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A probability distribution

  1. Jan 29, 2007 #1
    Here is a question that was in on of my exams a few months ago. It asks what is the value of k so that the function f(x;k) is a probability density.
    I didn't really answer it, but put an answer as [tex] k= \frac{1}{2\sqrt{\pi}}[/tex] because that somewhat resembled the Normal distribution.
    Does anyone know?

    [tex]f(x;k) = -(\frac{1}{2\sqrt{\pi}} - k)^2 + k.e^{(10x - 0.25x^2 - 100)}[/tex]

    (this is not a homework question)
    Last edited: Jan 29, 2007
  2. jcsd
  3. Jan 29, 2007 #2


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    What are the properties of a probability density function?
  4. Jan 29, 2007 #3
    Area under curve = 1
  5. Jan 29, 2007 #4


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    Right, i.e. [tex]\int_{-\infty}^{+\infty} f(x)dx =1[/tex]
  6. Jan 29, 2007 #5


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    You also need f(x)>=0 for all x for f(x) to be a probability density.
  7. Jan 29, 2007 #6
    But [tex]k.e^{(10x - 0.25x^2 - 100)}[/tex] isn't able to be integrated using elementary functions.
  8. Jan 29, 2007 #7

    D H

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    That doesn't matter, does it? The standard normal distribution integrates to 1:

    [tex]\int_{-\infty}^{\infty} \frac 1 {\sqrt{2\pi}} \exp\left(-\,\frac1 2 x^2\right) = 1[/tex]

    From this, you should be able to calculate the integral of [itex]k.e^{(10x - 0.25x^2 - 100)}[/itex] over all x.

    BTW, your guess was right because setting [itex] k= \frac{1}{2\sqrt{\pi}}[/itex] eliminates the first term and makes the latter term a normal PDF. Can you see why?
  9. Jan 29, 2007 #8
    Ahh, right. I guessed it for that reason (to eliminate the other term), but kind of by accident.

    I see how it goes now. Thanks for your responses.
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