CLT question without characteristic functions

Question

Let $(X_n)_n$ be i.i.d. of mean $0$ and positive and finite variance, and let $S_n=X_1+\cdots+X_n$. I want to see a proof of the following statement, without resorting to characteristic functions

$$\lim_n\mathbb P(S_n>0,S_{2n}<0)=\frac18$$

I've already seen a proof using characteristic functions, but I would like a proof that doesn't resort to the theory of characteristic functions.

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What I've tried is the following. Let $S'_{n}=X_{n+1}+\cdots+ X_{2n}$, then we want to find

$$\lim_n\mathbb P(S_n>0,S_n+S'_{n}<0)$$

Both $S_n/\sqrt n,S'_n/\sqrt n\xrightarrow{d}\mathcal N(0,\sigma^2)$, and if $X,Y\sim\mathcal N(0,\sigma^2)$ are independent, a simple symmetry argument shows

$$\mathbb P(X>0,X+Y<0)=\frac18$$

So I would love to "interchange the limit with the probability", of course, this is nonsense mathematically, but I would like to turn this intuition into a proof

Answer 1

Here's a proof I find satisfactory, following the ideas laid out in the other answer. Notice that by independence $(S_n/\sqrt n,S'_n/\sqrt n)\xrightarrow{d}(X,Y)$ where $X,Y\sim\mathcal N(0,\sigma^2)$ are independent. Now, by the continuous mapping theorem, using $g(x,y)=(x,x+y)$ we get \begin{align}(S_n/\sqrt n, S_{2n}/\sqrt n)&=g(S_n/\sqrt n,S'_n/\sqrt n)\\&\xrightarrow{d}g(X,Y)\\&=(X,X+Y)\end{align} and therefore \begin{align}\lim_n\mathbb P(S_n>0,S_n+S'_n<0)&=\lim_n\mathbb P\left(\frac{S_n}{\sqrt n}>0,\frac{S_n}{\sqrt n}+\frac{S'_n}{\sqrt n}<0\right)\\&=\mathbb P(X>0,X+Y<0)\\ &=\frac18\end{align}

Answer 2

This is not too difficult of an integral to compute, note that $$ \mathbb P(X +Y < 0, X > 0) = \mathbb P(Y < -X, X > 0) = \frac{1}{2 \pi \sigma ^2} \int_{0}^\infty \int_{-\infty}^{-x} \exp \left( \frac{-x^2 - y^2}{2 \sigma^2} \right) \, \mathrm d y \, \mathrm d x$$ Converting to polar coordinates gives us with $(x, y) = (r \cos \theta, r \sin \theta)$ changes the bounds to $\theta \in [-\pi/2, -\pi/4], r \in (0, \infty)$ hence we get $$ \mathbb P(X + Y < 0, X > 0) = \frac{1}{2 \pi} \int_{-\pi/2}^{- \pi/4} \int_0^\infty e^{-r^2/2}r \, \mathrm d r \mathrm d \theta = \frac{1}{2 \pi} \cdot \frac{\pi}{4} = \frac{1}{8}$$ The only difficult part is to get the bounds on $\theta$, note that $x > 0$ means $\theta \in (-\pi/2, \pi /2)$ and $y < -x \implies \tan(\theta) < -1$, hence $\theta \in (-\pi/2, \pi /2) \cap (- \pi, - \pi/4) = (-\pi/2, -\pi/4)$.

The only part of this that we rely on the fact that $(X_n, Y_n) \Rightarrow (X, Y)$ which is a very basic fact of weak convergence. Note that $$ \mathbb P((X_n, Y_n) \in A \times B) = \mathbb P(X_n \in A) \mathbb P(Y_n \in B) \to \mathbb P(X \in A) \mathbb P(Y \in B)$$ Which is of course $\mathbb P((X, Y) \in A \times B)$.

Edit: Using the work done here all you would need is 1. $X_n \Rightarrow X$ and $Y_n \Rightarrow Y$, independent of each other implies that $(X_n, Y_n) \to (X, Y)$ and 2. by continuous mapping theorem with $g:(x, y) \mapsto x + y$ we can get $$g((X_n, Y_n)) = X_n + Y_n \Rightarrow X + Y$$