# Cumulants of non central chi-square distribution

#### Jesmin

##### New Member
Cumulant generating function is defined by logarithm of Moment generating function.

$$K_X(t)=\log M_X(t)$$

let $$X$$ is a non central $$\chi^2$$ variate with parameters degrees of freedom,$$n$$ and non-centrality parameter,$$\lambda$$.

Moment generating function of $$X$$ is:

$$M_X(t)=(1-2t)^{-\frac{n}{2}}\exp(\frac{\lambda t}{1-2t})$$
$$K_X(t)=\log M_X(t)=\log [(1-2t)^{-\frac{n}{2}}\exp(\frac{\lambda t}{1-2t})]$$
$$K_X(t)={-\frac{n}{2}}\log (1-2t)+(\frac{\lambda t}{1-2t})$$
$$K_X(t)={-\frac{n}{2}}[-2t-\frac{(2t)^2}{2}-\frac{(2t)^3}{3}-\frac{(2t)^4}{4}-\ldots]+(\frac{\lambda t}{1-2t})\ldots\ldots (A)$$

$$r^{th}$$ cumulant is the coefficient of $$\frac{t^r}{r!}$$ in $$K_X(t)$$.

But i couldn't separate $$\frac{t^r}{r!}$$ in $$K_X(t)$$ because of the denominator $$(1-2t)$$.
Hence couldn't able to compute the cumulants to know the mean,variance, 3rd&4th central moment of the distribution.

Can you please rearrange the equation $$(A)$$ as $$\frac{t^r}{r!}$$ so that the four cumulants can be obtained easily.

#### BGM

##### TS Contributor
You just need the geometric series to expand the fraction.