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Math Help - Exponential Dist

  1. #1
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    Exponential Dist

    Hi guys,

    A sample of 5 iid is taken from an exponential distribution with λ =1.
    1) What is the P(smallest observation is < 0.5)?
    2) What is the P(largest observation is < 0.5)?

    I am thrown off by the smallest/largest condition.

    Many thanks in advance for any feedback.
    Linda
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  2. #2
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    P(smallest observation is < 0.5) = p(at least one < 0.5) = 1 - P(all ovservations >= 0.5)

    can you evaluate the RHS?

    The other question is similar
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  3. #3
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    Is it, using the cdf:

    =1 - e^{\lambda x}
    =1 - e^{1 x 0.5}
    =0.3934

    With the largest observation, would this be:
    P(largest observation is < 0.5) = 1 - P(all ovservations >= 0.5)
    because of the fact that the exponential distribution is memoryless??
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  4. #4
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    I did not check your arithmatic. Assuming you calculated properly 0.3934 would be the chance of one obvservation being greater than 0.5.

    the chance of 5 being great than 0.5 would be 0.3934 ^ 5

    so answer = 1-(0.3934^5)

    P(largest observation is < 0.5) = 1 - P(all ovservations >= 0.5)
    because of the fact that the exponential distribution is memoryless??
    I dont think that's right.

    p(largest < 0.5) = p(all are < 0.5)

    This would be true for any distribution. I dont think the memorylessness property is relevent here, but i could be wrong.
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  5. #5
    MHF Contributor matheagle's Avatar
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    i claim everything i say is inaccurate, but then this is inaccurate
    oh my head hurts
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  6. #6
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    Quote Originally Posted by matheagle View Post
    i claim everything i say is inaccurate, but then this is inaccurate
    oh my head hurts
    Mine too!!!
    Is it possible for you to elaborate?

    Quote Originally Posted by SpringFan25 View Post
    I did not check your arithmetic. Assuming you calculated properly 0.3934 would be the chance of one observation being greater than 0.5.
    I know you have not checked the arithmetic - the chance of one observation greater than 0.5 is 0.6065.
    So I am gathering the answer is

    so answer = 1-(0.6065^5)
    p(largest < 0.5) = p(all are < 0.5)

    This would be true for any distribution. I don't think the memoryless property is relevant here, but i could be wrong.
    If that is the case would this calculation then be:
    =(1-0.6065)^5

    Thank you for your patience
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