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Expotential distribution

  1. Nov 11, 2014 #1
    1. The problem statement, all variables and given/known data
    Let x be an exponential random variable with lamda = .5

    P(x=5|2<x<9} = 0 since exponential is continuous => probability of any single number = 0
    Calculate E[x| 2<x<9] = integral from 2 to 9 of (x* .5*exp(-.5*x)) dx ? is this true or is there something wrong because of the memoryless property of exponentials?

    3. The attempt at a solution
     
  2. jcsd
  3. Nov 11, 2014 #2

    haruspex

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    No, that equation is not quite right. Imagine reducing the given range to a very narrow one. The expected value should then be some value in that range, but your integral will tend to zero. What have you forgotten?

    I don't understand why you think it would violate the memorylessness though. The conditional is not saying anything about the outcome of a previous trial; it's giving you information about the outcome of the present trial.
     
  4. Nov 11, 2014 #3

    RUber

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    P(5|2<x<9) often is referring to the PDF evaluated at 5 divided by the (cdf at 9 - cdf at 2)
    The expected value integral looks okay to me.
     
  5. Nov 11, 2014 #4

    HallsofIvy

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    "Give that 2< x< 9" means that the probability that x is beween 2 and 9 is 1. The "conditional probability" that "[itex]a\le x\le b[/itex]' where, of course, [itex]2< a< b< 9[/itex], is [tex]\frac{\int_a^b e^{-.5t}dt}{\int_2^9 e^{-.5t}dt}[/tex]
     
  6. Nov 11, 2014 #5

    haruspex

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    Yes, the integral is ok in itself, but
    Exactly - the division is missing.
     
  7. Nov 11, 2014 #6

    Ray Vickson

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    Because of the upper bound ##X < 9## the memoryless property does not apply in full. However, it applies in part. Conditioned on ##X > 2##, the (conditional) distribution of ##X## is the same as that of ##2+Y##, where ##Y \sim \text{Expl}(\lambda = 0.5)##. We can thus write
    [tex] E(X | 2 < X < 9) = 2 + E(Y|Y < 7) = 2 + E(X | X < 7)[/tex]
    Whether or not you think this is useful is for you to decide.
     
    Last edited: Nov 12, 2014
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