Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Exercise: Suppose that whether or not it rains on a given day depends on the weather conditions of the previous day. If it rains today, then it will rain tomorrow with probability $0.7$; if it doesn't rain today, then it will rain tomorrow with probability $0.4$. Given that it rained today, compute the probability that it will rain the day after tomorrow.

An informal solution: there are two different "ways" in which it can rain the day after tomorrow. It either (A) rains tomorrow and then rains the day after tomorrow or (B) it doesn't rain tomorrow and then rains the day after tomorrow. Since these two possibilities are disjoint (i.e., they cannot be realized at the same time), it follows that the probability that it will rain the day after tomorrow is given by the sum of the probabilities that each of these two will happen: $$P(A\cup B) = P(A) + P(B) = (0.7)(0.7) + (0.3)(0.4) = 0.61$$

The problem: I have tried to formalize this solution by determining the probability space we're working with and formally writing exactly what the assumptions we are making are. I've been able to do this with other execises, but apparently this one isn't so straight-forward. One of my attempts was to put everything in terms of conditional probabilities, but that would force me to translate the assumptions of the exercise in a not-so-natural way:

My try (conditional probabilities): We can think of the sample space as $\Omega = \{f:\mathbb{Z}\to \{0, 1\}\}$. For all $n\in\mathbb{Z}$ the $n$-th number (i.e., $f(n)$) corresponds to the weather conditions on the $n$-th day (the $0$-th day is today, the first is tomorrow, the $-1$-th is yesterday, and so on): if it's zero, that means it didn't/won't rain that day; if it's one, that means it did/will rain that day. There's some unknown probability measure $P:\mathcal{P}(\Omega)\to\mathbb{R}$. Then $(\Omega, \mathcal{P}(\Omega), P)$ is a probability space.

If for each $n\in\mathbb{Z}$, $C_n$ is the event "it rained or will rain on the $n$-th day", then: $$P(C_2|C_0) = P(C_2\cap C_1|C_0) + P(C_2\cap C_1^C|C_0) = P(C_1|C_0)P(C_2|C_1\cap C_0) + P(C_1^C|C_0)P(C_2|C_1^C\cap C_0) =^{(!?)} (0.7)(0.7) + (0.3)(0.4) = 0.61$$

Note that in the second to last equality we must assume that $P(C_2|C_1\cap C_0) = 0.7$ and $P(C_2|C_1^C\cap C_0) = 0.4$, but I think this wouldn't be true to the information we were given, which is more like $P(C_{n+1}|C_n) = 0.7$ and $P(C_{n+1}|C_n^C) = 0.4$ for all $n\in\mathbb{Z}$.

Can anybody see a more faithful formalization for this exercise and its solution?

share|improve this question
add comment

2 Answers

up vote 2 down vote accepted

Your post in fact proposes a correct formalization of the question, and the only remaining question is whether one can use $P[A_{n+1}\mid A_n,A_{n-1}]=P[A_{n+1}\mid A_n]$ for every $n$, where $A_k$ depends on $f(k)$ only.

Rigorously speaking, this cannot be deduced from the hypotheses which were given to you and one could imagine models compatible with these hypotheses and such that the conditional probability of $A_{n+1}$ does depend on $A_{n-1}$, for example.

However, the formulation of the exercise corresponds to the implicit assumption that the probability of $A_{n+1}$ conditioned by any sequence $(A_k)_{k\leqslant n}$ of events such that $A_k$ depends on $f(k)$ only, depends on $A_n$ only. This is called the Markov property and, with this tacit convention, your solution is complete (and correct).

share|improve this answer
    
Hmm, this exercise is actually from a course on Stochastic Processes I'm taking. I wonder if our professor wants to surprise us later on saying we assumed this property without noticing. Won't be much of a surprise to me now, hehe. Thanks, Did! :) –  TuringMachine Apr 7 '13 at 18:45
add comment

Your intuitive approach can be formalized using a Bayesian Network

share|improve this answer
    
Or not. $ $ $ $ –  Did Apr 3 '13 at 9:09
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.