Probability distribution function proof

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[SOLVED] Probability distribution function proof

Homework Statement



Prove the function p(x) = [tex]\frac{3}{4b^{3}}[/tex](b[tex]^{2}[/tex]-x[tex]^{2}[/tex]) for -b [tex]\leq[/tex] x [tex]\leq[/tex] +b is a valid probability distribution function.

Homework Equations



I'm not sure if it's as simple as this, but I've been using [tex]\int[/tex] p(x) dx (between b and -b) = 1 if the function is valid


The Attempt at a Solution



[tex]\int[/tex] p(x) dx (between b and -b) = [tex]\frac{3x}{4b}[/tex] - [tex]\frac{x^{3}}{12b^{3}}[/tex] between b and -b = [tex]\frac{3}{4}[/tex] - [tex]\frac{1}{12}[/tex] + [tex]\frac{3}{4}[/tex] - [tex]\frac{1}{12}[/tex] = [tex]\frac{4}{3}[/tex]

I treated b as a constant as I was integrating with respect to x, but clearly [tex]\frac{4}{3}[/tex] is not equal to 1.... please help!
 

Answers and Replies

  • #2
HallsofIvy
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You seem to have forgotten the "3" in "3/4" when you multiplied [tex]\frac{x^3}{3}[/tex] by [tex]\frac{3}{4b^3}[/tex]. You should have
[tex]\frac{3}{4}- \frac{1}{4}+ \frac{3}{4}- \frac{1}{4}= 1[/tex]
 
  • #3
25
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Heh thanks, stupid mistake...
I assumed it was the integration I got wrong as opposed to an earlier stage, I'll check more thoroughly next time!
Thanks again :)
 

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