cumulants의 값을 알면이 확률 분포의 그래프가 어떻게 보이는지 알 수 있습니다. 분포의 평균과 분산은

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

왜도 및 초과 첨도 계수는

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

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

{0,1,...,m}

$\{0,1,...,m\}$

$\{0,1,...,m\}${p0,p1,...,pm},∑mk=0pk=1

$\{p_0,p_1,...,p_m\},\sum _{k=0}^m{p}_k=1$

$\{p_0,p_1,...,p_m\},\; \sum_{k=0}^mp_k =1$, 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

$m=5$

$m=5$ 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

$m$

$m$ to be equal to 5

$5$

$5$. But increasing gradually m

$m$

$m$ (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.

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