If

f(x,y) = 4xy, for 0<x<1, 0<y<1
f(x,y) = 0, elsewhere

What's the probability that X<Y?

It seems likely that the probability is 0.5, but how can I show it matematically?

 

If

f(x,y) = 4xy, for 0<x<1, 0<y<1
f(x,y) = 0, elsewhere

What's the probability that X<Y?

Use can use symmetry here. The problem is symmetrical in the sense that x and y can be swapped and the problem is the same. Hence, the probability that x<y must equal the probalibity that y<x.

P=0.5, by symmetry.

 

OK, thanks!

If(x,y)=cxy^2, what would I have to do?

 

OK, thanks!

If(x,y)=cxy^2, what would I have to do?

I can start the approach.

\( P=c\; \int^{1}_{0} x \; \int^{x}_{0} y^2 \; dy \;dx \)

Simply integrate overt the area of interest. The geometry is simple in this case, so establishing the limits of integration is easy.

 

Makes sense. Great stuff.