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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?

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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 $$\psi_Y(t) = \operatorname{E}[t^{aX + b}] = \operatorname{E}[(t^a)^X t^b] = t^b \psi_X(t^a).$$

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

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