There are some PGFs of common distributions you should be aware of. These are given in the formula booklet, but you are required to know their proof, and how to derive the mean and variance of their respective distribution. I'll show the four you need to know, and how to derive them, but I'll leave the derivation of the mean and variance to you, since it's just an exercise of differentiation like we did in the last section.
Poisson
You should be aware that: \[e^\lambda = \sum_{x = 0}^\infty \frac {\lambda^x} {x!}\] and that for \(X \sim \textrm{Po} \left({\lambda}\right)\), we have: \[\mathbb P \left({X = x}\right) = \frac {e^{-\lambda} \lambda^x} {x!}\] Remember that the PGF, \(G_X(t)\) is given by: \[G_X(t) = \sum_{x = 0}^\infty t^x \mathbb P \left({X = x}\right) = \sum_{x = 0}^\infty t^x \frac {e^{-\lambda} \lambda^x} {x!}\] This is not too hard to evaluate, taking \(e^{-\lambda}\) out of the sum, and writing \(\lambda^x t^x = (\lambda t)^x\), we have: \[G_X(t) = e^{-\lambda} \sum_{x = 0}^\infty \frac {(\lambda t)^x} {x!} = e^{-\lambda} e^{\lambda t} = e^{\lambda (t - 1)}\] Binomial
We know the binomial theorem: \[(p + q)^n = \sum_{k = 0}^n \binom n k p^k q^{n - k}\] and that for \(X \sim \textrm{B} \left({n, p}\right)\): \[\mathbb P \left({X = k}\right) = \binom n k p^k (1 - p)^{n - k}\] The PGF is then given by: \[G_X(t) = \sum_{k = 0}^n t^k \binom n k p^k (1 - p)^{n - k}\] This is, like the Poisson distribution, quite easy to deal with. We can just write \(t^k p^k = (tp)^k\), then we have: \[G_X(t) = \sum_{k = 0}^\infty \binom n k (tp)^k (1 - p)^{n - k} = (1 - p + tp)^n\] Geometric
We know that: \[\frac 1 {1 - t} = \sum_{x = 0}^\infty t^x\] for \(|t| < 1\), and that for \(X \sim \textrm{Geo} \left({p}\right)\): \[\mathbb P \left({X = x}\right) = p(1-p)^{x - 1}\] So the PGF is given by: \[G_X(t) = \sum_{x = 1}^\infty t^x p(1 - p)^{x - 1}\] Taking out a factor of \(tp\), and starting the sum from \(0\), we have: \[G_X(t) = tp \sum_{x = 0}^\infty (t - tp)^x\] Which is, by our previous fact, equal to: \[G_X(t) = \frac {tp} {1 - (t - tp)} = \frac {tp} {1 - (1 - p)t}\] Negative binomial
You'll probably be the least familiar with the negative binomial distribution. We start with the result that: \[\sum_{x = r}^\infty \binom {x - 1} {r - 1} q^{x - r} = \frac 1 {(1 - q)^r}\] which is just the result we used for the geometric distribution differentiated \(r - 1\) times. We know that for \(X \sim \textrm{NB} \left({r, p}\right)\), we have: \[\mathbb P \left({X = x}\right) = \binom {x - 1} {r - 1} (1-p)^{x - r} p^r\] So the pgf is given by: \[G_X(t) = \sum_{x = r}^\infty \binom {x - 1} {r - 1} (1-p)^{x - r} p^r t^x\] The \(p^r\) term is independent of the summation index, \(x\), so can be extracted as a constant. In a similar way we did before, we'll also write \(t^x = t^{x - r} t^r\). \(t^r\) is also independent of the summation index so that too can be taken out as constant. We can then write \((1-p)^{x - r} t^{x - r} = (t - tp)^{x - r}\), leaving us with: \[G_X(t) = (pt)^r \sum_{x = r}^\infty \binom {x - 1} {r - 1} (t-tp)^{x - r} = \frac {(pt)^r} {(1 - (1 - p)t)^r} = \left({\frac {pt} {1 - (1 - p)t}}\right)^r\]
