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Math Help - laplace cdf integration

  1. #1
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    laplace cdf integration and kurtosis

    Hello,
    given the laplace distribution. f(x)=0.5*e^(-|x|)
    How do you integrate in order to find the cdf for x>0 (I have no problem finding the solution for x<0)

    The solution is 1-0.5*e^(-x)

    I would be very grateful for a detailed solution or a link to a website were I could find it.

    Thanks in advance!
    Last edited by Weezphili; June 19th 2011 at 06:15 AM.
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  2. #2
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    Re: laplace cdf integration

    Hello,

    Well the best way is to see that if X follows a Laplace distribution, so does -X, since the pdf is symmetric over 0... Then for x>0, P(X\leq x)=P(X<x)=P(-X>-x)=P(X>-x)=1-P(X\leq -x)=1-F_X(-x)
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    Re: laplace cdf integration

    Thanks a lot for your help, your answer is very clear,

    another thing: (I don't know whether I should start a new thread for this, I changed the title of the original post to make it clear that there is a 2nd part to the original question)

    how do I derive the kurtosis of the laplace function using an integral of the pdf? I will have to integrate by parts, but how?
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    Re: laplace cdf integration

    Quote Originally Posted by Weezphili View Post
    Thanks a lot for your help, your answer is very clear,

    another thing: (I don't know whether I should start a new thread for this, I changed the title of the original post to make it clear that there is a 2nd part to the original question)

    how do I derive the kurtosis of the laplace function using an integral of the pdf? I will have to integrate by parts, but how?
    Well the kurtosis is known to be E\left[\left(\frac{X-\mu}{\sigma}\right)^4\right]
    Here, the "location" is \mu=0 and the "scale" is b=1. But we know that the variance is \sigma^2=2b^2=2, hence \sigma=\sqrt{2}

    So the kurtosis will be \frac12 E\left[X^4\right]=\frac12 \int_{\mathbb R} x^4 f_X(x) ~ dx=\frac14 \int_{\mathbb R} x^4 e^{-|x|} ~dx=\frac 12 \int_0^\infty x^4 e^{-x} ~dx (symmetry over 0).

    You can do 3 (I think) successive integrations by parts or you can simply recognize the gamma function, which is much easier :P
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    Re: laplace cdf integration

    Quote Originally Posted by Moo View Post
    You can do 3 (I think) successive integrations by parts or you can simply recognize the gamma function, which is much easier :P
    It's exactly with these integrations by parts that I am having trouble
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    Re: laplace cdf integration

    Let u'=e^{-x} and v=x^4
    Every time, let u'=e^{-x}
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