Questions on probability generating functions

Question 1

Define the probability generating function of a discrete random variable X, and state the value of G(1).

Accepted Answer

G(t) = E(t^X) = Σ t^xP(X = x). Setting t = 1 makes every power 1, so G(1) is the total probability, which is always 1. B1 for the definition, B1 for G(1) = 1. It is the quickest check on any generating function you write down.

Question 2

X has G(t) = (0.6 + 0.4t)⁸. Name the distribution and find E(X).

Accepted Answer

This is the binomial form (q + pt)ⁿ, so X ~ B(8, 0.4). Differentiating, G'(t) = 8(0.4)(0.6 + 0.4t)⁷, and at t = 1 the bracket is 1, so E(X) = G'(1) = 3.2. B1 for naming the binomial, M1 for differentiating and setting t = 1, A1 for 3.2. That agrees with np = 8 × 0.4.

Question 3

Write down the probability generating functions of X ~ Po(3.5) and of Y ~ B(5, 0.2).

Accepted Answer

G_X(t) = e^3.5(t − 1) and G_Y(t) = (0.8 + 0.2t)⁵. B1 B1. Both give 1 at t = 1: e⁰ and (0.8 + 0.2)⁵. In the binomial form the constant term is q = 1 − p, and the coefficient of t is p.

Question 4

X has probability generating function G(t) = e^2(t − 1). Find P(X = 0) and E(X).

Accepted Answer

P(X = 0) is the constant term of the series, which is G(0) = e⁻² = 0.1353. G'(t) = 2e^2(t − 1), so E(X) = G'(1) = 2. M1 for G(0) as the constant term, A1 for 0.1353, B1 for 2. The generating function is that of Po(2), and e⁻² is the Poisson probability of zero.

Question 5

X has G(t) = (0.6 + 0.4t)⁸, with E(X) = 3.2. Find Var(X) using the generating function.

Accepted Answer

G''(t) = 8(7)(0.4)²(0.6 + 0.4t)⁶, so G''(1) = 56 × 0.16 = 8.96. Then Var(X) = G''(1) + G'(1) − [G'(1)]² = 8.96 + 3.2 − 10.24 = 1.92, which matches npq = 8(0.4)(0.6). M1 for the second derivative, A1 for G''(1) = 8.96, M1 for the variance formula, A1 for 1.92.

Question 6

Derive the probability generating function of X ~ Po(λ), and use it to confirm E(X) = λ.

Accepted Answer

G(t) = Σ t^xe^−λλ^x/x! = e^−λΣ(λt)^x/x! = e^−λe^λt = e^λ(t − 1). Differentiating: G'(t) = λe^λ(t − 1), so G'(1) = λ. M1 for the defining sum, A1 for pulling out e^−λ, A1 for e^λ(t − 1), M1 for differentiating, A1 for G'(1) = λ. Checking G(1) = e⁰ = 1 confirms the derivation.

Question 7

X and Y are independent with X ~ Po(2) and Y ~ Po(7). Use generating functions to identify the distribution of X + Y.

Accepted Answer

G_X+Y(t) = e^2(t − 1) × e^7(t − 1) = e^9(t − 1), which is the generating function of Po(9). Since generating functions determine distributions uniquely, X + Y ~ Po(9). M1 for multiplying the two generating functions, A1 for e^9(t − 1), A1 for Po(9).

Question 8

The discrete random variable X has probability generating function G(t) = k(2 + t)², where k is a constant. Find k, write down the probability distribution of X in a table, and find E(X).

Accepted Answer

G(1) = 1 forces 9k = 1, so k = 1/9. Expanding, G(t) = (4 + 4t + t²)/9, and the coefficient of t^x is P(X = x), so P(X = 0) = 4/9, P(X = 1) = 4/9 and P(X = 2) = 1/9, which total 1. Differentiating, G'(t) = 2(2 + t)/9, so E(X) = G'(1) = 6/9 = 2/3. M1 for applying G(1) = 1, A1 for k = 1/9, B1 for the three probabilities read off the coefficients, M1 for differentiating and setting t = 1, A1 for 2/3.

