Limit of $\frac{S_n}{n}$, for $S_n \sim N(0, n)$ - prerequisite to convergence in probability

Question

Here $Z \sim N(0,1), S_n = \sum_{k=1}^{n} Z_k$, and all $Z_n$ are iid.

To prove convergence in probability of $\frac{S_n}{n} \to_n 0$ it is shown that $P\left(\left|\frac{S_n}{n}\right|> \varepsilon\right) \to_n 0$. I then realized that the fact that $L=0$ is the limit is not proven. I decided to show that.

Assume $$ \frac{S_n}{n} \to_n L $$ with $L \neq 0$. Wlog $L >0$. I get: $$ P\left(\left|\frac{S_n}{n}- L \right|>\varepsilon\right)= 2 P(S_n > n(L +\varepsilon)) = 2 \times \frac{1}{\sqrt{2 \pi n}} \int_{n(L+\varepsilon)}^{\infty} e^{-\frac{s^2}{2n}}ds $$ by symmetry of Gaussian rvs and taking $\varepsilon < L$ to make $L-\varepsilon>0$. Obviously this probability $\to_n 0$, so $L$ is a limit, too? Clearly if the limit exists, it is unique, so I got a contradiction. Could somene pls show where I made a mistake in my logic.

Answer 1

1. By definition, $P\left(\left\lvert \frac{S_n}{n}-0\right\rvert > \varepsilon \right)=P\left(\left\lvert \frac{S_n}{n} \right\rvert > \varepsilon\right) \to 0$ means that $\frac{S_n}{n}$ converges to $0$ in probability. So I am not sure why you are trying to prove that $\frac{S_n}{n}$ cannot converge in probability to some other limit $L$.
2. The problem with your computation is the symmetry argument. The density of $S_n$ is symmetric with respect to the $y$-axis but not with respect to the line $x=L$. So you have to compute each probability separately: $$P\left(\left\lvert \frac{S_n}{n}- L\right\rvert > \varepsilon \right) = P(S_n > n(L+\varepsilon)) + P(S_n < n(L-\varepsilon)).$$ As you showed, the first one goes to $0$. On the other hand, making the change of variable $t = s/\sqrt{n}$, $$\begin{align}P(S_n < n(L-\varepsilon)) &= \frac{1}{\sqrt{2\pi n}}\int_{-\infty}^{n(L-\varepsilon)}e^{-s^2/(2n)}\, ds\\ &= \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\sqrt{n}(L-\varepsilon)}e^{-t^2/2}\, dt \to 1 \end{align}$$ if we choose $\varepsilon >0$ small enough such that $L-\varepsilon >0$.