The following problems is from Durrett 3.4.9:
Suppose $X_i$ are independent and $S_n = X_1 + ... + X_n$. Assume that $$ \begin{split} P(X_m = m) &= P(X_m = -m) = m^{-2}/2, \text{ and }\\ P(X_m = 1) &= P(X_m = -1) = (1 - m^{-2})/2. \end{split} $$ Show that $$\frac{S_n}{\sqrt{n}} \to \chi.$$
My attempt:
The expectation is $0$, and $EX_m^2 = 2 - 1/m^2 < \infty$. Thus, $\varphi(t) = 1 - (2 - 1/m^2) \frac{t^2}{2} + o(t^2)$.
Sum $X_m$ up and divided by $\sqrt{n}$, we get:
$$
\varphi_\frac{S_n}{\sqrt{n}}(t)
= \prod_{m = 1}^n 1 - \left(1 - \frac{1}{2m^2}\right) \frac{t^2}{n} + o(n^{-1})
$$
Using the lemma: If
$$
\max_{1\le j\le n}\left|c_{j,n}\right|\to 0,
\sum_{j=1}^n c_{j,n} \to \lambda, \text{ and }
\sup_n \sum_{j=1}^n \left|c_{j,n}\right| < \infty
$$
then
$$\prod_{j=1}^n (1 + c_{j,n}) \to e^\lambda.$$
The characteristic function becomes:
$$\varphi_\frac{S_n}{\sqrt{n}}(t) = e^{-t^2}. $$
But then, it converges weakly to ${\frac{1}{\sqrt{2}}} \chi$ rather than $\chi$.
Can any one help pointing out the mistake in the above reasoning?
Thank you!
The issue seems to be in your asymptotic expansion of $\phi_m(t)$ (for $t\to 0$), that you then use to approximate the product $\prod_{m=1}^n \phi_m\left(\frac{t}{\sqrt{n}}\right)$.
Namely, you write $$ \phi_m(t) = 1 - \left(1-\frac{1}{2m^2}\right)t^2 +o(t^2). $$ This is correct, for $m$ fixed and as $t\to 0$. But now, you use it for $\frac{t}{\sqrt{n}}$ (which indeed goes to $0$) in a product from $m$ ranging from $1$ to $n$: $m$ is not a fixed parameter anymore, there is a relation between the asymptotics of $\frac{t}{\sqrt{n}}$ and $m$...
Let us look at the actual expression of $\phi_m$: after an easy computation, $$ \phi_m(t) = \left(1 - \frac{1}{m^2}\right) \cos t + \frac{1}{m^2}\cos(mt) $$ and when doing a Taylor expansion (with respect to $t$), the fourth order term has a quadratic dependence on $m$: $$ \phi_m(t) = 1-\left(1 - \frac{1}{2m^2}\right) t^2+\frac{1}{24}\left(m^2+1-\frac{1}{m^2}\right)t^4 + o(t^5). $$ While this does look innocuous, it is not: recall that you use then this expression evaluated on $\frac{t}{\sqrt{n}}$, so that the fourth order term brings in $\frac{t^4}{n^2}$. But the product $\prod_{m=1}^n \phi_m\left(\frac{t}{\sqrt{n}}\right)$ is for $m$ up to $n$, and when $m=n$ (and similar when $m$ is roughly close to $n$) we get $$ \frac{1}{24}\left(m^2+1-\frac{1}{m^2}\right)\frac{t^4}{n^2} = \Theta(t^4). $$ This is a constant/term that you neglected; but cannot neglect...
As a side note: looking at the exercise statement you mention (Exercise 3.4.9, p.113 in the edition linked), there is this note: "The trouble here is that $X_{n,m}=\frac{X_m}{\sqrt{n}}$ does not satisfy (ii) of Theorem 3.4.5." Checking (the proof of) this theorem, (ii) is the hypothesis used to guarantee that the product of the characteristic functions is indeed well-approximated by the product of their second-order Taylor expansion.