Question 9

Derive the probability generating function of X ~ B(n, p) from the definition G(t) = E(t^X), and use it to confirm E(X) = np.

Accepted Answer

G(t) = Σ t^x ⁿC_x p^xq^n−x = Σ ⁿC_x(pt)^xq^n−x, which the binomial theorem collapses to (q + pt)ⁿ. Differentiating, G'(t) = np(q + pt)ⁿ⁻¹, and at t = 1 the bracket is q + p = 1, so E(X) = G'(1) = np. M1 for the defining sum with t^x absorbed into (pt)^x, A1 for (q + pt)ⁿ, M1 for differentiating, A1 for np. Checking G(1) = (q + p)ⁿ = 1 confirms the form before any differentiation is attempted.

Question 10

X ~ B(3, 0.5) and Y = 2X + 1. Find the probability generating function of Y, and use it to find P(Y = 3).

Accepted Answer

G_Y(t) = E(t^2X+1) = t × E((t²)^X) = t G_X(t²) = t(0.5 + 0.5t²)³. Expanding the cube, (0.5 + 0.5t²)³ = 0.125 + 0.375t² + 0.375t⁴ + 0.125t⁶, so G_Y(t) = 0.125t + 0.375t³ + 0.375t⁵ + 0.125t⁷ and P(Y = 3), the coefficient of t³, is 0.375. M1 for writing t^2X+1 as t times a power of t², A1 for t(0.5 + 0.5t²)³, M1 for expanding to the t³ term, A1 for 0.375. Check: Y = 3 exactly when X = 1, and P(X = 1) = 3 × 0.5³ = 0.375. Only odd powers of t appear, because Y takes only odd values.

Question 11

X ~ B(4, 0.3) and Y ~ B(6, 0.3) are independent. Use probability generating functions to identify the distribution of X + Y, and find P(X + Y = 1). Explain why the same method would not give a binomial distribution if Y were B(6, 0.5).

Accepted Answer

G_X+Y(t) = (0.7 + 0.3t)⁴ × (0.7 + 0.3t)⁶ = (0.7 + 0.3t)¹⁰, which is the generating function of B(10, 0.3); since a generating function determines the distribution, X + Y ~ B(10, 0.3). Then P(X + Y = 1) = 10 × 0.3 × 0.7⁹ = 10 × 0.3 × 0.04035 = 0.1211.
With Y ~ B(6, 0.5) the product would be (0.7 + 0.3t)⁴(0.5 + 0.5t)⁶, and two brackets with different p do not combine into a single power of one bracket, so the sum is not binomial: the trials do not share a common success probability. M1 for multiplying the generating functions, A1 for (0.7 + 0.3t)¹⁰, A1 for B(10, 0.3), A1 for 0.1211, B1 for the explanation about unequal p.

Question 12

A geometric distribution has G(t) = pt/(1 − qt) with q = 1 − p. Show that its variance is q/p². The number of trials up to and including the third success is the sum of three independent such variables. Write down its probability generating function and deduce its mean and variance.

Accepted Answer

By the quotient rule G'(t) = p/(1 − qt)², so G'(1) = p/p² = 1/p. Differentiating again, G''(t) = 2pq/(1 − qt)³, so G''(1) = 2pq/p³ = 2q/p².
Var = G''(1) + G'(1) − [G'(1)]² = 2q/p² + 1/p − 1/p² = (2q + p − 1)/p². Since p − 1 = −q the numerator is 2q − q = q, giving q/p² as required.
For independent variables the generating functions multiply, so the sum of three has G(t) = [pt/(1 − qt)]³. Differentiating a product of three identical factors gives three times the single mean, so the mean is 3/p, and independence lets the variances add, giving 3q/p².
M1 for G'(t) by the quotient rule, A1 for 1/p, M1 for G''(t), A1 for 2q/p², M1 for the variance formula, A1 for q/p², B1 for the cubed generating function, B1 for the mean and variance of the sum. With p = 0.25 the single distribution has mean 4 and variance 12, and the third success is expected on trial 12 with variance 36. The multiplication rule holds only because the trials are independent.

