Characteristic function of exponential and geometric distributions

Question

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:

Characteristic function of Exponential distribution:

$\phi(t) = E[e^{itX}]$ 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 variable, we have:

$\phi(t) = E[e^{itX}] = \sum_{j = 0}^{\infty} e^{itj} (1 - P)^j P$ my issue is I don't know how to get a closed form for the characteristic function?

Thanks for your help.

Answer 1

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.

Answer 2

• 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 $$\phi(t):=\sum_{k=0}^{+\infty}e^{itk}P\{X=k\}.$$