Poisson
You should be aware that: \[e^\lambda = \sum_{x = 0}^\infty \frac {\lambda^x} {x!}\] and that for \(X \sim \textrm{Po} \left({\lambda}\right)\), we have: \[\mathbb P \left({X = x}\right) = \frac {e^{-\lambda} \lambda^x} {x!}\] Remember that the PGF, \(G_X(t)\) is given by: \[G_X(t) = \sum_{x = 0}^\infty t^x \mathbb P \left({X = x}\right) = \sum_{x = 0}^\infty t^x \frac {e^{-\lambda} \lambda^x} {x!}\] This is not too hard to evaluate, taking \(e^{-\lambda}\) out of the sum, and writing \(\lambda^x t^x = (\lambda t)^x\), we have: \[G_X(t) = e^{-\lambda} \sum_{x = 0}^\infty \frac {(\lambda t)^x} {x!} = e^{-\lambda} e^{\lambda t} = e^{\lambda (t - 1)}\] Binomial
We know the binomial theorem: \[(p + q)^n = \sum_{k = 0}^n \binom n k p^k q^{n - k}\] and that for \(X \sim \textrm{B} \left({n, p}\right)\): \[\mathbb P \left({X = k}\right) = \binom n k p^k (1 - p)^{n - k}\] The PGF is then given by: \[G_X(t) = \sum_{k = 0}^n t^k \binom n k p^k (1 - p)^{n - k}\] This is, like the Poisson distribution, quite easy to deal with. We can just write \(t^k p^k = (tp)^k\), then we have: \[G_X(t) = \sum_{k = 0}^\infty \binom n k (tp)^k (1 - p)^{n - k} = (1 - p + tp)^n\] Geometric
We know that: \[\frac 1 {1 - t} = \sum_{x = 0}^\infty t^x\] for \(|t| < 1\), and that for \(X \sim \textrm{Geo} \left({p}\right)\): \[\mathbb P \left({X = x}\right) = p(1-p)^{x - 1}\] So the PGF is given by: \[G_X(t) = \sum_{x = 1}^\infty t^x p(1 - p)^{x - 1}\] Taking out a factor of \(tp\), and starting the sum from \(0\), we have: \[G_X(t) = tp \sum_{x = 0}^\infty (t - tp)^x\] Which is, by our previous fact, equal to: \[G_X(t) = \frac {tp} {1 - (t - tp)} = \frac {tp} {1 - (1 - p)t}\] Negative binomial
You'll probably be the least familiar with the negative binomial distribution. We start with the result that: \[\sum_{x = r}^\infty \binom {x - 1} {r - 1} q^{x - r} = \frac 1 {(1 - q)^r}\] which is just the result we used for the geometric distribution differentiated \(r - 1\) times. We know that for \(X \sim \textrm{NB} \left({r, p}\right)\), we have: \[\mathbb P \left({X = x}\right) = \binom {x - 1} {r - 1} (1-p)^{x - r} p^r\] So the pgf is given by: \[G_X(t) = \sum_{x = r}^\infty \binom {x - 1} {r - 1} (1-p)^{x - r} p^r t^x\] The \(p^r\) term is independent of the summation index, \(x\), so can be extracted as a constant. In a similar way we did before, we'll also write \(t^x = t^{x - r} t^r\). \(t^r\) is also independent of the summation index so that too can be taken out as constant. We can then write \((1-p)^{x - r} t^{x - r} = (t - tp)^{x - r}\), leaving us with: \[G_X(t) = (pt)^r \sum_{x = r}^\infty \binom {x - 1} {r - 1} (t-tp)^{x - r} = \frac {(pt)^r} {(1 - (1 - p)t)^r} = \left({\frac {pt} {1 - (1 - p)t}}\right)^r\]
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