I'm trying to learn some probability theory atm and got stuck with the following exercise in Durrett's Probability: Theory and Examples:
Exercise 3.3.17. Let $X_1, X_2, \ldots $ be i.i.d. with characteristic function $\varphi$.
- If \varphi'(0) = ia and $S_n = X_1 + \dots + X_n$, then $S_n/n\to a $ in probability.
- If $S_n/n\to a$ in probability then $\varphi(t/n)^n\to e^{iat}$ as $n\to \infty$.
- Use 2. and the uniform continuity of $\varphi$ to show that $(\varphi(h)-1)/h \to -ia$ as $h\to 0$. Thus the weak law holds if and only if \varphi'(0) exists.
I would really appreciate some help with the third part of this exercise (I don't quite see the connection between $\varphi(t/n)^n$ and $(\varphi(h)-1)/h$, yet). Thanks for your help! =)
My thoughts on 1, 2:
I managed to prove 1. using the inequality $\mu\{x\, : \, |x|>u/2\} \le u^{-1} \int_{-u}^u (1-\varphi(t)) \, dt$, where $\mu$ is the pushforward measure of a random variable $X$ and $\varphi$ is its characteristic function. Using the fact that the ch.f. of $S_n/n - a$ is given by $e^{-iat}\varphi(t/n)^n$, this leads to
$P\left[\left|\frac{S_n}n - a\right| > 2/u \right]\le u^{-1} \int_{-u}^u (1-e^{-iat}\varphi(t/n)^n) \, dt$
Now 1. implies $\varphi(t/n)^n \to e^{iat}$, so the RHS goes to zero as $n\to \infty$ for every fixed $u$.
2.: Using $|e^{i\epsilon t} - 1| \le 2\epsilon |t|$ for small enough $\epsilon>0$:
\begin{align} \left|\varphi (t/n)^n - e^{iat}\right| &= \left| E\left[e^{iS_n/nt} - e^{iat}\right]\right| \\ &\le E\left|e^{i(S_n/n - a)t} - 1\right| \\ &\le 2 \epsilon |t| + 2P[|S_n/n - a|> \epsilon] \end{align}
So $\limsup_{n\to\infty}\, \left|\varphi (t/n)^n - e^{iat}\right| \le 2\epsilon |t|$ and since $\epsilon>0$ was arbitrary (apart from being small) this proves $\varphi(t/n)^n\to e^{iat}$.