In calculus one learns that $$ p + \frac{p^2} 2 + \frac{p^3} 3 + \frac{p^4} 4 + \cdots = -\log(1-p). \tag 1 $$ Thus a discrete probability distribution on the set $\{1,2,3,\ldots\}$ is given by $$ \Pr(X=x) = \frac{-p^x}{ x\log(1-p)} \text{ for } x = 1,2,3,\ldots\, \tag 2 $$ with $0<p<1.$

I seem to recall that this probability distribution was written about by Ronald Fisher in the '30s, and called the $\text{“}$logarithmic distribution$\text{''}$ presumably because of line $(1)$ above. It is known (but I don't know where it first appeared) that if $X_1,X_2,X_3,\ldots$ are i.i.d. with this distribution, and $N\sim\operatorname{Poisson},$ then $$ \sum_{n=1}^N X_n $$ has a negative binomial distribution (on the set $\{0,1,2,3,\ldots\}$).

It is readily seen that, with $X$ as in line $(2)$ above, $$ \operatorname E(X) = \frac{-p}{(1-p)\log(1-p)} =: \mu. $$ As $p$ increases from $0$ to $1,$ then $\mu$ increases from $1$ to $+\infty.$ So my question is:

How do we interpret this parameter $p\text?;$ i.e. is it perhaps the probability of some event related the random variable $X$ or the expected value of some function of $X$ or some sort of rate, or what?

$\big($And while we're at it, can anything further of interest be said about $\mu$ as a function of $p$ or about $p$ as a function of $\mu\text{?}\big)$


1 Answer 1


When you have a parameterised distribution, the best way to get an interpretation for a parameter is to try to derive an equation that equates that parameter value to some other properties of the distribution. In the case of the logarithmic distribution, the simplest way to get an equation for $p$ is to look at the ratio of successive values of the probability mass function. In particular, if $X \sim \text{logarithmic}(p)$ then we have:

$$\frac{p_X(x+1)}{p_X(x)} = \frac{p^{x+1}}{x+1} \bigg/ \frac{p^x}{x} = p \cdot \frac{x}{x+1}.$$

Re-arranging this gives us an explicit equation for the parameter:

$$p = \frac{x+1}{x} \cdot \frac{p_X(x+1)}{p_X(x)}.$$

This equation (or any other equation for $p$) can be used to give a valid interpretation of the meaning of the parameter. Putting the interpretation into words is a little strained, but basically you just want to describe what is being stated in the equation.

  • 1
    $\begingroup$ It's unclear what this interpretation would be or whether it would be insightful. I believe what the OP seeks is a reasonably simple stochastic process that yields, in a natural way, a random variable with the logarithmic distribution. The hope is that $p$ might be a readily interpretable property of that process. $\endgroup$
    – whuber
    Commented Sep 4, 2019 at 12:48
  • 1
    $\begingroup$ I am certainly open to alternative equations for $p$ that give a more insightful interpretation. The challenge in these problems is to find an equation that has a nice simple interpretation. $\endgroup$
    – Ben
    Commented Sep 4, 2019 at 13:02
  • $\begingroup$ @Ben : One could give a similar characterization of $\lambda$ in the usual parameterization of the family of Poisson distributions: $$ \left. \frac{\lambda^{x+1} e^{-\lambda}}{(x+1)!} \right/ \frac{\lambda^x e^{-\lambda}}{x!} = \frac \lambda {x+1}. $$ But that is not the best interpretation of the role of $\lambda,$ since one can also say that $\lambda$ is the expected value. And also the variance. $\qquad$ $\endgroup$ Commented Sep 4, 2019 at 20:40
  • $\begingroup$ @MichaelHardy: Indeed, it is not the best characterisation, since the equation $\lambda = \mathbb{E}(X)$ is much simpler in that case. However, if you have a look at the moments of the logarithmic distribution, you will see that they do not give rise a simple equation for $p$. Like I said, I am open to simpler equation giving rise to a better interpretation, if someone wants to offer one. It is no good criticising the interpretation I have given without any attempt to offer an improvement. With respect, I'm afraid this is a case where people need to "put up, or [that other thing]". $\endgroup$
    – Ben
    Commented Sep 4, 2019 at 21:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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