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 $\varphi$.

  1. If $\varphi'(0) = ia$ and $S_n = X_1 + \dots + X_n$, then $S_n/n\to a $ in probability.
  2. If $S_n/n\to a$ in probability then $\varphi(t/n)^n\to e^{iat}$ as $n\to \infty$.
  3. 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 $\varphi(t/n)^n\to e^{iat}$.


For part 3, the connection is this: take logs. Using this, you can get that $n (\varphi(t/n)-1) \to iat$: write it as $$\left(\frac{\varphi(t/n)-1}{\log \varphi(t/n)}\right)( n \log \varphi(t/n))$$ and use the differentiability of $\log z$ at $z=1$ to show that the first factor goes to 1 as $n \to \infty$.

To finish, one needs to work a little harder. A previous exercise in Durrett shows that if a sequence of random variables converges in distribution, then the corresponding characteristic functions converge uniformly on compact subsets of $\mathbb{R}$. Using this fact in the first part of this problem, we can show that $n (\varphi(t/n)-1) \to iat$ locally uniformly in $t$. For sufficiently small $h$, we can write $h = t/n$ where $n$ is suitably large and $t$ is close to 1 (or -1, when $h < 0$); I will let you fill in the details.

  • $\begingroup$ Thank you for your answer! I very much appreciate your help. =) $\endgroup$
    – Sam
    Mar 26 '12 at 4:29
  • $\begingroup$ You said 'Using this fact in the first part of this problem'. Is this necessary? I think just locally uniformity is implied by the previous exercise in Durrett. $\endgroup$
    – kayak
    May 1 '17 at 11:26
  • $\begingroup$ You are correct, @kayak. Exercise 3.3.13 in the 5th edition. $\endgroup$
    – Chris Wang
    Oct 22 '20 at 1:45

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