TITLE: Limit of Probability and Probability of Limit
QUESTION [5 upvotes]: Let $\{x_k\}$ and $x^*$ be a sequence and a point in $\mathbb{R}^n$, respectively. Can we conclude that
$$\lim_{k\to\infty} \mathrm{Prob}(x_k=x^*)=1$$
and
$$\mathrm{Prob}(\lim_{k\to\infty} x_k=x^*)=1$$
are equivalent or that one implies the other?
I think the first one implies the second, but not vice versa since $x^*$ might not be part of the sequence.

REPLY [3 votes]: Imagine you had a bunch of sample trajectories of $x_k$—an infinite bunch, in fact—and further imagine that we treat $k$ as a kind of time index.
Intuitively, the first statement says that eventually (i.e., as $k$ increases without bound), almost all of the trajectories at any given moment will be at $x^*$—not just arbitrarily close to $x^*$, but equal to it.  However, it is possible for the exceptions to change from one $k$ to the next, with each trajectory momentarily leaving $x^*$ infinitely often, so that none of the trajectories actually has a limiting value.
$$
\begin{array}{|c|c|c|c|c|c|c|c|c|c|c|} \hline
\text{Trajectory} & k = 1 & k = 2 & k = 3 & k = 4 & k = 5 & k = 6 & k = 7 & k = 8 & k = 9 & \ldots \\ \hline
A & 1 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & \ldots \\ \hline
B & 1 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & \ldots \\ \hline
C & 1 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & \ldots \\ \hline
D & 1 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & \ldots \\ \hline
E & 1 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & \ldots \\ \hline
F & 0 & 1 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & \ldots \\ \hline
G & 0 & 1 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & \ldots \\ \hline
H & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & \ldots \\ \hline
I & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 1 & \ldots \\ \hline
J & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & \ldots \\ \hline
\end{array}
$$

In contrast, the second statement says that almost all of the trajectories will eventually get arbitrarily close to $x^*$.  However, it may happen that none of the trajectories ever reaches $x^*$.
$$
\begin{array}{|c|c|c|c|c|c|c|c|c|c|c|} \hline
\text{Trajectory} & k = 1 & k = 2 & k = 3 & k = 4 & k = 5 & k = 6 & k = 7 & k = 8 & k = 9 & \ldots \\ \hline
A & 1 & 1/2 & 1/3 & 1/4 & 1/5 & 1/6 & 1/7 & 1/8 & 1/9 & \ldots \\ \hline
B & 1 & 1/2 & 1/4 & 1/8 & 1/16 & 1/32 & 1/64 & 1/128 & 1/256 & \ldots \\ \hline
C & 1 & -1/3 & 1/5 & -1/7 & 1/9 & -1/11 & 1/13 & -1/15 & 1/17 & \ldots \\ \hline
D & 1 & 1/2 & 1/2 & 1/4 & 1/4 & 1/4 & 1/4 & 1/8 & 1/8 & \ldots \\ \hline
E & 1 & 1/2 & 1/3 & 1/5 & 1/7 & 1/11 & 1/13 & 1/17 & 1/19 & \ldots \\ \hline
F & 1 & 1/2 & 1/3 & 1/5 & 1/8 & 1/13 & 1/21 & 1/34 & 1/55 & \ldots \\ \hline
G & 1 & 2/3 & 1/2 & 1/3 & 1/4 & 1/6 & 1/8 & 1/12 & 1/16 & \ldots \\ \hline
H & 1 & 1/2 & 1/6 & 1/12 & 1/20 & 1/30 & 1/42 & 1/56 & 1/72 & \ldots \\ \hline
I & 1 & 1/4 & 1/10 & 1/20 & 1/35 & 1/56 & 1/84 & 1/120 & 1/165 & \ldots \\ \hline
J & 1 & 1/4 & 1/9 & 1/16 & 1/25 & 1/36 & 1/49 & 1/64 & 1/81 & \ldots \\ \hline
\end{array}
$$

So neither of them implies the other.