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if I have the following dataset and I'm interested to find the probability of it raining today given the last 2 days were raining. It seems like I could use Bayesian model to get the probability to forecast that.

I'm not sure how do I start.

1) the event that A|B --> 3 occurrences, day 3, day 4 and day 15. Does that mean that, $P(A|B) = \frac{3}{20}$? If it is, given the dataset that I have do I still need the following Bayes formula to obtain $P(A|B) $?

$$P(A|B) =\frac{P(B| A)\cdot P(A)}{P(B)}$$

A = rain on 3rd day

B = rain previous 2 days

P(A|B) = probability of 3rd day raining given the last 2 days raining

P(B|A) = probability of the last 2 days raining given the 3rd day raining, (this doesn't make intuitive sense) If I'm to obtain this probability from the dataset, how do I do that?

P(A) = probability of 1 day raining

P(B) = probability of 2 consecutive days raining

2) I don't know if the above approach is the right way to forecast the chances of 3rd day raining given that the previous 2 days were raining. Having said that, is there any other stochastic models that I could use?

Note: I'm currently studying Kai Lai Chung's A Course in Probability Theory, and bought the book, Probability and Stochastic Modeling by Vladimir Rotar to self-study

$$\begin{array}{|c|c|} \hline Day & Rain \\ \hline 1&rain\\ \hline 2&rain\\ \hline 3&rain\\ \hline 4&rain\\ \hline 5&no\\ \hline 6&rain\\ \hline 7&no\\ \hline 8&no\\ \hline 9&no\\ \hline 10&rain\\ \hline 11&rain\\ \hline 12&no\\ \hline 13&rain\\ \hline 14&rain\\ \hline 15&rain\\ \hline 16&no\\ \hline 17&no\\ \hline 18&rain\\ \hline 19&rain\\ \hline 20&no\\ \hline \end{array}$$

tkj80
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    I think the simplest approach is to consider the fact that that there were 7 instances of rain 2 days in a row and that 3 of those instances were followed by a third day of rain while the other 4 did not. I would say $P(A|B)=3/7$. – Laars Helenius Mar 13 '18 at 23:41

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The Bayes' approach of finding $\mathsf P(A\mid B)$ by calculating it as $\mathsf P(B\mid A)\mathsf P(A)/\mathsf P(B)$ is only helpful if you have ways to evaluate $\mathsf P(B\mid A)$, $\mathsf P(A)$, and $\mathsf P(B)$ that are easier than just evaluating $\mathsf P(A\mid B)$ directly.

All you have is the data.   You can use a frequentist approximation to evaluate $\mathsf P(B\mid A)$, $\mathsf P(A)$, and $\mathsf P(B)$, but you may as well just use it to evaluate $\mathsf P(A\mid B)$.

There are 19 blocks of three consecutive days.   10 of these have rain on their third day (event $A$).   7 of them have rain on the first and second days (event $B$).   3 of them have rain on all three days (event $A\cap B$).

Graham Kemp
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