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Is this is a Poisson process?
A Special Flood Hazard Area at high risk for flooding has a 1 in 4 chance of flooding during a 30-year mortgage.
What is the probability of flooding for during a general duration, t, say 5 years or 70 years?
Hopefully someone more familiar with probability can help me out.
Thanks.
That's the expectation value you're showing, right?
To find the probability of at least one occurrence, I need sum the individual cumulative distribution functions for n=1,2, . . . inf, right?
If so, I'll just sum over the few lowest n to get a general idea of risk.
Cogito: You've blocked me so I can't comment underneath your answer, but that use of the compliment was what I was originally hoping to do. I didn't realize that P(0) gave a valid probability! Thanks! I'll give you BA when I can.
Actually, apparently I can't award you BA because I'm blocked. That's a curious choice on Yahoo's part, but I will at least refrain from awarding Betty BA in the hopes that you will get BA eventually . . .
2 Answers
- 4 years agoFavorite Answer
Yes, this is a Poisson process. You can always recognize them easily. Whenever a discrete event (number of flooding events, for example) occur over a continuous domain variable (in this case time). That should immediately jump out to you.
That doesnt mean you cant approximate this process with a discrete binomial or bernoilli process, or a continuous normal process, etc. But these are only approximations for the truth, albeit very good approximations and usually much easier to solve.
During a 30-year mortgage, you have a 1 in 4 chance of flooding. That means your expected number of floods during that time is lambda = 0.25.
The expected number of floods in a 5 year period is 5/30 of lambda = 0.0416666, and the expected number of floods in a 70 year period is 70/30 of lambda = 0.58333.
The Poisson process has a probability of P(k) = e^-lambda * lambda^k / k!
We want to find the probability of no flooding events. k=0.
P(0) = e^-(0.0416666) * (0.0416666)^0 / 0! = 0.959189
P(0) = e^-(0.58333) * (0.58333)^0 / 0! = 0.558035
Now we just find the compliment of these values, 1-p, for the probability of any amount of flooding.
So in a 5-year period there is a 1 - 0.959189 = 0.040811, or roughly 4% chance of flooding.
In a 70-year period there is a 1 - 0.558035 = 0.441965, or roughly 44% chance of flooding.
If you do the same line of reasoning for a 30 year period, you get a roughly 22% chance of flooding, not the 25% chance implied by the 1 in 4 statement. Its counter intuitive.
Really, I dont like the "1 in 4" statement because its misleading, and it begs the question how the statistic/paramter was figured in the first place. 1 in 4 what? Houses in the area? Was it really counted over a 30 year period? Has the house flooded once in the last 120 years, because you might have bigger problems than flood damage at this point. What data is this "1 in 4" based on because it might be a completely different, completely irrelevant statistic that has nothing to do with that particular houses chances of flooding. Had they simply said lambda, the mean expected number of floods was 0.25 over the 30-year period, Id be less inclined to ask about the number.
- Anonymous4 years ago
a