累积量由

是否存在有关给出第个累积量的分布的任何信息？累积量生成函数的形式为我已经将其作为某些随机变量的极限分布来进行研究，但是我无法找到有关它的任何信息。 $n$ $\frac 1 n$

κ (t) = \int_{0}^{1} \frac{e^{t x} - 1}{x} d x .

$\kappa(t) = \int_0 ^ 1 \frac{e^{tx} - 1}{x} \ dx.$

distributions mgf cumulants

— 家伙
source

我看不到您给定的函数

κ (t)

$\kappa(t)$ 具有要求的属性！您应该修改您的工作。用

逼近接近零的整数n的指数，接近零的整数n

1 + t x

$1+tx$ 变为

t / x

$t/x$ ，因此是发散的。因此该积分不能代表累积量生成函数。

— kjetil b halvorsen 2013年

@kjetilbhalvorsen不确定我是否遵循。近似

e^{t x}

$e^{tx}$ 与

1 + t x

$1 + tx$ 给出

\frac{t x}{x} = t

$\frac{tx}{x} = t$ 被积分数。另外，根据该我给函数在双曲正弦和余弦积分并已知积分。为了证明

κ (t)

$\kappa(t)$ 具有所要求的性质，只需对

做一个大约为

的完整泰勒级数，并通过积分求和，得出

在

附近的泰勒级数。

0

$0$

e^{t x}

$e^{tx}$

κ (t)

$\kappa(t)$

0

$0$

— 家伙

sympy说积分是发散的（以它自己的偏心方式！）。但是sympy一定是错误的，我现在看到了，尝试了一些数值积分，并且效果很好。会再试一次。

— kjetil b halvorsen 2013年

查看Wolphram alphas结果，它也不是正确的，当t接近零时它具有非零极限，而

显然。

κ (0) = 0

$\kappa(0)=0$

— kjetil b halvorsen 2013年

我相信

上是绝对连续的。它被实现为复合泊松随机变量的极限；如

一个与速率化合物泊松

(0, \infty)

$(0, \infty)$

n \to \infty

$n \to \infty$

和跳跃分布密度

\int_{1 / n}^{1} \frac{1}{x} d x

$\int_{1/n}^1 \frac 1 x \ dx$

收敛到该分布。

f_{n} (x) \propto \frac{1}{x} I (1 / n < x < 1)

$f_n(x) \propto \frac 1 x I(1/n < x < 1)$

— 人

知道累积量的值可以使我们了解这种概率分布的图形。分布的均值和方差为

E [Y] = κ_{1} = 1, Var [Y] = κ_{2} = \frac{1}{2}

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

而其偏度和过量峰度系数为

γ_{1} = \frac{κ_{3}}{(κ_{2})^{3 / 2}} = \frac{(1 / 3)}{(1 / 2)^{3 / 2}} = \frac{2 \sqrt{2}}{3}

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

γ_{2} = \frac{κ_{4}}{(κ_{2})^{2}} = \frac{(1 / 4)}{(1 / 2)^{2}} = 1

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

因此，这可能是显示正偏度的正随机变量的熟悉外观图。至于发现的概率分布，一个工匠的方法可以是指定通用离散概率分布，在取值，具有相应的概率 $\{0,1,...,m\}$ ，然后使用累积量来计算原始矩，目的是形成一个线性方程组，其概率为未知数。累积量通过与原始矩 $\{p_0,p_1,...,p_m\},\; \sum_{k=0}^mp_k =1$ 解决前五个原始矩这给出（在最后的数值是特定于在我们的情况下，累积量）

κ_{n} = μ_{n}^{'} - \sum_{i = 1}^{n - 1} (\binom{n - 1}{i - 1}) κ_{i} μ_{n - i}^{'}

$\kappa_n=\mu'_n-\sum_{i=1}^{n-1}{n-1 \choose i-1}\kappa_i \mu_{n-i}'$

\begin{aligned} μ_{1}^{'} = & κ_{1} = 1 \\ μ_{2}^{'} = & κ_{2} + κ_{1}^{2} = 3 / 2 \\ μ_{3}^{'} = & κ_{3} + 3 κ_{2} κ_{1} + κ_{1}^{3} = 17 / 6 \\ μ_{4}^{'} = & κ_{4} + 4 κ_{3} κ_{1} + 3 κ_{2}^{2} + 6 κ_{2} κ_{1}^{2} + κ_{1}^{4} = 19 / 3 \\ μ_{5}^{'} = & κ_{5} + 5 κ_{4} κ_{1} + 10 κ_{3} κ_{2} + 10 κ_{3} κ_{1}^{2} + 15 κ_{2}^{2} κ_{1} + 10 κ_{2} κ_{1}^{3} + κ_{1}^{5} = 243 / 15 \end{aligned}

$\begin{align} \mu'_1=&\kappa_1 =1\\ \mu'_2=&\kappa_2+\kappa_1^2=3/2\\ \mu'_3=&\kappa_3+3\kappa_2\kappa_1+\kappa_1^3=17/6\\ \mu'_4=&\kappa_4+4\kappa_3\kappa_1+3\kappa_2^2+6\kappa_2\kappa_1^2+\kappa_1^4=19/3\\ \mu'_5=&\kappa_5+5\kappa_4\kappa_1+10\kappa_3\kappa_2+10\kappa_3\kappa_1^2+15\kappa_2^2\kappa_1+10\kappa_2\kappa_1^3+\kappa_1^5=243/15\\ \end{align}$ If we (momentarily) set

m = 5

$m=5$ we have the system of equations

\begin{aligned} \sum_{k = 0}^{5} p_{k} = & 1, \sum_{k = 0}^{5} p_{k} k = 1 \\ \sum_{k = 0}^{5} p_{k} k^{2} = & 3 / 2, \sum_{k = 0}^{5} p_{k} k^{3} = 17 / 6 \\ \sum_{k = 0}^{5} p_{k} k^{4} = & 19 / 3, \sum_{k = 0}^{5} p_{k} k^{5} = 243 / 15 \\ s . t . p_{k} \geq 0 \forall k \end{aligned}

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

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.

— Alecos Papadopoulos
source

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)

$(0, a)$ and then on

(a, \infty)

$(a, \infty)$ it decays with something like an exponential tail (it's been a long time since I did the simulation).

— guy

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

$a$ ?

— Alecos Papadopoulos

Dug up my old code and found

a \approx 1

$a \approx 1$ . If

Y \sim κ (t)

$Y \sim \kappa(t)$ then

[Y ∣ Y < 1]

$[Y \mid Y < 1]$ is approximatey

U (0, 1)

$\mathcal U(0, 1)$ and

[Y - 1 ∣ Y > 1]

$[Y - 1\mid Y > 1]$ is approximately gamma distributed with shape

1.4

$1.4$ and mean

0.64

$0.64$ .

— guy

What do you mean by

Y \sim κ (t)

$Y \sim \kappa(t)$ ?

— Alecos Papadopoulos

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?

— wolfies