Multiple Applications of Bayes' Theorem If I calculate $\Pr(X \mid E_1, E_2, E_3, \dots)$ using Bayes' Theorem all at once (applying Bayes' Theorem with observation $E_1, E_2, E_3, \dots$) vs. one update at a time (i.e. starting with my prior, updating my prior with $E_1$, updating it with $E_2$, etc.), will I get the same result? How might I mathematically demonstrate this?
 A: I'll just do the case of two updates. You can generalize the argument by induction without much difficulty. 
I assume that by an "update" of the prior $P$ on an event $E$, you mean the probability measure $P_E$ defined by $P_E(\cdot) = P(\cdot \mid E)$.
We want to show that $P_{E_1 \cap E_2} = (P_{E_1})_{E_2}$, where the left-hand side is the "all at once" update, and the right-hand side is the "one at a time" update. There's no need to use Bayes' Theorem; using just the definition of conditional probability we have
$$P_{E_1 \cap E_2}(X) = P(X \mid E_1 \cap E_2) = \frac{P(X \cap E_1 \cap E_2)}{P(E_1 \cap E_2)}.$$
Now, for the one at a time case, we have
$$(P_{E_1})_{E_2}(X) = P_{E_1}(X \mid E_2) = \frac{P_{E_1}(X \cap E_2)}{P_{E_1}(E_2)} = \frac{P(X \cap E_2 \mid E_1)}{P(E_2 \mid E_1)} = \frac{P(X \cap E_2 \cap E_1)/P(E_1)}{P(E_2 \cap E_1)/P(E_1)} = \frac{P(X \cap E_2 \cap E_1)}{P(E_2 \cap E_1)} = P_{E_1 \cap E_2}(X).$$
So all at once and one at a time yield the same probability measure.
