# Characteristic function of exponential and geometric distributions

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?

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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.

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• If $X$ is a random variable with density $f$ (with respect to Lebesgue measure on the real line) and $G$ is a well-behaved function, then $$E[G(X)]=\int_{\Bbb R}G(x)f(x)d\lambda(x).$$

In the particular case of the exponential law, this gives $$\phi(t)=\int_0^{+\infty}e^{itx}e^{-\lambda x}\lambda dx.$$

• If $X$ is a random variable with values in the set of non-negative integers, then its characteristic function is given by $$\phi(t):=\sum_{k=0}^{+\infty}e^{itk}P\{X=k\}.$$

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Hi Davide, in my specific case, I wanted to see whether for the geometric or exponential characteristic functions, I can get a closed form. Is it enough to write the summation and not to get the closed form? –  user48405 Dec 7 '12 at 15:57
A closed form can be found, and I think it's what is expected. –  Davide Giraudo Dec 7 '12 at 16:08