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Find the Probability: P(X<1/2  Y=1) 
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#1
Dec908, 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)=6x^{2}y, 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) 1/2 =∫ f_{XY}(xy=1) dx ∞ My computations: Marginal density of Y: fY(y)=2y^4, 0<y<1 fY(y)=2y(2y)^3, 1<y<2 Condition density of X given Y=y: Case 1: For given/fixed 0<y<1, f_{XY}(xy)=3x^2 / y^3, 0<x<y Case 2: For given/fixed 1<y<2, f_{XY}(xy)=3x^2 / (2y)^3, 0<x<2y 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 f_{XY}(xy) should I use? Thanks for any help! 


#2
Dec908, 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!



#3
Dec1008, 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?



#4
Dec1008, 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.



#5
Dec1108, 12:06 AM

P: 1,270

um...Any proof about it?



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