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I'm trying to derive the characteristic function for exponential distribution and geometric distribution. Can you please 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.

Rodrigo de Azevedo
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2 Answers2

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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.

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

Davide Giraudo
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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? –  Dec 07 '12 at 15:57
  • A closed form can be found, and I think it's what is expected. – Davide Giraudo Dec 07 '12 at 16:08