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Math Help - Show that Y has an exponential distribution from f(x)??

  1. #1
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    Question Show that Y has an exponential distribution from f(x)??

    hey guys, I'm just wondering if you guys can help me with this question, thank you in advance.

    A random variable, X, has a probability density function (p.d.f.) given by

    f(x)=2xe^(-x^2)
    where x>0

    Y is a random variable such that Y = X^2.

    Show that Y has an exponential distribution and state its mean.

    thanks.
    Last edited by hazeleyes; January 3rd 2011 at 03:50 PM.
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  2. #2
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    Hi there hazeleyes,

    Here's a kick off...

    \displaystyle f(Y) = f(X^2) = 2(X^2)e^{-(X^2)^2}

    for the mean \displaystyle E(Y) = \int_{-\infty}^{\infty}Y\times f(Y)~dY
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  3. #3
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    Quote Originally Posted by pickslides View Post
    Hi there hazeleyes,

    Here's a kick off...

    \displaystyle f(Y) = f(X^2) = 2(X^2)e^{-(X^2)^2}

    for the mean \displaystyle E(Y) = \int_{-\infty}^{\infty}Y\times f(Y)~dY
    Sorry, but that's not correct.

    There are several approaches that can be taken.

    The OP has carelessly neglected to mention that the support of X is x \geq 0 (so that f(x) = 0 for x < 0). The support of Y will be y \geq 0.

    The most basic approach to finding the pdf of Y is to calculate the cdf of Y and then recall that the pdf is the derivative.

    \displaystyle cdf = G(y) = \Pr(Y \leq y) = \Pr(X^2 \leq y) = \Pr(-\sqrt{y} \leq X \leq \sqrt{y}) = \int_0^{\sqrt{y}} 2 x e^{-x^2} \, dx.

    Differentiating this is simple - either integrate (use a substitution) directly and then differentiate, or use the chain rule and the Fundamental Theorem of Calculus.

    You should get \displaystyle pdf = \frac{dG}{dy} = g(y) = e^{-y} for y \geq 0 and zero elsewhere. Calculating the mean of Y is trivial.


    Some trivia: The random variable X follows a Weibull distribution (http://en.wikipedia.org/wiki/Weibull_distribution) with parameters k = 2 and \lambda = 1. Since Y = X^2 = X^k it is well known that Y follows an exponential distribution.
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  4. #4
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    Thanks alot guys!!
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  5. #5
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    Re: Show that Y has an exponential distribution from f(x)??

    Quote Originally Posted by mr fantastic View Post

    The most basic approach to finding the pdf of Y is to calculate the cdf of Y and then recall that the pdf is the derivative.

    \displaystyle cdf = G(y) = \Pr(Y \leq y) = \Pr(X^2 \leq y) = \Pr(-\sqrt{y} \leq X \leq \sqrt{y}) = \int_0^{\sqrt{y}} 2 x e^{-x^2} \, dx.

    Differentiating this is simple - either integrate (use a substitution) directly and then differentiate, or use the chain rule and the Fundamental Theorem of Calculus.

    You should get \displaystyle pdf = \frac{dG}{dy} = g(y) = e^{-y} for y \geq 0 and zero elsewhere. Calculating the mean of Y is trivial.
    Apologies in advance as I am not yet familiar with the Weibull distribution, and also I haven't yet learned to use LaTex...

    I follow mr. fantastic's logic up to the point of G(y) = P(-y1/2 <= X <= y1/2)

    But shouldn't this mean that G(y) = the integral from 0 to y1/2 of 2e-2x dx = 1 - e-2y1/2?

    and therefore, g(y) = dG/dy = y-1/2 e-2y1/2?
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  6. #6
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    Re: Show that Y has an exponential distribution from f(x)??

    Sorry, regarding my last post I realized I did not read the original post carefully enough - I thought the original pdf was f(x) = 2e-2x. In the process however, I think I learned that where X~exp(lambda), Y= X2 is also exp(lambda). Is that correct?
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