Stackexchange isn't getting really excited about this, so here it is.

The $n$th cumulant of the uniform probability distribution on the interval $[-1,0]$ is $B_n/n$, where $B_n$ is the $n$th Bernoulli number.

And $-\zeta(1-n)=B_n/n$, where $\zeta$ is Riemann's function.

Those two facts can be derived, but is there some argument that shows, without doing that, that you'd expect the cumulants of the uniform distribution to be those values of the $\zeta$ function?

Might this shed some light either on the $\zeta$ function or the uniform distribution? Or anything else?

Appendix: Cumulants are like moments, but better. The $n$th cumulant of a probability distribution is homogeneous of degree $n$ (like the $n$th moment). If $n=1$ it is shift-equivariant; if $n>1$ it is shift-invariant (like the $n$th central moment). If $X_1,X_2,X_3,\ldots$ are independent random variables, then the $n$th cumulant of the distribution of their sum is just the sum of the $n$th cumulants of their distributions. That last property is shared by central moments only in the cases $n=2,3$ (where the cumulant is just the central moment). Simplest nontrivial example: the 4th cumulant is the fourth central moment minus 3 times the square of the second central moment.

Later addition (should this be a separate question?): The number of independent Bernoulli trials strictly preceding the first success, with probability $1/(1+c)$ of success on each trial, is a geometrically distributed random variable with expected value $c$, taking values in the set $\lbrace 0,1,2,\ldots \rbrace$. The first several cumulants of that distribution are these $$ \begin{align} & c \\ & c+c^2 \\ & c + 3c^2 + 2c^3 \\ & c + 7c^2 + 12c^3 + 6c^4 \\ & c + 15c^2 + 50c^3 + 60c^4 + 24c^5 \\ & c + 32c^2 +180c^3+390c^4+360c^5+120c^6 \end{align} $$ The probability distribution of the number of successes in just one trial, with probability $c$ of success on each trial, has cumulants $$ \begin{align} & c \\ & c-c^2 \\ & c - 3c^2 + 2c^3 \\ & c - 7c^2 + 12c^3 - 6c^4 \\ & c - 15c^2 + 50c^3 - 60c^4 + 24c^5 \\ & c - 32c^2 +180c^3-390c^4+360c^5-120c^6 \end{align} $$ They're the same except for alternating signs.

If it were possible for the probability to be $-c$, where $c>0$, this sequence would be just $-1$ times the sequence of cumulants of the geometric distribution. Likewise, the sequence of values of $\zeta(1-n)$ is $-1$ times the sequence of cumulants of the probability distribution specified above. Could $\zeta(1-n),\quad n=1,2,3,\ldots$ be the sequence of cumulants of a distribution that would exist if we allowed negative probabilities (but still required the measure of the whole probability space to be $1$)?

If so, do we have two instances of a phenomenon that can be stated in a general way?


1 Answer 1


$X$ is the uniform distribution on $[-1,0]$, so the moment generating function of $X$ is $$E(e^{tX})=\lim_{k\to\infty} \frac{1}{k}\left(1+e^{-t/k}+\dots+e^{-(k-1)t/k}\right)=\lim_{k\to\infty} \frac{1}{k}\frac{1-e^{-t}}{1-e^{-t/k}}=\frac{1-e^{-t}}{t}$$ The cumulant generating function is the logarithm of this, call it $g(t)$. It'll be easier to compute the Taylor series of $tg'(t)+1$: $$tg'(t)+1=-\frac{t}{1-e^t}=-\sum_{k=0}^\infty te^{kt}=-t\sum_{k=0}^\infty \sum_{n=0}^\infty \frac{k^nt^{n}}{n!}$$ Rearrange the order of summation: $$tg'(t)+1=-t\sum_{n=0}^\infty \zeta(-n)\frac{t^n}{n!}$$ Now compare coefficients on either side: the $n$th cumulant of $X$ is plainly $-\zeta(1-n)$.

What? Why are you looking at me like that?

  • 2
    $\begingroup$ But does this really answer the OP's question? $\endgroup$
    – Suvrit
    Sep 18, 2011 at 9:43
  • $\begingroup$ Can anyone tell me the meaning or apropos of the last line (two questions)? I need to develop my sense of humor. $\endgroup$
    – GH from MO
    Sep 19, 2011 at 0:09
  • 3
    $\begingroup$ I had in my mind a scene where I am presenting this bogus argument on a blackboard to a skeptical, wincing audience. $\endgroup$ Sep 19, 2011 at 1:14

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