Let $X$ be a discrete random variable with factorial moment generating function $\psi_X(t)$ and define $Y=aX+b$, where $a$ and $b$ are constants. Express the factorial moment generating function for $Y$ in terms of $\psi_X(t)$

The factorial generating function is defined as $\psi_X(t) = E[t^X]$

My answer is $\psi_Y(t) = at^b\psi_X(t)$

My reasoning is as follows:

$\psi_Y(t) = E[t^Y] = E[t^{aX+b}] = E[e^{ln(t^{aX+b})}] = E[e^{aXln(t)}e^{ln(t^b)}] = t^bm_X(aln(t)) = at^bm_X(ln(t)) = at^b\psi_X(t)$

where $m_X$ is the moment generating function defined as $m_X(t) = E[e^{Xt}]$. Simple algebra will show that $m_X(ln(t)) = \psi_X(t)$

I can find a lot of proofs for the linear relationship of the moment generating function, but not for the factorial generating function. Does anyone see any flaws in my logic?


The relationship you appear to use in the penultimate step $$m_X(a \log t) = a m_X(\log t)$$ is not correct. For example, if $X \sim \operatorname{Gamma}(n,\lambda)$, then $$m_X(t) = (1- \lambda t)^{-n}.$$ Then $$m_X(a \log t) = (1 - \lambda a \log t)^{-n} \ne a (1 - \lambda \log t)^{-n} = a m_X (\log t).$$

I think the best you can do is to simply state $$\psi_Y(t) = \operatorname{E}[t^{aX + b}] = \operatorname{E}[(t^a)^X t^b] = t^b \psi_X(t^a).$$

  • $\begingroup$ Makes sense to me. Thanks $\endgroup$ – lstbl Feb 2 '16 at 22:24

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