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[PROBABILITY] Conditional probability for random variable

  • Thread starter libelec
  • Start date
  • #1
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Homework Statement



X and Y two independent random variables with distribution U(0, 1/2). Find the density of (X + Y)2|X - Y > 0

The Attempt at a Solution



I was hoping this would be simpler, but somehow I always end up with nothing.

The only thing I can work out just fine is that P(X - Y > 0) = 1/2.

Then I tried to find the conditional probability with P((X + Y)2|X - Y > 0) = P((X + Y)2, X - Y > 0)/P(X - Y > 0), and then stop dead.

Any guidance here? Because I have to do many of these problems.
 

Answers and Replies

  • #2
378
2
X and Y are uniformly distributed between 0 and 0.5

density of x-y>0, x>y is just 1/2*uniform density which is 1/0.5*0.5

for finding density of (x+y)^2 AND x-y>0
define one random variables
h = (x+y)^2
g = (x-y)^2

Using Jocbian find f_hg, and then using appropriate limits (x-y>0 and 0 and 0.5) find f_h

The conditional will be just be f_h/2

I am hoping that should work.
 
Last edited:
  • #3
176
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It isn't so easy to find the Jacobian, since the inverse transformation is quite complicated. Unless I can just factor like this:

g = (X+Y)2 ------> g1/2 = X+Y.

But I would lose information, wouldn't I?
 
  • #4
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OK, if I can do what I just did, I find that

fhg = fxy((g1/2+h1/2)/2, (g1/2-h1/2)/2)/|-1/(8(hg)1/2)| = 8 (hg)1/2, if 0<(g1/2+h1/2)/2<1/2 and 0<(g1/2-h1/2)/2<1/2.

Now, I don't understand what you say about taking apropriate limits considering x-y>0 to find f_h.
 
  • #5
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Also, why did you choose those two random variables to do the Jacobian? Why is g that way?
 
  • #6
378
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Two multiple random variables you need a pseudo variable to take Jacobian (pg 7 and 8):
http://www.hss.caltech.edu/~mshum/stats/lect3.pdf

That's one the methods to solve. G(x,y) can be anything. In the end you will get same f_H.

Also, Note that |J_gh| = 1/|J_xy|

When I said limits, I meant to find f_H you need to integrate f_GH and you need limits for that definite integral. That's what I meant by those limits.

There are other approaches but I only use above for majority of the times.
 
  • #7
176
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No, I know about the Jacobian method. I meant, how did you come up with h = (X-Y)2 to make the transformation one-to-one?
 
  • #8
378
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No, I know about the Jacobian method. I meant, how did you come up with h = (X-Y)2 to make the transformation one-to-one?
Randomly picked, h =x+y might have been easier
 
  • #9
176
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But I can't randomly pick any given variable, I could end up with a 2-to-1 or 3-to-1 function.

Anyway, I'm having troubles setting up the limits of the function f_HG (that is, where it isn't zero).

I have that h is in [0,1] and g is in [0,1/4]. Then, with the inverse transformation and considering that both X and Y are uniform in (0, 1/2):

0< (g1/2 + h1/2)/2 < 1/2
0< (g1/2 - h1/2)/2 < 1/2

Then:

0< g1/2 + h1/2 < 1
0< g1/2 - h1/2 < 1

-h1/2< g1/2 < 1 - h1/2
h1/2< g1/2 < 1 + h1/2

I can see that this ultimately tells me that

-h1/2< g1/2 < 1 + h1/2

But how do I get rid of the sqrt?
 
  • #10
378
2
Do you have the answer?
I managed to get
1/4 (1/sqrt(u)-1)
I found it bit simpler to solve without any complicated equations..

I don't understand
But I can't randomly pick any given variable, I could end up with a 2-to-1 or 3-to-1 function.
 
  • #11
176
0
Do you have the answer?
I managed to get
1/4 (1/sqrt(u)-1)
I found it bit simpler to solve without any complicated equations..
How did you do that?

This is what I did: I need the inverse transformation first (the one that turns h and g into x and y).

g = (X+Y)2 ----------> g1/2 = X+Y
h = (X-Y)2 ----------> h1/2 = X-Y

g1/2 + h1/2 = 2X ----------> X = (g1/2 + h1/2)/2
g1/2 - h1/2 = 2Y ----------> Y = (g1/2 - h1/2)/2

Which ends this. The Jacobian is then -1/(8h1/2g1/2) and so:

fHG(h,g) = fXY ((g1/2 + h1/2)/2, (g1/2 - h1/2)/2)*(1/(8h1/2g1/2)).

Since fXY is uniform (4), all I have is 1/(2h1/2g1/2).

And I couldn't work the limits out.

How did you get to that result so fast?
 
  • #12
378
2
How did you do that?

This is what I did: I need the inverse transformation first (the one that turns h and g into x and y).

g = (X+Y)2 ----------> g1/2 = X+Y
h = (X-Y)2 ----------> h1/2 = X-Y

g1/2 + h1/2 = 2X ----------> X = (g1/2 + h1/2)/2
g1/2 - h1/2 = 2Y ----------> Y = (g1/2 - h1/2)/2

Which ends this. The Jacobian is then -1/(8h1/2g1/2) and so:

fHG(h,g) = fXY ((g1/2 + h1/2)/2, (g1/2 - h1/2)/2)*(1/(8h1/2g1/2)).

Since fXY is uniform (4), all I have is 1/(2h1/2g1/2).

And I couldn't work the limits out.

How did you get to that result so fast?
Using the assumption that other random variable can be defined as x.
h = (x+y)^2
g = x

Jacobian:
|J_xy| = |-2 (x+y)|
|J_hg| = 1/ (2(x+y))

Joinit distribution:
f_gh = f_xy/[2(x+y)]
f_gh = 2/[2(x+y)] {Using 2 for f_xy, it should be 8 I believe 1/(0.5*0.5*0.5) given x>y distribution doubles}
= 1/(x+y)
= 1/sqrt(h)

Marginal distribution:
Limits:
x = v
y = sqrt(u)-v

x>y
2v>sqrt(u)
v>sqrt(u)/2
f_h = integrate f_gh from sqrt(u)/2 to 1/2
 
  • #13
176
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Marginal distribution:
Limits:
x = v
y = sqrt(u)-v

x>y
2v>sqrt(u)
v>sqrt(u)/2
f_h = integrate f_gh from sqrt(u)/2 to 1/2
That I didn't understand. What did you mean with that?
 
  • #14
378
2
That I didn't understand. What did you mean with that?
You are not interested in f_gh but only in f_h. So, I found f_h knowing that
x = ..
y = ..
and x>y
or
in terms of u and v (used u for h and v for g)
v>sqrt(u)/2
 

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