Let $P(k,X)=p(1-p)^{k-1}$.
When deriving the moment generating function I start off as follows:
$E[e^{kt}X]=\sum\limits_{k=1}^{\infty}e^{kt}p(1-p)^{k-1}$.
How I end up rearranging this is as follows:
$\frac{p}{1-p}\sum\limits_{k=1}^{\infty}e^{kt}(1-p)^k=\frac{p}{1-p}\sum\limits_{k=1}^{\infty}(e^{t}(1-p))^k=\frac{p}{1-p}\frac{1}{1-e^t(1-p)}$
I'm obviously not arriving at the correct answer, but I want to know why this derivation is wrong.
$\endgroup$1 Answer
$\begingroup$Formula: Let $|q|<1$ then we have $$(\star) \ \ \sum_{k=1}^{\infty} q^k = \frac{q}{1-q}.$$ We use this fact for the calculations of MGF $$\mathbb{E}[e^{tX}] = \frac{p}{1-p}\sum_{k=1}^{\infty}e^{tk}(1-p)^k=\frac{p}{1-p}\sum_{k=1}^{\infty}\left(e^{t}(1-p)\right)^k $$ now let us apply $(\star)$ to obtain $$\frac{p}{1-p}\sum_{k=1}^{\infty}\left(\underbrace{e^{t}(1-p)}_{=:q}\right)^k = \frac{p}{1-p}\frac{\overbrace{e^t(1-p)}^{q}}{\underbrace{1-e^t(1-p)}_{1-q}} = \frac{pe^t}{1-e^t(1-p)}.$$
It is well-defined for all $t< -\ln(1-p)$.
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