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.


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 Oct 10 '14 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 Oct 10 '14 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 Dec 15 '14 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 Dec 15 '14 at 10:32
  • $\begingroup$ The formulation $\phi(t/n)=1+ia(t/n)+o(t/n)$ is a direct application of Taylor's theorem. $\endgroup$ – svendvn Dec 10 '18 at 14:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.