A probability distribution

  • #1
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Main Question or Discussion Point

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:

Answers and Replies

  • #2
radou
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What are the properties of a probability density function?
 
  • #3
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Area under curve = 1
 
  • #4
radou
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Area under curve = 1
Right, i.e. [tex]\int_{-\infty}^{+\infty} f(x)dx =1[/tex]
 
  • #5
mathman
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You also need f(x)>=0 for all x for f(x) to be a probability density.
 
  • #6
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But [tex]k.e^{(10x - 0.25x^2 - 100)}[/tex] isn't able to be integrated using elementary functions.
 
  • #7
D H
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But [tex]k.e^{(10x - 0.25x^2 - 100)}[/tex] isn't able to be integrated using elementary functions.
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?
 
  • #8
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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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