I have realizations of two different random vectors, where one is a subset (is that proper terminology here?) of the other $$s^t = (x_1,x_2,x_3,\dots x_\tau, x_{\tau +1},\dots x_t)$$ and $$ s^\tau = ({x_1,x_2,x_3,\dots ,x_\tau})$$ clearly $\tau \leq t$. Now Bayes' rule (for sets, at least) says $$ P(A\mid B) = P(B\mid A)\frac{P(A)}{P(B)}$$ My question is, would $$ P(s^t\mid s^\tau) = \frac{P(A)}{P(B)}$$ since $P(B\mid A) = P(s^\tau \mid s^t)$ and $s^\tau \text{ is a subset(?) of } s^t$? If I think of $s^t$ and $s^\tau$ as information then doing so makes (intuitive) sense to me.
Note: $P(A\mid B)$ is used to indicate conditional probabilities.
Can I even use Bayes' rule as above, here?
Here is an example since I am not sure that my terminology is correct: Let us have a coin-flipping experiment. Let the two random variables be flipping a coin $t$ times and flipping the coin $\tau \leq t$ times. So we have $X_1 = (x_1, x_2, \dots x_t)$ and $Y_2=(x_1,x_2,\dots, x_\tau)$, where $x_n$ is the result of the $n$th coin flip. Let $0$ denotes tails, $1$ heads.
Let $\tau = 3$, $t=5$, Lets say that we get $$X_1 = (0,0,1,1,1), X_2=(0,0,1)$$
I am asking, although in more general terms, whether $$ P(X_1 \mid X_2) = \frac{P(X_1)}{P(X)}$$ since I believe $P(X_2\mid X_1) =1$
My confusion on the terminology I think is because I'm not sure what to refer to the realization of the coin flips (the random variables). They are no longer random variables since they are realizations, not functions. I was thinking set, but order matters (since $x_n$ refers to the $nth$ coin flip), so perhaps sequence, but I don't know if probabilities on sequences make sense? Thanks
Let's be more explicit and things will be clearer. There is a probability distribution over outcomes on $t$ flips. For each possible outcome of the first $\tau$ flips, $x$, there is a set of outcomes on $t$ flips whose first $\tau$ flips are $x$. This set is an event, which you may denote $s^\tau$ (which depends on $x$). For any outcome on $t$ flips, $y$, $\{y\}$ is also an event; denote it $s^t$. Since in your case $y$ starts with $x$, $s^\tau \cap s^t = s^t$ so $\Pr[s^\tau \wedge s^t] = \Pr[s^t]$.
Whenever we have two events, we may use Bayes' law on them. Because all events have nonzero probability, we have:
$$\Pr[s^t | s^\tau] = \frac{\Pr[s^\tau \wedge s^t]}{\Pr[s^\tau]} = \frac{\Pr[s^t]}{\Pr[s^\tau]}.$$