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Find the Probability: P(X<1/2 | Y=1)

by kingwinner
Tags: probability, px<1 or 2
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Dec9-08, 08:22 AM
P: 1,270
1. The problem statement, all variables and given/known data
Suppose X and Y are jointly continuous random variables with joint density function
f(x,y)=6x2y, 0<x<y, x+y<2
f(x,y)=0, otherwise
Find P(X<1/2 | Y=1).

2. Relevant equations
3. The attempt at a solution
By definition,
P(X<1/2 | Y=1)
=∫ fX|Y(x|y=1) dx

My computations:
Marginal density of Y:
fY(y)=2y^4, 0<y<1
fY(y)=2y(2-y)^3, 1<y<2

Condition density of X given Y=y:
Case 1: For given/fixed 0<y<1,
fX|Y(x|y)=3x^2 / y^3, 0<x<y

Case 2: For given/fixed 1<y<2,
fX|Y(x|y)=3x^2 / (2-y)^3, 0<x<2-y

I hope these are correct. Now P(X<1/2 | Y=1) is the troublesome case because we are given Y=1, which formula for fX|Y(x|y) should I use?

Thanks for any help!
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Dec9-08, 10:38 AM
Sci Advisor
P: 1,232
Your two formulas are the same at y=1, so it doesn't matter which one you use!
Dec10-08, 07:09 AM
P: 1,270
OK, but in general will they always be the same? What should we do in such a case in general?

Dec10-08, 11:02 AM
Sci Advisor
P: 1,232
Find the Probability: P(X<1/2 | Y=1)

If correctly derived from a given joint density function, yes, they must be the same.
Dec11-08, 12:06 AM
P: 1,270
um...Any proof about it?

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