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Thread: Probability of Exceed and Return Period

  1. #1
    Super Member sakonpure6's Avatar
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    Probability of Exceed and Return Period

    I am in disagreement with a statement that a part time teacher made. This is more of a theoretical question (no calculations).

    Suppose I am analyzing a data set of wind speed of monthly data.
    I fit a distribution to my data and I determine the CDF function.

    Now, I am asked what is the 50-year design wind speed?
    The question is: how do you translate the given return period?


    Usually, a 50 year return period corresponds to P(X>x) = 1/50 = 0.02 (regardless of measurement units, it is the definition).
    However, my part-time professor argues that since the data we are working with has units of month, we need to find: P(X>x)=1/(50*12)





    I do not agree with the latter. Once you analyse your data and devlop the CDF, units become meaningless. The concept of return period is in years and simply conveys a probability of exceedance (there are no units attached to it.)

    Can someone please advise?

    Thank you

    EDIT:
    What I am trying to say is that the frequency of the data measurement has nothing to do in determining the probability of exceedance for a given return period. The frequency (monthly, hourly, yearly) only affects how many data points you have and as a result how accurate your CDF is.
    Last edited by sakonpure6; Jul 17th 2019 at 03:09 PM.
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  2. #2
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    Re: Probability of Exceed and Return Period

    I am puzzled at your saying "Usually, a 50 year return period corresponds to P(X>x) = 1/50 = 0.02 (regardless of measurement units, it is the definition)." Where did you get that "50" from? Was it not from "50 years"? If so, then 50 does have units, of years, and so 1/50 has the units "per year". Suppose the problem had said "600 month return period". Would you have used P(X>x)= 1/600, even though 600 months and 50 years are exactly the same time?
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