How can we write a non-central chi-squared distribution as gamma distribution?

Question

Consider a random variable that has a non-central chi-squared distribution \begin{eqnarray*} L & = & \chi_{1}^{2}(b^{2}), \end{eqnarray*} where$\chi_{1}^{2}(b^{2})$ represents a non-central chi-squared with one degree of freedom. In fact $\chi_{1}^{2}(b^{2})$ is the square of $\mathcal{N}(b,1)$. How can we write $L$ as a Gamma distribution please? I know that if $b=0,$ we can write \begin{eqnarray*} L & \sim & \Gamma(\frac{1}{2},2). \end{eqnarray*} What happens when $b\neq0$ please? Thanks.

Answer 1

When the noncentrality parameter is nonzero, the distribution of $L$ is a mixture of infinitely-many central Chi-square distributions (which are Gamma distributions with appropriate parameters). Specifically, the density function of $L$ is $$f(x) = \sum_{n=0}^\infty \frac{e^{-b^2/2} (b^2/2)^n}{n!} g_n(x) $$ where $g_n(x)$ is the density function for a Gamma distribution with parameters $k=n+{1\over 2},\theta=2:$ $$g_n(x)=\frac{x^{(n+{1\over 2})-1}e^{-\frac{x}{2}}}{2^{n+{1\over 2}}\Gamma(n+{1\over 2})}[x > 0]. $$

A sketch of the derivation of this density is given here (in sections 3, 3.1).

Note: The derivation uses the Taylor series $$\cosh x = \sum_{n=0}^\infty \frac{x^{2n}}{(2n)!}$$ and the fact that $${\Gamma(n+{1\over 2})\over\Gamma(2n+1)}={\sqrt{\pi}\over 4^n\,n!}.$$