} . \end{align}\]

The final equality follows from the Maclaurin expansion of \(e^x\) where \(x=4t\). Equivalently you can use the exponential summation:

\[\sum_{k=0}^{\infty} \frac{a^k}{k!} =e^a.\]

The probability of a seed will germinate is \(0.35\). Let the random variable \(X\) denote the number of seeds that have germinated out of \(4\) planted seeds.

a) Show, from first principles, that the probability generating function for \(X\) is

$$G_X(t)=(0.65+0.35t)^4.$$

b) Using your answer to a), determine the mean of \(X\).

Solution:

a) Observe that \(X ∼ \text{Bin}(4, 0.35)\), so

\[\mathbb{P}(X=x)= \binom{6}{x}0.35^x(1-0.35)^{6-x}\]

for \(x=0,\dots ,6\).

Using the formula for the probability generating function:

\[\begin{align} G_X(t)&=\sum_{x=0}^{4}t^x\mathbb{P}(X=x) \\ &=\sum_{x=0}^{4}t^x\binom{4}{x}0.35^x(1-0.35)^{6-x} \\ &=(0.65)^4+4(0.35)(0.65)^3t+6(0.35)^2(0.65)^2t^2+4(0.35)^3(0.65)t^3+(0.35)^4t^4 \\ &=(0.65)^4+4(0.65)^3(0.35t)+6(0.65)^2(0.35t)^2+4(0.65)(0.35t)^3+(0.35t)^4 \\ &=(0.65+0.35t)^4 ,\end{align}\]

where the last equality follows from the binomial formula:

\[(a+b)^n=\sum_{k=0}^{n}\binom{n}{k}a^k b^{(n-k)}.\]

Hence, the probability generating function for \(X\) is:

$$G_X(t)=(0.65+0.35t)^4.$$

b) Using property 4 above, you have that \(G_X'(t)=\mathbb{E}(X)\), so

\[\begin{align} G_X'(t)&=1.4(0.65+0.35t)^3 \\ G_X'(1)&=1.4 \end{align}\]

Becky rolls a fair six-sided dice. Let the random variable \(X\) denote the number of rolls it takes for her to get a multiple of \(2\). Given that each roll is independent, find the probability generating function of \(X\).

Solution:

Let \(p\) be the probability that Becky rolls an even number. Then \(p=0.5\) and the random variable \(X\sim \text{Geo}(0.5).\)

Therefore, using the formula given above, the probability generating function of \(X\) is

\[\begin{align} G_X(t)&=\frac{pt}{1-(1-p)t} \\ &=\frac{0.5t}{1-0.5t}. \end{align}\]

Let the random variable \(X\sim \text{Geo}(p)\), use the PGF of \(X\) to show that \(\mathbb{E}(X)=\dfrac{1}{p}\) and \(\text{Var}(X)=\dfrac{1-p}{p^2}.\)

Solution:

Using properties 4 and 7 you have that \(G_X'(1)=\mathbb{E}(X)\) and

\[\text{Var}(X)=G_X''(1)+G_X'(1)-(G_X'(1))^2.\]

From the definition above, a random variable \(X\sim \text{Geo}(p)\) has the PGF given by

\[G_X(t)=\frac{pt}{1-(1-p)t}.\]

So, using the quotient rule, you have that,

\[\begin{align} G_X'(t)&=\frac{(1-(1-p)t)(p)-(pt)(-(1-p))}{(1-(1-p)t)^2} \\ &=\frac{p(1-(1-p)t+(1-p)t)}{(1-(1-p)t)^2} \\ &=\frac{p}{(1-(1-p)t)^2} \\ G_X'(1)&=\frac{p}{(1-(1-p))^2} \\ &=\frac{p}{p^2} \\ &=\frac{1}{p}, \end{align} \]

therefore

\[\mathbb{E}(X)=\frac{1}{p} .\]

Using the chain rule, you have that,

\[\begin{align} G_X''(t)&=\frac{-2(-(1-p))p}{(1-(1-p)t)^3} \\ &=\frac{2p(1-p)}{(1-(1-p)t)^3} \\ G_X''(1)&=\frac{2p(1-p)}{p^3} \\ &=\frac{2(1-p)}{p^2} .\end{align}\]

Therefore,

\[\begin{align} \text{Var}(X)&=G_X''(1)+G_X'(1)-(G_X'(1))^2 \\ &=\frac{2(1-p)}{p^2}+\frac{1}{p}-\left(\frac{1}{p}\right)^2 \\ &=\frac{2(1-p)+p-1}{p^2} \\ &=\frac{1-p}{p^2}.\end{align}\]