Question 13

X ~ B(2, 0.3) and Y ~ Po(2) are independent. Write down the probability generating function of X + Y, and use it to find P(X + Y = 1) and E(X + Y).

Accepted Answer

Independence multiplies the generating functions: G(t) = (0.7 + 0.3t)² e^2(t − 1), and G(1) = 1² × e⁰ = 1 checks it. P(X + Y = 1) is the coefficient of t, so expand each factor far enough: (0.49 + 0.42t + 0.09t²) × e⁻²(1 + 2t + 2t² + …). The t terms come from 0.49 × 2t and 0.42t × 1, so P(X + Y = 1) = (0.98 + 0.42)e⁻² = 1.4e⁻² = 0.189 to three significant figures. For the mean, differentiate the product or add the means: E(X + Y) = G'(1) = 0.6 + 2 = 2.6, and independence lets the variances add too, giving 0.42 + 2 = 2.42. B1 for the product generating function, M1 for expanding e^2(t − 1) as a series, M1 for collecting the coefficient of t, A1 for 1.4e⁻², A1 for 0.189, M1 for the mean from G'(1) or from adding means, A1 for 2.6. The coefficient extraction is the convolution P(X = 0)P(Y = 1) + P(X = 1)P(Y = 0) done by algebra instead of by cases.

Question 14

X is the number of the trial on which the first success occurs, with P(X = x) = pq^x−1 for x = 1, 2, 3, … and q = 1 − p. Derive from the definition G(t) = E(t^X) that the probability generating function is pt/(1 − qt), stating the values of t for which the derivation is valid. Y = X − 1 is the number of failures before the first success. Find the probability generating function of Y and hence E(Y).

Accepted Answer

G(t) = Σ t^x pq^x−1 summed over x = 1, 2, 3, … = pt + pqt² + pq²t³ + …, a geometric series with first term pt and common ratio qt. It converges when |qt| < 1, that is for |t| < 1/q, which includes t = 1 since q < 1. Its sum is pt/(1 − qt).
Y = X − 1, so G_Y(t) = E(t^X−1) = G_X(t)/t = p/(1 − qt). Check: at t = 1 this is p/(1 − q) = 1.
Differentiating, G_Y'(t) = pq/(1 − qt)², so E(Y) = pq/p² = q/p, which is 1/p − 1, one less than E(X), as it must be. M1 for the defining sum with the terms written out, M1 for recognising a geometric series with ratio qt, A1 for pt/(1 − qt), B1 for the condition |t| < 1/q, M1 for dividing by t to get G_Y, A1 for q/p. Dividing a generating function by t shifts every value down by one, which is why the first-term factor t disappears.

Question 15

X has probability generating function G(t). Show that P(X is even) = [1 + G(−1)]/2. Use this to find P(X is even) when X ~ Po(λ), and evaluate it for λ = 2.

Accepted Answer

G(−1) = Σ (−1)^x P(X = x) = P(X even) − P(X odd), because the sign is plus for even x and minus for odd x. Also G(1) = P(X even) + P(X odd) = 1. Adding the two equations, 1 + G(−1) = 2 P(X even), so P(X even) = [1 + G(−1)]/2.
For Po(λ), G(t) = e^{λ(t − 1)}, so G(−1) = e^−2λ and P(X even) = (1 + e^−2λ)/2. With λ = 2 this is (1 + e⁻⁴)/2 = (1 + 0.0183)/2 = 0.5092. M1 for G(−1) as a signed sum, A1 for the identity, M1 for substituting t = −1 into e^{λ(t − 1)}, A1 for (1 + e^−2λ)/2, A1 for 0.5092. Even values are favoured for every λ, and the excess e^−2λ/2 shrinks as λ grows; adding the even Poisson terms directly would take many lines.