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If $N \in Fs(p)$ and $X_i \in Ge(\alpha)$ and we define $Y=\sum_{i=1}^N X_i$ then show that the distribution of $Y$ is $Ge(\beta)$.
Clearly the characteristic functions are required, as $\varphi_Y (t) = \varphi_N(\varphi_{X_i}(t))$

$\varphi_N (t)= \frac{p*\exp(i*t)}{1-(1-p)\exp(i*t)}: 0 \le p \le 1$

$\varphi_X (t)= \frac{\alpha}{1-(1-\alpha)\exp(i*t)}: 0 \le \alpha \le 1$

So I need to show that there is SOME $\beta$ such that

$\varphi_Y (t)= \frac{\beta}{1-(1-\beta)\exp(i*t)}: 0 \le \beta \le 1$

I'm getting stuck pretty early with the algebra.... maybe a hint would be greatly appreciated as I want to solve this myself. Thank you all.

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    No.... you are wrong. Read the question, $N$ is NOT deterministic, it is randomly distributed so it will NOT be a negative binomial distribution but a geometric distribution, still. Fs is "first success".2012-10-19

1 Answers 1

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You are confusing characteristic functions and generating functions.

If $Y=\sum\limits_{k=1}^NX_k$, as in your question, then $g_Y=g_N\circ g_X$ where, for every integer valued random variable $Z$ and every real number $s$ in $[0,1]$, $g_Z(s)=\mathbb E(s^Z)$.

Here, $g_N(s)=\dfrac{ps}{1-(1-p)s}$ and $g_X(s)=\dfrac{\alpha}{1-(1-\alpha)s}$ hence you should reach the result that $g_Y(s)=\dfrac{r}{1-(1-r)s}$ for some $r$ depending on $p$ and $\alpha$ whose value I will let you discover.