Is there any information out there about the distribution whose $n$th cumulant is given by $\frac 1 n$? The cumulant-generating function is of the form $$ \kappa(t) = \int_0 ^ 1 \frac{e^{tx} - 1}{x} \ dx. $$ I've run across it as the limiting distribution of some random variables but I haven't been able to find any information on it.

  • $\begingroup$ I cannot see that this function $\kappa(t)$ you have given has the claimed property! You should revise yoiur work. Approximating the exponential n the integrand close to zero with $1+tx$, the integrand close to zero becomes $t/x$, so is divergent. So that integral cannot represent a cumulant generating function. $\endgroup$ – kjetil b halvorsen Jul 11 '13 at 15:32
  • $\begingroup$ @kjetilbhalvorsen not sure I follow. Approximating $e^{tx}$ with $1 + tx$ gives $\frac{tx}{x} = t$ for the integrand. Also, according to this the function I gave has a known integral in terms of hyperbolic cosine and sine integrals. To show that $\kappa(t)$ has the claimed property just do a full Taylor series around $0$ for $e^{tx}$ and push the integral through to sum to get the Taylor series for $\kappa(t)$ around $0$. $\endgroup$ – guy Jul 11 '13 at 15:54
  • $\begingroup$ sympy says the integral is divergent (in its own eccentric way!). But sympy must be wrong, I see it now, experimented with some numerical integration, and it works just well. Will try again. $\endgroup$ – kjetil b halvorsen Jul 11 '13 at 16:36
  • $\begingroup$ Looking at Wolphram alphas result, it cannot be correct either, it haves a non-zero limit when t approaches zero, while $\kappa(0)=0$ clearly. $\endgroup$ – kjetil b halvorsen Jul 11 '13 at 19:10
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    $\begingroup$ I believe it is absolutely continuous on $(0, \infty)$. It is realized as a limit of compoud Poisson random variables; as $n \to \infty$ a compound Poisson with rate $\int_{1/n}^1 \frac 1 x \ dx$ and jumping distribution density $f_n(x) \propto \frac 1 x I(1/n < x < 1)$ converges weakly to this distribution. $\endgroup$ – guy Jul 11 '13 at 22:21

Knowing the values of the cumulants permits us to get an idea of how the graph of this probability distribution will look like. The mean and variance of the distribution is

$$E[Y] = \kappa_1 =1, \;\; \text{Var}[Y] = \kappa_2 = \frac 12$$

while its skewness and excess kurtosis coefficients are

$$\gamma_1 = \frac{\kappa_3}{(\kappa_2)^{3/2}} = \frac{(1/3)}{(1/2)^{3/2}} = \frac{2\sqrt 2}{3}$$

$$\gamma_2 = \frac{\kappa_4}{(\kappa_2)^{2}} = \frac{(1/4)}{(1/2)^{2}} = 1$$

So this could be a familiar looking graph of a positive random variable exhibiting positive skewness. As for finding the probability distribution, a craftsman's approach could be to specify a generic discrete probability distribution, taking values in $\{0,1,...,m\}$, with corresponding probabilities $\{p_0,p_1,...,p_m\},\; \sum_{k=0}^mp_k =1$, and then use the cumulants to calculate the raw moments, with the purpose of forming a system of linear equations with the probabilities being the unknowns. Cumulants are related to raw moments by $$\kappa_n=\mu'_n-\sum_{i=1}^{n-1}{n-1 \choose i-1}\kappa_i \mu_{n-i}'$$ Solved for the first five raw moments this gives (the numerical value at the end is specific to the cumulants in our case) $$\begin{align} \mu'_1=&\kappa_1 =1\\ \mu'_2=&\kappa_2+\kappa_1^2=3/2\\ \mu'_3=&\kappa_3+3\kappa_2\kappa_1+\kappa_1^3=17/6\\ \mu'_4=&\kappa_4+4\kappa_3\kappa_1+3\kappa_2^2+6\kappa_2\kappa_1^2+\kappa_1^4=19/3\\ \mu'_5=&\kappa_5+5\kappa_4\kappa_1+10\kappa_3\kappa_2+10\kappa_3\kappa_1^2+15\kappa_2^2\kappa_1+10\kappa_2\kappa_1^3+\kappa_1^5=243/15\\ \end{align} $$ If we (momentarily) set $m=5$ we have the system of equations

$$\begin{align} \sum_{k=0}^5p_k=&1,\qquad \sum_{k=0}^5p_kk=1\\ \sum_{k=0}^5p_kk^2=&3/2,\qquad \sum_{k=0}^5p_kk^3=17/6\\ \sum_{k=0}^5p_kk^4=& 19/3 ,\qquad \sum_{k=0}^5p_kk^5= 243/15\\ &s.t. p_k\ge 0 \;\;\forall k\\ \end{align} $$

Of course we do not want $m$ to be equal to $5$. But increasing gradually $m$ (and obtaining the value of the subsequent moments), we should eventually reach a point where the solution for the probabilities stabilizes. Such an approach cannot be done by hand -but I have neither the software access, nor the programming skills necessary to perform such a task.

  • $\begingroup$ This is cool. Maybe I could do some kind of Edgeworth expansion as well? Actually, I have an idea of what the density looks like already (assuming it exists) since I can simulate directly from it. It is very strange - it looks uniform over some range $(0, a)$ and then on $(a, \infty)$ it decays with something like an exponential tail (it's been a long time since I did the simulation). $\endgroup$ – guy Feb 25 '14 at 22:48
  • $\begingroup$ Thanks. Of course you can always perform an Edgworth expansion based on the cumulants, but I wonder how well it will perform, given the strange shape you describe. It would be interesting to contrast the two.Can you tell me the value for $a$? $\endgroup$ – Alecos Papadopoulos Feb 25 '14 at 22:58
  • $\begingroup$ Dug up my old code and found $a \approx 1$. If $Y \sim \kappa(t)$ then $[Y \mid Y < 1]$ is approximatey $\mathcal U(0, 1)$ and $[Y - 1\mid Y > 1]$ is approximately gamma distributed with shape $1.4$ and mean $0.64$. $\endgroup$ – guy Feb 26 '14 at 0:04
  • $\begingroup$ What do you mean by $Y \sim \kappa(t)$? $\endgroup$ – Alecos Papadopoulos Feb 28 '14 at 14:33
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    $\begingroup$ So what does the pdf look like then? As for fitting by moments, is the fit 'robust' and 'stable' as one increases the number of moments used (4, 5, 6, 7 or 8 etc), or is it all over the place? $\endgroup$ – wolfies Feb 28 '14 at 15:31

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