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Probability (density function)

  • Thread starter forty
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  • #1
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A point is selected at random and uniformly from the region

R = {(x,y): |x| + |y| <= 1 }

Find the probability density function of the x-coordinate of the point selected at random.



By definition f(x) = the integral of f(x,y) over all y values.

after this I'm pretty much stuck. does f(x,y) = 1/4? I mean the definition is simple I think it's just me not knowing how to deal with the modulus signs.

(Is the region a square(diamond) intersecting at (0,1)(1,0)(-1,0)(0,-1))

Any help always appreciated.

Thanks.
 

Answers and Replies

  • #2
392
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Your diamond is right. You are right that [tex]f_{XY}(x,y)[/tex] is constant, but you have the wrong constant, because the area of that diamond is not 4.

You could split the diamond into two halves, the case where x>0 and the case where x<0. Then you can obtain equations for the boundary lines without using absolute value. Your [tex]f_X(x)[/tex] could then be piecewise defined.

When you are done, you could try to combine the piecewise defined [tex]f_X(x)[/tex] into one formula using absolute value, if you want, or simply leave it piecewise defined.
 
  • #3
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f(x,y) = 1 (was looking at the diamond thinking it had side lengths of 2 :S, hope this is right now)

as for the integrals im still having trouble...

for x > 0

f(x) = integral(from x-1 to -x+1) of 1 dy


for x < 0

f(x) = integral(from -x-1 to x+1) of 1 dy

are these even remotely correct?

I end up with 2-2x for x>0
and 2+2x for x<0

which if i combined them i get 2(1-|x|)

the answer in the book is just 1-1|x|.... when i combine them do i drop the 2 :P
 
Last edited:
  • #4
392
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Your f(x,y) is still wrong. What is the area of that diamond? When you fix it, you will get what they got.
 
  • #5
135
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Thanks for the help Billy! Once i fixed up f(x,y) (dont ask me why it took me so long to realize that the area wasn't 1 :S) it all worked out!

Thanks again!
 

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