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I'm trying to derive the probability generating function (pgf) for the logarithmic series distribution, and not getting the expected form $\frac{\log{(1-qs)}}{\log{(1-q)}}$.

It seems that pgfs are usually written for discrete distributions on $k\geq 0$, whereas for this distribution the support is $k\geq1$ in the form usually written. So I parameterise the distribution as follows to get to the support $k\geq0$:

$$p(k) = \frac{-1}{\log{(1-q)}} \frac{q^{k+1}}{k+1}, k \geq0.$$

So then to find the pgf, I have

$$G_K(s) = \sum_{k=0}^\inf{p(k) s^k}$$

$$G_K(s) = \frac{-1}{\log{(1-q)}} \sum_{k=0}^\inf{\frac{q^{k+1}}{k+1} s^k}$$

which I can reparametrise "back" to get

$$G_K(s) = \frac{-1}{\log{(1-q)}} \sum_{k=1}^\inf{\frac{q^{k}}{k} s^{k-1}}.$$

By rearranging and using the fact that $\sum_{n=1}^{\inf}{x^n / n} = -\log{(1-x)}$, I end up with

$$G_K(s) = \frac{\log{(1-qs)}}{s \log{(1-q)}}$$

This doesn't match what I see elsewhere, and it's merely the presence of that $\frac{1}{s}$ factor that is the problem, which clearly comes directly from my reparametrisation. Have I derived the correct pgf, or have I misunderstood how to handle the support of a probability distribution when constructing the pgf? Where is my error?

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    $\begingroup$ i think your result is correct, it is expected that you'll get that form because instead of computing $\sum_{k=1}^{\infty}P_ks^k$, you are calculating $\sum_{k=0}P_{k+1}s^k=1/s\sum_{k=1}P_{k}s^k$. But there is no harm in it. $\endgroup$
    – Samrat Mukhopadhyay
    Jul 29, 2013 at 14:19
  • $\begingroup$ The tweak you get at the end is the tweak you put at the beginning, namely, unnecessarily shifting the index. $\endgroup$
    – Did
    Jul 29, 2013 at 14:33
  • $\begingroup$ @Did: so when writing a pgf one is not constrained in the support used? And the range of the sum, I guess, would match the support? $\endgroup$
    – Dan Stowell
    Jul 29, 2013 at 14:48
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    $\begingroup$ In other words, one can describe the distribution of any integer-valued random variable $X$ bounded below by its GF $g_X:s\mapsto E[s^X]$. $\endgroup$
    – Did
    Jul 29, 2013 at 14:52

1 Answer 1

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For the logarithmic series pgf sum from k=1 to infinity since p(0) would be undefined (division by zero (0)). Your result is still correct though because of the reparametrisation.

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