Time-ordering in Bayes' formula

Question

From a physicist's point of view, I was just curious if Bayes' formula has a "time-ordering" / causality built into it? I.e., if I wanted to calculate the following using Bayes' theorem: $P(A(t_2) \mid A(t_1))$, that is, the probability of observing event $A$ at $t_2$ given that it was observed at $t_1$, Bayes' theorem would say:

$$P(A(t_2) \mid A(t_1)) = \frac{P(A(t_1)\mid A(t_2)) P(A(t_2))}{P(A(t_1))},$$

but this would violate causality, wouldn't it, since it is requiring one to know what happened at $t_2$ without observing the event at $t_1$ first. Or, rather, at least the likelihood factor is saying the probability of observing $A(t_1)$ given $A(t_2)$, where $t_2 > t_1$. Is this implying that Bayesian "events" are time-symmetric? How would one interpret something like this?

Thanks.

Answer 1

There is no causation.   Probability is a measure of expecation for an event's occurance.   This expectation may be predictive or forensic; "will it occure" or "has it occured".   Either way, it is just a question of what information we have about it.

If I roll a die the probability that it is a six is $1/6$; but wait, I have already rolled the die ... so what is the probability that it was a six?

There is symmetry because Bayes' rule is based on the definition of conditional probability.   Conditional probability is the measure under additional information.

$$\begin{align}\mathsf P[A(t_1)\mid A(t_2)]~\mathsf P[A(t_2)] &= \mathsf P[A(t_2)\mid A(t_1)]~\mathsf P[A(t_1)] \\[1ex] & = \mathsf P[A(t_1)\cap A(t_2)]\end{align}$$

Answer 2

There is no time involved in the definition of conditional probability, and in problems that do involve time, there is no "violation of causality" in observation of an event without observing an earlier event. We observe dinosaur skeletons without observing dinosaurs, and that does not violate causality.

Suppose an urn contains $4$ red and $6$ green marbles. Draw one at random.

• What is the conditional probability that the second one is red, given that the first one was red? We have $$ \Pr(\text{2nd is red}\mid\text{1st is red}) = \frac{\Pr(\text{1st is red & 2nd is red)}}{\Pr(\text{2nd is red})} = \frac{\binom42/\binom{10}{2}}{4/10} = \frac13 $$

• What is the conditional probability that the second is red?
  
  Here I have heard students say "Doesn't it depend on whether the first is read?". The conditional probability that the second one is red, given the color of the first one, does depend on whether the first is red. But this is not that conditional probability. If one insists on conditioning on the color of the first one, then one has \begin{align} & \Pr(\text{2nd is red}) \\[10pt] = {} & \Pr(\text{1st is red})\Pr(\text{2nd is red}\mid\text{1st is red}) \\ & {} + \Pr(\text{1st is green})\Pr(\text{2nd is red}\mid\text{1st is green}) \\[10pt] = {} & \text{etc.} \end{align} But there is no need for that. The probability that the second one is red is the same as the probability that the first is red, for the same reason.

• What is the conditional probability that the second one is red, given that the first one is red? This can be done by the same method that is used if the roles of "first" and "second" were the opposite of what they are here. However, if one insists (needlessly) of reducing the problem to conditioning on earlier events, one can write \begin{align} & \Pr(\text{1st is red}\mid\text{2nd is red}) \\[10pt] = {} & \frac{\Pr(\text{1st is red})\Pr(\text{2nd is red}\mid\text{1st is red})}{\Pr(\text{1st is red}) \Pr(\text{2nd is red}\mid\text{1st is red}) + \Pr(\text{1st is green})\Pr(\text{2nd is red}\mid\text{1st is green})}. \end{align} The result is the same.