Is the probability vector of a random walk deterministic? We have a Lazy random walk on the n-hypercube. i.e. the graph's vertices are $\{0, 1\}^n$ and and two vertices are adjacent if they differ in exactly one coordinate. At each iteration w.p. $\frac{1}{2}$ we either stay in the same vertex or choose uniformly a neighbor to move to. Denote by $X_t$ the random variable with the location at time $t$ and by $x_t$ the distribution vector of $X_t$.
We can view this process as choosing each iteration at random an entry in $X_t$ and then w.p.  $\frac{1}{2}$  flipping its value. Let $T$ be the RV with the iteration number where all entries have been selected (a coupon collector problem), then we are guaranteed that for all $t>T$ we have that $X_t$ is distributed uniformly, and therefore $x_t=(\frac{1}{n}, \ldots, \frac{1}{n})$.
Here's my problem. If we set $x_0$ as the initial distribution, then $x_t$ is deterministic (if $P$ is the transition matrix then $x_t=x_0P^t$). How does that fit with the analysis above? (where $x_t$ is dependent on a random variable $T$)

Indeed, conditional on $t\gt T$, the distribution is uniform, and indeed, without this conditioning, it is $x_0P^t$. There's no contradiction because the one distribution is conditional and the other one isn't.
Perhaps the impression of a contradiction arises because when you write “for all $t\gt T$” you think of $T$ as a certain time after which the distribution changes. But $T$ isn't a certain time; it's a random variable.
