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Math Help - Functions of one random variable question

  1. #1
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    Functions of one random variable question

    I have a problem as follows:
    Let X have the pdf f(x)= θ * e( - θ x), 0 < x < infinity


    find the pdf of Y = ex

    I've gone about this the way I normally do for these problems.
    I have
    G(y) = P(X < ln y) = integral of
    θ * e( - θ x) dx from 0 to ln y = 1 - y-θ


    So then I need g(y) = G'(y) of Y? which would be
    θ * e( - θ (ln y)??

    or am I missing something super-duper obvious?

    Thanks
    Last edited by SpiffWilkie; November 8th 2012 at 10:30 AM.
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  2. #2
    MHF Contributor Siron's Avatar
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    Re: Functions of one random variable question

    Note that the random variable X \sim \mbox{exp}(\theta) (exponential distribution) where the distribution function F is given by F_{X}(x) = 1-e^{-\theta x} (for x \geq 0). We can compute the distribution function of the random variable Y=e^X as follows: F_{Y}(x) = P(Y \leq x) = P(e^X \leq x) = P(X \leq \ln(x)) = F_{X}[\ln(x)] therefore
    F_{Y}(x) = 1-e^{-\theta \ln(x)}. We can find the pdf of Y (let's call it g(x)) by calculating the first derivative of the distribution function, hence g(x) = \frac{dF_{Y}(x)}{dx} = \frac{d}{dx}\left[1-e^{-\theta \ln(x)}\right] = \left(\frac{\theta}{x}\right)e^{\theta \ln(x)}
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