I am trying to prove the following:
If $(X_1+...+X_n)/n \xrightarrow{n \to \infty} m < \infty$ in probability then the characteristic function $\phi$ is differentiable in zero and $\phi'(0)=im$, where $X_1,X_2,...$ are i.i.d. random variables and $\phi$ is the characteristic function of $X_1$.
My strategy is to get a contradiction by using the Levy's continuity theorem in this way:
- $(X_1+...+X_n)/n \xrightarrow{n \to \infty} m$ in probability and so in distribution to a constant $m$.
- By Levy's continuity theorem, and the property of the characteristic function of a sum of i.i.d. random variables, $\lim_{n\to\infty} (\phi(t/n))^n = e^{imt}=\lim_{n\to\infty}(1+\frac{imt}{n})^n$. (The central term is the characteristic function of a constant random variable equal to $m$).
- If $\phi'(0)$ is not finite, then I cannot approximate it with a Taylor's polynomial of order 1 around zero: $\phi(t/n)\ne1+\phi'(0)\frac{t}{n}+o(t/n)$, $o(t)\xrightarrow{t\to0}0$.
- Concluding, $e^{\phi'(0)t}=\lim_{n\to\infty}(1+\phi'(0)\frac{t}{n})^n \ne\lim_{n\to\infty} (\phi(t/n))^n=e^{imt}=\lim_{n\to\infty}(1+\frac{imt}{n})^n.$
Clearly 4. is not a contradiction: how can I adjust my proof? Is this the right strategy to prove the claim?
Thanks.
