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I'm currently stuck on the following exercise:

If $\{X_n\}$ is independent and $X_i$ are have the same distribution with finite first moment, then $n^{-1}S_n \to E[X_1]$ with probability 1 (via strong law of large numbers), so that $n^{-1}S_n \Rightarrow E[X_1]$ (weak convergence). Prove the latter fact by characteristic functions. Hint: use $|z_1\cdots z_m - w_1\cdots w_m|\leq \sum_{k=1}^m|z_k-w_k|$ where the norm of $w_i,z_i$ is 1 or less.

I've been trying to write out the characteristic function of $n^{-1}S_n$ and trying to relate it to $E[X]$ but have been unable to. Can anyone help me with this? thanks in advance!!

Mittens
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Jord van Eldik
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1 Answers1

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Let $X=X_1$. As the $(X_n:n\in\mathbb{N})$ are i.i.d., $$E[\exp(itS_n/n)]=\big(E[e^{iX/n}]\big)^n=(\phi_X(t/n)\big)^n=\Big(1+\frac{n(\phi_X(t/n)-1)}{n}\Big)^n$$

Since $E[|X|]<\infty$, $\phi_X$ is differentiable and $$\phi'_X(t)=\lim_{h\rightarrow0}\frac{\phi_X(t+h)-\phi_X(t)}{h}=E[iX\exp(iXt)]$$ In particular, for $t=0$, $\phi'(0)=iE[X]$.

Notice that $\lim_nn(\phi_X(t/n)-1)\xrightarrow{n\rightarrow\infty}t\phi'(0)=itE[X]$. Putting things together $$\lim_nE[\exp(itS_n/n)]=e^{itE[X]}$$

The hint, I gather, is to prove that for any complex sequence $(a_n:n\in\mathbb{N})$ with $a_n\xrightarrow{n\rightarrow\infty}a$, $$\lim_n\Big(1+\frac{a_n}{n}\Big)^n=e^a$$ A proof along the lines of the hint in the OP is discussed the following posting in MSE. This also appears in several textbooks of Probability, for example Durrett, R., Probability Theory and Examples, 5th edition, Cambridge University. Press.

Mittens
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