Let $(\xi_t)_{t \geq 0}$ be random variables (not necessarily independent) defined on a probability space such that $\xi_t \sim N(0, t)$. Is it necessarily true that $\xi_t \rightarrow 0$ almost surely as $t \rightarrow 0$?
I want to know that because I am trying to convince myself of continuity at 0 of time inverted Brownian motion, so please don't rely on properties of Brownian motion when answering this question.
Thanks in advance.
Edit: I am happy that they must converge in distribution to 0 and therefore in probability (since 0 is constant). I don't know if this helps.
Not without more assumptions.
Suppose the random variables were all independent. Fix $N > 0$ and choose countably many distinct times $t_n \in [1/N, 2/N]$. Then for each $\xi_{t_n}$, we have $$P(\xi_{t_n} > 50) = 1 - \Phi(50/\sqrt{t_n}) \ge 1-\Phi(50 \sqrt{N})$$ where $\Phi$ is the standard normal cdf. By the Borel zero-one law, almost surely, there are infinitely many $t_k \in [1/N, 2/N]$ with $\xi_{t_k} > 50$. This is true for every $N$, so we have shown that $\limsup_{t \to 0} \xi_t \ge 50$ almost surely. Indeed, you could similarly show that $\limsup_{t \to 0} \xi_t = +\infty$ and $\limsup_{t \to 0} \xi_t = -\infty$.
To prove that time inverted Brownian motion is continuous at 0, you are going to have to use not only the one-dimensional distributions of the random variables $B_t$, but also their joint distributions, as well as the fact that they depend continuously on $t$.
To see that the latter will be needed, let $W_t$ be a Brownian motion, and suppose $U_n$ is an iid sequence of $U(0,1)$ random variables on the same probability space, independent of $W$. Set $$\xi_t(\omega) = \begin{cases} 50, & t = U_n(\omega) \text{ for some $n$} \\ W_t(\omega), & \text{otherwise.} \end{cases}$$ For any finite set $\{t_1, \dots, t_k\}$, there is zero probability that any of the $t_i$ equal any of the $U_n$, so we have $P(\xi_{t_1} = W_{t_1}, \dots, \xi_{t_k} = W_{t_k}) = 1$; that is, $\xi$ is a modification of the Brownian motion $W$. In particular $(\xi_{t_1}, \dots, \xi_{t_k})$ has the same joint distribution as $(W_{t_1}, \dots, W_{t_k})$. However, it is also true that almost surely, the set $\{U_n(\omega)\}$ is dense in $[0,1]$ and in particular has 0 as an accumulation point, so $\limsup_{t \to 0} \xi_t = 50$ almost surely.