1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    radou

    User Avatar
    Homework Helper

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

    radou

    User Avatar
    Homework Helper

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

    mathman

    User Avatar
    Science Advisor
    Gold Member

    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

    User Avatar
    Staff Emeritus
    Science Advisor

    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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: A probability distribution
Loading...