I'm new here so please criticize errors!

My moment generating function looks like this (after some tidying):

$E[e^{wY(t)}] = \frac{1}{\left(1-2\theta(t) w\right)^{k/2}} \exp{\left(\frac{\lambda(t)Y(0) w}{1-2\theta(t) w}\right)}$

It's very close to a non-central $\chi^2$ distribution but it's not quite there. My supervisor has suggested that from this I can immediately write down that Y scaled by a factor is non-central $\chi^2$ but I don't understand how this is true.

If anyone could suggest what the actual distribution is, or how to scale Y, it would be much appreciated! Thanks!

For completeness, the full expression before tidying is:

$E[e^{wY(t)}] = {\left[1-\tfrac{\alpha_Y}{\beta_Y}(e^{\beta_Yt}-1) w \right]^{-\frac{b_Y}{\beta_Y}}} \exp\left(\frac{ e^{\beta_Y t} Y(0) w}{1-\tfrac{\alpha_Y}{\beta_Y} (e^{\beta_Y t}-1) w }\right)$

The process I am considering is time-homgeneous.

  • $\begingroup$ In your last expression ("before tidying") there is dependence on $Y_t$ (assuming that is the same as $Y(t)$, which cannot be right, since this is the expectation of a function of $Y_t$? Can you correct? $\endgroup$ Sep 1, 2015 at 11:30
  • 1
    $\begingroup$ Apologies, it was silly notation. What I have is an affine stochastic process, $Y$, I am trying to determine the distribution at a time $T$ given the an initial condition for $Y$ at time = $t$. This little $t$ can be set to zero without loss of generality because the process is time-homogeneous. $\endgroup$ Sep 1, 2015 at 11:41

1 Answer 1


Writing $u = \theta(t)w$ and $\lambda = \lambda(t)Y(0)/\theta(t)$, you have already observed that

$$E\left[\exp\left(u \frac{Y(t)}{\theta(t)}\right)\right] = E[\exp(w\,Y(t))] = \frac{1}{(1 - 2u)^{k/2}}\exp\left(\frac{\lambda u}{1-2u}\right)$$

which is the MGF for a non-central $\chi^2$ distribution with $k$ degrees of freedom and non-centrality parameter $\lambda$. Therefore $Y(t)/\theta(t)$ has this distribution, whence $Y(t)$ is the same distribution scaled by $\theta(t)$.

  • $\begingroup$ Ah, this is exactly what I needed. Thank you very much! $\endgroup$ Sep 1, 2015 at 13:38

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