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Homework Help: Find k for this probability density function to be valid.

  1. Jan 18, 2013 #1
    1. The problem statement, all variables and given/known data

    Find k such that the function [itex]f(x)=ke^{-\frac{x-\mu}{\theta}}[/itex] is a probability density function (pdf), for [itex] x > \mu, \mu[/itex] and [itex]\theta[/itex] are constant.

    2. Relevant equations

    The property of a pdf says that the integral of f(x) from [itex]-\infty[/itex] to [itex]\infty[/itex] equals 1, that is [itex]\int\limits_{-\infty}^\infty f(x)dx=1[/itex]

    3. The attempt at a solution

    [itex]\int\limits_{-\infty}^\infty ke^{-\frac{x-\mu}{\theta}}dx[/itex]
    [itex]=k\int\limits_{-\infty}^\infty e^{-\frac{x-\mu}{\theta}}dx[/itex]
    Let [itex]t=-\frac{x-\mu}{\theta} => -\theta dt=dx[/itex]

    [itex]=> -k\int\limits_\infty^{-\infty} e^t(-\theta)dt[/itex]

    [itex]=>k\theta\int\limits_{-\infty}^\infty e^tdt[/itex]

    But [itex]e^t[/itex] would diverge going towards infinity? How could this integral be equal to 1. I am not sure what I'm doing wrong.
  2. jcsd
  3. Jan 18, 2013 #2


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    It's supposed to be a density function only for x>μ. You don't want to integrate from -infinity to +infinity. You want to integrate from μ to +infinity. The pdf will be zero for x<μ.
  4. Jan 18, 2013 #3
    Ok I see what you mean. But even if I do that, the upper bound of infinity is giving me problems because e^(infinity) is infinity :S
  5. Jan 18, 2013 #4
    Never mind!! I finally figured it out... I had a -infinity instead. All is good :) I got k = 1/theta if anyone was interested
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