Let $\phi(t)$ be a characteristic equation of a discrete r.v. X. Show that the function $e^{\mu (\phi(t) -1)}$ is a characteristic function. (Hint: Consider the sum of iid R.v.)

Attempt: I was told that the function could be shown to be a the sum of characteristic function of iid Poisson rv. I know the Poisson characteristic for the sum of 2 rv to be $e^{(\mu + \lambda)(e^{it}-1)}$. I am having trouble splitting this function $e^{\mu (\phi(t) -1)}$ to $e^{(\mu + \lambda)(e^{it}-1)}$. I was trying $e^{\mu (\phi(t) -1)} = \sum \frac{(\mu (\phi (t)-1))^x}{x!}$. But I got stuck as that's an infinite sum.


1 Answer 1


Let $\{X_i\}$ be i.i.d with characteristic function $\phi$. Let $N$ be a Poisson random variable with parameter $\mu$ independent of $X_i$'s. Then $Ee^{it(X_1+X_2+...+X_N)}=\sum_{n=0}^{\infty} Ee^{it(X_1+X_2+...+X_n)}e^{-\mu} \frac {\mu ^{n}} {n!}=\sum_{n=0}^{\infty} \phi (t)^{n}e^{-\mu} \frac {\mu ^{n}} {n!}=e^{\mu (\phi(t)-1)}$.

  • $\begingroup$ What is E here? The characteristic function? Shouldn't that be Phi? $\endgroup$
    – Dom
    Commented Aug 9, 2018 at 23:47
  • $\begingroup$ Sorry for typos. I have edited the answer. $\endgroup$ Commented Aug 10, 2018 at 1:56

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