I was reading a textbook about showing the following Weak Law of Large Number but I stuck in some intermediate steps.

Here is the statement I work with

Let $\{X_i\}$ be i.i.d. random variables with same characteristic function $\phi$, and $\phi(0) = 1$, $\phi'(0)=ai$. Then, $$\frac{S_n}{n} \rightarrow a \;\;\text{ in probability}$$ where $S_n := X_1 + X_2 + ... + X_n.$

My approach :

If I can show

  1. $\phi_{S_n/n}(t) \rightarrow e^{iat} : = \phi_\infty(t)$ as $n \rightarrow \infty$ for all $t$, and,

  2. $\lim_{t \rightarrow 0}\phi_\infty(t) =1$,

then the associated sequence of distributions $\mu_n$ (whose characteristic function is $\phi_{S_n/n}$) converges to $\mu$ (whose characteristic function is $\phi_\infty$); Then I can infer that $S_n/n \rightarrow a$ in distribution and then since $a$ is constant, can further conclude $S_n/n \rightarrow a$ in probability.

Therefore, I first compute the characteristic function and using the i.i.d. fact to get $$ {\phi _{{S_n}/n}} = E[{e^{it\frac{{{S_n}}}{n}}}] = E[{e^{i\frac{t}{n}{S_n}}}] = E[{e^{i\frac{t}{n}\left( {{X_1} + ...{X_n}} \right)}}] = \prod\limits_{i = 1}^n {\underbrace {E[{e^{i\frac{t}{n}{X_i}}}]}_{ = \phi \left( {\frac{t}{n}} \right)}} = {\left[ {\phi \left( {\frac{t}{n}} \right)} \right]^n} $$

But I stuck to go further and show $$\phi_{S_n/n}(t) = {\left[ {\phi \left( {\frac{t}{n}} \right)} \right]^n} \rightarrow e^{iat}...$$

Any thought is appreciated.


1 Answer 1


Lemma: For every complex number, $(1+z/n)^n\to\mathrm e^z$. For every sequence of complex numbers $(z_n)$ such that $z_n\to z$, $(1+z_n/n)^n\to\mathrm e^z$.

Thus,$$\phi(t/n)^n=\left(1+\mathrm ia(t/n)+o(1/n)\right)^n\to\mathrm e^{\mathrm iat}.$$

  • $\begingroup$ Thank you, @Did, I know $\lim_{n \rightarrow \infty}(1 + ia(t/n))^n = e^{iat}$ but whats going on for the little oh part? if we apply the lemma you gave, we ignore the little oh part? I know the definition about little oh : if $f(n) = o(g(n))$, then $\lim \frac{f(n)}{g(n)}=0$, but still not quite understand how to use it correctly. $\endgroup$
    – Fianra
    Commented Oct 10, 2014 at 15:14
  • 1
    $\begingroup$ One applies the second part of the lemma with $z_n/n=iat/n+o(1/n)$, that is, $z_n\to iat$. $\endgroup$
    – Did
    Commented Oct 10, 2014 at 17:42
  • $\begingroup$ Hello, @Did, I went back and revisited your approach again, should the last little-oh term of the characteristic function be $\phi {(t/n)^n} = {\left( {1 + {\rm{i}}a(t/n) + o(t/n)} \right)^n}$? Thanks $\endgroup$
    – Fianra
    Commented Dec 15, 2014 at 8:39
  • 2
    $\begingroup$ No--or rather, yes if one wishes to, but this is irrelevant since one uses this expansion for every fixed $t$ when $n\to\infty$. $\endgroup$
    – Did
    Commented Dec 15, 2014 at 10:32
  • 1
    $\begingroup$ The formulation $\phi(t/n)=1+ia(t/n)+o(t/n)$ is a direct application of Taylor's theorem. $\endgroup$
    – svendvn
    Commented Dec 10, 2018 at 14:57

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