# Finding $G_{cX}(t)$ given $G_{X}(t)$

I have recently been studying probability generating functions, and have seen the proof that the sum of two independent Poisson random variables has a Poisson distribution also. This used the fact that $G_{X+Y}(t)=G_{X}(t)G_{Y}(t)$ for independent random variables $X$ and $Y$.

I then thought 'What would the generating function of $X-Y$ be?". Since $X-Y=X+(-Y)$, we can use the above result if we can find the p.g.f. of $-Y$ (or, more generally, $cY$ for some constant $c$).

Since $-Y$ takes values $0,-1,-2\ldots$ with respective probabilities $p_{0},p_{1} \ldots$, substituting this into the definition of a generating function (or at least the definition I have) gives $$G_{-Y}(t)=\sum_{k=0}^{\infty}p_{k}t^{-k}=G_{Y}\left(\frac{1}{t}\right)$$
And a similar argument shows $G_{cY}(t)=G_{Y}(t^{c})$. Is this correct? My textbook doesn't say anything about this, and I can't find it on the internet either.

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In case that the series is convergent for $t$ and $\frac{1}{t}$ or $t^c$ then your argument holds. But note that in general the probability generating function is only defined for $t\in [-1,1]$ (here the series is absolute convergent). And for $t\in (-1,1)$ we have $\left|\frac{1}{t}\right|>1$ and hence $G_Y\left(\frac{1}{t}\right)$ may not exist. – Stefan Hansen Jan 29 '13 at 20:09

Note that $G_{cX}(t)=E(t^{cX})=\sum_{k=0}^{\infty}t^{cx}\cdot p_k$.

Now let $t^c=m$.

Then we have $G_{cX}(t)=\sum_{k=0}^{\infty}m^x\cdot p_k=G_{X}(m)$.

Again $m=t^c$.

So $G_{cX}(t)=G_{X}(t^c)$

Clearly your argument is right and here it is proved.

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How can you be sure that $\sum_{k=0}^\infty t^{cx}\cdot p_k$ is convergent? – Stefan Hansen Jan 29 '13 at 20:33
@StefanHansen : If $E(t^{(cX)})$ exists then $\sum_{k=0}^{\infty}t^{cx}\cdot p_k$ is convergent. And your comment under the question is correct and shows when it converge. – A.D Jan 30 '13 at 3:22
Thank you for your answer! – Daniel Littlewood Jan 30 '13 at 19:28