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Math Help - simulation variable

  1. #1
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    simulation variable

    HI please can someone cheeck this problem I tried to do?

    Suppose X is the waiting time, in seconds, for a certain particle's emmission,
    f(x)=0.1e^(-0.1x), x>0
    0 elsewhere.

    You are simualting a random sample of 2 values of X. You have simulated the following 2 values of Y, where Y has a (0,1) uniform pdf; y1=0.1942, y2=0.6209.
    Find the 2 simulated values of X, x1 and x2.

    Here what I did.
    f(x)=0.1e^(-0.1x)

    --> F(x)=1-e^(-0.1x)
    let u =F(x)=1-e^(-0.1x)

    x= -[ln(1-u)]/0.1 (1)
    then for x1 take y1=0.1942=u , replace in the (1)
    x1=2.1591

    I do the same for x2. you get x2=9.6996

    Is it the correct way to do it?
    Thank you.
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by braddy View Post
    HI please can someone cheeck this problem I tried to do?

    Suppose X is the waiting time, in seconds, for a certain particle's emmission,
    f(x)=0.1e^(-0.1x), x>0
    0 elsewhere.

    You are simualting a random sample of 2 values of X. You have simulated the following 2 values of Y, where Y has a (0,1) uniform pdf; y1=0.1942, y2=0.6209.
    Find the 2 simulated values of X, x1 and x2.
    Here what I did.
    f(x)=0.1e^(-0.1x)

    --> F(x)=1-e^(-0.1x)
    let u =F(x)=1-e^(-0.1x)

    x= -[ln(1-u)]/0.1 (1)
    then for x1 take y1=0.1942=u , replace in the (1)
    x1=2.1591

    I do the same for x2. you get x2=9.6996

    Is it the correct way to do it?
    Thank you.
    Yes (though there is some ambiguity in this as you are using more operations
    than necessary, as x=ln(u)/0.1 will also give a negative exponential random
    number with the right rate constant, but different from what you have even
    though it starts with the same u).

    RonL
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