Is there any information out there about the distribution whose nth cumulant is given by 1n? The cumulant-generating function is of the form

κ(t)=∫10etx−1x 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.

**Answer**

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]=κ1=1,Var[Y]=κ2=12

while its skewness and excess kurtosis coefficients are

γ1=κ3(κ2)3/2=(1/3)(1/2)3/2=2√23

γ2=κ4(κ2)2=(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 {p0,p1,...,pm},∑mk=0pk=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.

**Attribution***Source : Link , Question Author : guy , Answer Author : Alecos Papadopoulos*