Differentiability in zero of characteristic function and expected value of i.i.i random variables

Question

I would like to prove the following implication:

If the characteristic function $\phi$ is differentiable in zero and $X_1\geq0$ a.s., then $E(X_1)=i\phi'(0)<\infty$.

(This is Exercise 15.4.4 iii) of Probability Theory by A. Klenke, 3rd version).

My steps:

1. If $\phi$ is differentiable in zero, then I could write it as $\phi'(0)=im$, for some $m\in\mathcal{R}$.

2. Thus, by exercise 15.4.4 ii) of the same book, $(X_1+...+X_n)/n \xrightarrow{n \to \infty } m$ in probability.

3. By the weak law of large numbers we know that $(X_1+...+X_n)/n \xrightarrow{n \to \infty } E(X_1)$ in probability, where $X_1,X_2,...$ are i.i.d. random variables.

4. Concluding, by the uniqueness of the limit, $\phi'(0)=im=iE(X_1)$.

I am not sure about my use of the uniqueness of the limit and the fact that I didn't use the non negativity of the random variable.

What am I doing wrong?

Thank you.

Answer 1

There is a version of SLLN's for non-negative r.v's with infinte mean. It says $\frac 1 n(X_1+X_2+\cdots+X_n) \to \infty$ almost surely. Together with the result of exercise 15.4.4. ii) you get a contradiction. Hence, $EX_1$ has to be finite and it must be $i\phi'(0)$.

Ref. Law of large numbers for nonnegative random variables