2
$\begingroup$

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?

$\endgroup$
4
  • 2
    $\begingroup$ 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. $\endgroup$ 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? $\endgroup$ Jul 29, 2013 at 14:48
  • 1
    $\begingroup$ 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

0
$\begingroup$

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.

$\endgroup$
1

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.