Probability generating functions — practice questions

By definition, $G_X(t) = \sum P(X=x)t^x$. We sum over the possible values of $X$: 1, 2, and 3. $G_X(t) = P(X=1)t^1 + P(X=2)t^2 + P(X=3)t^3$ Substituting the given probabilities: $G_X(t) = 0.2t + 0.5t^2 + 0.3t^3$ This is the PGF for the random variable $X$. As a check, note that $G_X(1) = 0.2(1) + 0.5(1)^2 + 0.3(1)^3 = 0.2+0.5+0.3 = 1$.

(i) To find $P(X=2)$, we need the coefficient of $t^2$ in the expansion of $G_X(t) = (0.8 + 0.2t)^5$. This is a binomial expansion. The term in $t^2$ is given by $\binom{5}{2}(0.8)^{5-2}(0.2t)^2$. Coefficient = $\binom{5}{2}(0.8)^3(0.2)^2 = 10 \times 0.512 \times 0.04 = 0.2048$. So, $P(X=2) = 0.2048$.

(ii) To find the mean, we need $E(X) = G_X'(1)$. First, find the derivative: $G_X'(t) = 5(0.8 + 0.2t)^4 \times 0.2 = (0.8 + 0.2t)^4$. Now, substitute $t=1$: $G_X'(1) = (0.8 + 0.2(1))^4 = (1)^4 = 1$. So, the mean $E(X) = 1$.

(iii) To find the variance, we use $Var(X) = G_X''(1) + G_X'(1) - [G_X'(1)]^2$. We already have $G_X'(1)=1$. First, find the second derivative: $G_X''(t) = \frac{d}{dt}((0.8 + 0.2t)^4) = 4(0.8 + 0.2t)^3 \times 0.2 = 0.8(0.8 + 0.2t)^3$. Now, substitute $t=1$: $G_X''(1) = 0.8(0.8 + 0.2(1))^3 = 0.8(1)^3 = 0.8$. Now, use the variance formula: $Var(X) = G_X''(1) + G_X'(1) - [G_X'(1)]^2 = 0.8 + 1 - (1)^2 = 0.8$. So, the variance is 0.8. (Note: We could have recognised $G_X(t)$ as the PGF of $X \sim B(5, 0.2)$. Then $E(X)=np=5(0.2)=1$ and $Var(X)=npq=5(0.2)(0.8)=0.8$, which confirms our results.)

First, we recall the PGF for a Poisson distribution. If $W \sim Po(\lambda)$, its PGF is $G_W(t) = e^{\lambda(t-1)}$.

For $X \sim Po(2)$, the PGF is $G_X(t) = e^{2(t-1)}$. For $Y \sim Po(3)$, the PGF is $G_Y(t) = e^{3(t-1)}$.

Since $X$ and $Y$ are independent, the PGF of their sum $Z=X+Y$ is the product of their individual PGFs: $G_Z(t) = G_X(t) \times G_Y(t)$ $G_Z(t) = e^{2(t-1)} \times e^{3(t-1)}$ Using the laws of indices: $G_Z(t) = e^{2(t-1) + 3(t-1)}$ $G_Z(t) = e^{(2+3)(t-1)}$ $G_Z(t) = e^{5(t-1)}$

We recognise this resulting PGF, $e^{5(t-1)}$, as the PGF of a Poisson distribution with parameter $\lambda = 5$. Therefore, the distribution of $Z$ is Poisson with a mean of 5. $Z \sim Po(5)$.