0
$\begingroup$

Practical situation: I’ve got 120 days of data collected during rainy season. On an average it rained on 52.5 of those days.

  1. What is the probability of it raining at least once in 30 days?
  2. If it rains at least once in 30 days, then what is the probability that it continues to rain for 4 days in a stretch in those 30 days?

My Answer for (1): I considered a Poisson process with $\lambda = 52.5/120 = 0.4375$, calculated the non-occurrence of an event (in this case "no rain") in 30 days and subtracted that value from 1 to get .9999 as the probability of it raining in 30 days.

Please advise if I'm on the right track and also how do we go about part (2) of the question.

$\endgroup$
  • 1
    $\begingroup$ How do you get rain on 0.5 days? Do you have multiple 120 day periods, or are you counting the number of rainfall events per day? $\endgroup$ – jbowman May 31 '12 at 12:57
5
$\begingroup$

This is not a complete answer, but also too long for a comment and I do hope it answer at least parts of the question.

The Poisson distribution seems to be the wrong way to answer a, since it would theoretically allow for it to rain more than 30 times in 30 days and you seem to have just collected whether it rained at least once for each day. A binomial distribution would be more appropiate.

Another problem is that you basically assume that the weather on each day is independent of the day before. Weather has complex cycles and is certainly autocorellated.

I don't see any way to get more than an approximate answer, but it might help to at least condition the probability of rain on the weather of the day prior. This would be straightforward to implement (ask if necessary) for question 1 and more difficult for question since runs of events are always tricky.

$\endgroup$
  • 1
    $\begingroup$ hmmmm you think it would do any good if we had data from previous year around the same time period as current??? To be frank I feel there is a lot of uncertainity involved and we are making a lot of assumptions here $\endgroup$ – Vinay May 31 '12 at 11:58
  • $\begingroup$ @jbowman: a day is divided into 4 intervals of 6 hrs each and then rainfall is recorded. $\endgroup$ – Vinay May 31 '12 at 14:44
2
$\begingroup$

Erik is right that Poisson is not the right model to answer your question. If the estimate of 52.5% can be assumed to be the probability of rain on any given day and tomorrow's chance of rain is independent of today and the past then the binomial distribution with parameter n = 30 and p=0.525 gives you the distribution for the number of days of rain in the next 30. Regarding your question the answer is then p=1- P(X=0) where X is the number of days of rain in the next 30. However as Erik points out this depends on the assumption that P( rain tomorrow | it rains today) = P( rain tomorrow | it does not rain today). In reality because storm systems can stay in your area for more than one day these two probabilities are likely not going to be equal. Given that you have collected data on rainy days over a large number of successive days means that you can estimate both of these probabilities to check the independence assumption and if these are different you can use these estimates to calculate the probability that you are seeking.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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