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Math Help - distribution function

  1. #1
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    distribution function

    Hi people, I am new here and need a help !

    Actually this question is taken from home work in the "statistical inference" course, but it's the second week of the course, so the question is still about probability, so I hope it's ok in this forum...

    Let X be a random variable with a probability function f:

    f(x;alpha)=alpha*e^x*e^(-alpha*e^x)

    ( yes, it's e^(e^x).... )

    A. find the distribution of the random variable Y=exp(X).
    B. Let X1,...Xn be a random sample from this distribution.
    calculate the distribution of the random variable:
    2alpha*sigma(exp(xi))

    I failed in both section A and B....
    I have passed both "introduction to probability" and "distribution theory", so it's not an easy question, I think.
    I hope anyone here can help me, I am trying to solve it for 2 days now !

    It's great to find this forum, I will do my best to help other people as well.

    is anyone here know a good website with question and answers about statistical inference ? ( advanced inference, what's called statistical theory )

    Thanks a lot to all of you that try to help !!
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by WeeG
    Let X be a random variable with a probability function f:

    f(x;alpha)=alpha*e^x*e^(-alpha*e^x)

    ( yes, it's e^(e^x).... )

    A. find the distribution of the random variable Y=exp(X).
    Let X be a RV with pdf p(x), and let g(x) be a monotonic increasing function
    of x. Then the pdf q(y) of RV Y=g(X) is:

    <br />
q(y)=p(g^{-1}(y))\frac{dx}{dy}<br />
.

    This can be quite easily demonstrated by comparing

    \mbox{prob}(y_0<y<y_0+\delta y) \approx q(y_0) \delta y

    with

    \mbox{prob}(x_0<x<x_0+\delta x) \approx p(x_0) \delta x.

    In this case y=\exp(x), so \frac{dy}{dx}=y which maps the real line
    onto the positive half line, and

    <br />
f(x,\alpha)=\alpha e^x e^{-\alpha e^x}<br />
.

    So:

    <br />
q(y)=p(\ln(y))/y=\alpha e^{-\alpha y},\ y \ge 0<br />
,

    which is the (negative)exponential distribution.

    RonL
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  3. #3
    Grand Panjandrum
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    Quote Originally Posted by WeeG

    B. Let X1,...Xn be a random sample from this distribution.
    calculate the distribution of the random variable:
    2alpha*sigma(exp(xi))
    What we have shown in part A is that Y has an exponential distribution.
    So:

    <br />
Z=2 \alpha \sum_1^n e^{X_i}<br />

    where the X_is are iid RVs with the given distribution, then:

    <br />
Z=2 \alpha \sum_1^n Y_i<br />

    where the Y_is are iid RV with an exponential distribution, so
    the sum has a Gamma distribution with appropriate parameters (standard
    result).

    RonL
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