I'm trying to derive the characteristic function for exponential distribution and geometric distribution. Can you guide me on getting them?
Here is my solution so far:
Exponential Dist Characteristic function:
$\phi(t) = E[e^{itX}]$ where $X$ has exponential distribution , then by definition, expectation can be written as : $$ \int_0^{\infty} e^{itx} \lambda e^{-\lambda x } dx = \frac{\lambda}{it – \lambda} e^{(it – \lambda)x}\bigg|_0^{\infty} = \frac{\lambda}{ \lambda- it}$$
For the geometric random variables, assume $\Pr(X = 0) = P$, how can we find the characteristic function of $X$?
Here since $X$ is a discrete random variables, we have:
$\phi(t) = E[e^{itX}] = \sum_{j = 0}^{\infty} e^{itj} (1 – P)^j P$ my issue is I don't know to get a closed form for the characteristic function?
Thanks for your help.
Best Answer
You are almost there.
$$\phi(t) = E[e^{itX}] = \sum_{j = 0}^{\infty} e^{itj} (1 - P)^j P = P \sum_{j = 0}^{\infty} [e^{it} (1 - P) ]^j $$
And now, if you don't know about the geometric series ( $\sum_{k=0}^\infty a^k$ ), it's time to learn about it.