Let X be a discrete random variable with sample space ΩX, containing only non-negative numbers. We define the PGF of X, often written GX, to be equal to the power series:
GX(t)=∑x∈ΩXP(X=x)tx
for t≥0. Where x and t are both zero, we take tx=1 as is the usual convention with power series. It turns out that this function completely describes the distribution of X. From this we can extract the probability that an observation of X will take a particular value, as well as the mean and variance of X. In this section we will just look at how to calculate these, we'll leave PGFs of common distributions until later.
We know that: ∑x∈ΩXP(X=x)=P(X∈ΩX)=1 because the value of X will always fall within ΩX. This means that we must have GX(1)=1. It might be the case that you're given GX(t)=kf(t) for some normalising constant k, and this property means you can substitute t=1 to quickly get k out.
With that noted, we can start deducing stuff about X. By comparing the coefficient of tx with Taylor's theorem, we have: GX(x)(0)x!=P(X=x) By a similar process we can go on to compute E[X] in terms of GX(t) for some specific t. Remember that: E[X]=∑x∈ΩXxP(X=x)] Using a standard trick, we can differentiate GX with respect to t, [NB 1] to find: G′X(t)=∑x∈ΩXxP(X=x)tx−1 Substituting t=1 then gives us: G′X(1)=∑x∈ΩXxP(X=x)=E[X] Now, if we are to calculate the variance, we also need E[X2].
Differentiating G′X would give: G″X(t)=∑x∈ΩXx(x−1)P(X=x)tx−2 and so: G″X(1)=∑x∈ΩXx(x−1)P(X=x) Remember that: E[g(X)]=∑x∈ΩXg(x)P(X=x) So, we can split this sum up: G″X(1)=∑x∈ΩXx2P(X=x)−∑x∈ΩXxP(X=x)=E[X2]−E[X] and so: E[X2]=G″X(1)+E[X]=G″X(1)+G′X(1) We know that: var(X)=E[X2]−(E[X])2=G″X(1)+G′X(1)−(G′X(1))2 So we have a way to find the mean and variance of a random variable, just by constructing the PGF and differentiating it. Let's move on to a common example.
S5 2005 Question 3
Let: GX(t)=k[t3(2+3t)+(t+1)4]
For the second, we use: E[X]=G′X(1) Write the PGF as: GX(t)=121[2t3+3t4+(t+1)4] Differentiating: G′X(t)=121[6t2+12t3+4(t+1)3] And substituting t=1 gives: E[X]=121[6+12+4⋅8]=5021
All we need to now find the variance is to evaluate the second derivative of the pgf at 1: G″X(t)=121[12t+36t2+12(t+1)2] Substituting t=1 gives: G″X(1)=121[12+36+48]=9621 And so: var(X)=5021+9621−(9621)2=566441 There are two ways to approach the last question. First is to look at the third derivative, but in this case it's unnecessary. The more straightforward approach is to just look at the t3 term. The coefficient of the t3 term of (1+t)4 is (41)=(43)=4 (depending on which way you expand it), so the coefficient of the t3 of the PGF is: 121(2+4)=621 and by the definition of a pgf, we know that this is P(X=3).
[NB 1: For finite ΩX, termwise differentiation is obviously fine. For infinite ΩX, it is not trivial, but we are allowed to differentiate convergent sums of polynomials.]
We know that: ∑x∈ΩXP(X=x)=P(X∈ΩX)=1 because the value of X will always fall within ΩX. This means that we must have GX(1)=1. It might be the case that you're given GX(t)=kf(t) for some normalising constant k, and this property means you can substitute t=1 to quickly get k out.
With that noted, we can start deducing stuff about X. By comparing the coefficient of tx with Taylor's theorem, we have: GX(x)(0)x!=P(X=x) By a similar process we can go on to compute E[X] in terms of GX(t) for some specific t. Remember that: E[X]=∑x∈ΩXxP(X=x)] Using a standard trick, we can differentiate GX with respect to t, [NB 1] to find: G′X(t)=∑x∈ΩXxP(X=x)tx−1 Substituting t=1 then gives us: G′X(1)=∑x∈ΩXxP(X=x)=E[X] Now, if we are to calculate the variance, we also need E[X2].
Differentiating G′X would give: G″X(t)=∑x∈ΩXx(x−1)P(X=x)tx−2 and so: G″X(1)=∑x∈ΩXx(x−1)P(X=x) Remember that: E[g(X)]=∑x∈ΩXg(x)P(X=x) So, we can split this sum up: G″X(1)=∑x∈ΩXx2P(X=x)−∑x∈ΩXxP(X=x)=E[X2]−E[X] and so: E[X2]=G″X(1)+E[X]=G″X(1)+G′X(1) We know that: var(X)=E[X2]−(E[X])2=G″X(1)+G′X(1)−(G′X(1))2 So we have a way to find the mean and variance of a random variable, just by constructing the PGF and differentiating it. Let's move on to a common example.
S5 2005 Question 3
Let: GX(t)=k[t3(2+3t)+(t+1)4]
- (a) Show that k=121
- (b) Find E[X]
- (c) Find var(X)
- (d) Find P(X=3)
For the second, we use: E[X]=G′X(1) Write the PGF as: GX(t)=121[2t3+3t4+(t+1)4] Differentiating: G′X(t)=121[6t2+12t3+4(t+1)3] And substituting t=1 gives: E[X]=121[6+12+4⋅8]=5021
All we need to now find the variance is to evaluate the second derivative of the pgf at 1: G″X(t)=121[12t+36t2+12(t+1)2] Substituting t=1 gives: G″X(1)=121[12+36+48]=9621 And so: var(X)=5021+9621−(9621)2=566441 There are two ways to approach the last question. First is to look at the third derivative, but in this case it's unnecessary. The more straightforward approach is to just look at the t3 term. The coefficient of the t3 term of (1+t)4 is (41)=(43)=4 (depending on which way you expand it), so the coefficient of the t3 of the PGF is: 121(2+4)=621 and by the definition of a pgf, we know that this is P(X=3).
[NB 1: For finite ΩX, termwise differentiation is obviously fine. For infinite ΩX, it is not trivial, but we are allowed to differentiate convergent sums of polynomials.]
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