Noncentral chi² with a noncentral chi² noncentrality parameter Denote by $\chi^2(\nu,\lambda)$ the noncentral chi-square distribution with degrees of freedom $\nu$ and noncentrality parameter $\lambda$.
If $\Lambda \sim \chi^2(\nu, 2\theta)$ and $(X \mid \Lambda) \sim \chi^2(\nu, \Lambda)$, then $X/2 \sim \chi^2(\nu, \theta)$.
This result is derived from the theory of Cox-Ingersoll-Ros process. I don't find any proof of it. Could you prove it?
> sims <- rchisq(50000, df = 2, ncp = rchisq(50000, df = 2, ncp = 2))/2
> curve(pchisq(x, df = 2, ncp = 1), to=10)
> curve(ecdf(sims)(x), add=TRUE, col="red", lty=2, lwd=2)


 A: Let's do it with characteristic functions.  We'll start out with the definition of the characteristic function for an arbitrary distribution $F(x)$:
$$\phi_x(it) = \int e^{itx}dF(x)$$
The ch.f. of a noncentral $\chi^2(\nu, \Lambda)$ is:
$$\phi_{X|\Lambda}(it) = \frac{\exp\{\frac{it\Lambda}{1-2it}\}}{(1-2it)^{\nu/2}}$$
In this case, this is the ch.f. of $X|\Lambda$.  Now, if we have an arbitrary distribution of $\Lambda$, label it $F$, we can find the ch.f. of $X$ by integrating out $\Lambda$ from the ch.f. of $X|\Lambda$.  Note, however, that in this case the ch.f. of $X|\Lambda$ can be rearranged as:
$$\phi_{X|\Lambda}(it) = \frac{1}{(1-2it)^{\nu/2}}\exp\{[it/(1-2it)]\Lambda\}
$$
Looking carefully at the $\exp$ term and comparing to our initial expression for the characteristic function, we can see that integrating out $\Lambda$ is essentially the same integral as required to find the ch.f. of $F$, but with $it$ replaced by $it/(1-2it)$.  Consequently, we can see that:
$$\phi_X(it) = \frac{\phi_{\Lambda}(it/(1-2it))}{(1-2it)^{\nu/2}}$$
Substituting the appropriately-parameterized ch.f. of $\Lambda$ gives us:
$$\phi_X(it) = \frac{\exp\left\{\frac{2it\theta}{(1-2it)\left(1-\frac{2it}{1-2it}\right)}\right\}}{(1-2it)^{\nu/2}(1-\frac{2it}{1-2it})^{\nu/2}}$$
We rewrite the term $(1-2it/(1-2it))$ as $(1-4it)/(1-2it)$ by replacing the leading "1" with $(1-2it)/(1-2it)$ and working through the resultant algebra.  Substituting gives us the resulting mess:
$$\phi_X(it) = \frac{\exp\left\{\frac{2it\theta}{(1-2it)(1-4it)/(1-2it)}\right\}}{(1-2it)^{\nu/2}(1-4it)^{\nu/2}/(1-2it)^{\nu/2}}$$
The obvious cancellations result in:
$$\phi_X(it) = \frac{\exp\left\{\frac{2it\theta}{(1-4it)}\right\}}{(1-4it)^{\nu/2}}$$
Almost there! What we have is the characteristic function of $X$; what we want is the characteristic function of $X/2$.  A basic property of characteristic functions is that the ch.f. of $aX$ is the same as the ch.f. of $X$ with $ait$ substituted everywhere for $it$.  (If you look at the definition of the characteristic function at the beginning of the answer, you may be able to see why this is so.)  Substituting $it/2$ everywhere for $it$ and performing the resulting divisions $(2/2)$ and $(4/2)$ gives us:
$$\phi_{X/2}(it) = \frac{\exp\left\{\frac{it\theta}{(1-2it)}\right\}}{(1-2it)^{\nu/2}}$$
which, by comparing to our initial characteristic function for $X|\Lambda$, we can see is the characteristic function of a noncentral $\chi^2(\nu, \theta)$ distribution.
