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This post was updated on .
The below solution is not correct. The final and correct solution (verified by Amit Sir) is in the 2nd last post of this thread.I dont think U and V will be independent since both depend on X and Y.
F(u,v)= P(U<u, V<v)= P( U<u, V<v  X<Y) + P(U<u, V<v  X>Y)
since X can be either greater than Y or less than Y
=P ( X<u,Y<v) + P (Y<u,X<v)
=P( X<u)*P(Y<v) + P(Y<u)*P(X<v) since X and Y are independent.
= Integral (x=0 to u) 2x^2 dx * Integral(y=0 to v) 3y^3 dy + Integral(y=0 to u) 3y^3 dy * Integral (x=0 to v) 2x^2 dx
=u^2*v^3 + v^2*u^3
f(u,v)= ∂²F(u,v)/∂u∂v
= 6uv(u+v) 0<u<v<1

"You don't have to believe in God, but you should believe in The Book." Paul Erdős


Uffff. Forgot to make the two cases. x_x
Thanks a bunch!!


Hey Devika,
If possible can you please re post the question? :)


u shudnt have deleted that. because ur method of calculating f(u) and f(v) was correct. But I cant think of a way of going from marginal pdfs to joint pds if the random variables concerned are not independent.

"You don't have to believe in God, but you should believe in The Book." Paul Erdős


Let X and Y denote independent random variables, with the following pdf
f(x) = 2x 0<x<1
= 0 otherwise
f(y)= 3y^2
= 0 otherwise
Let U=Min {X,Y} and V=Max{X,Y}
Find the joint pdf of X and Y.


I deleted because I thought my solution would mislead and baffle others. Also, I took U and V to be independent, and hence multiplied the marginal densities to get the joint pdf.


@Sinistral
F(u,v)= P(U<u, V<v)= P( U<u, V<v  X<Y) + P(U<u, V<v  X>Y) ????ye kaise strike kiya..grt way

Administrator

This is incorrect:
F(u,v)= Pr(U<u, V<v)= Pr( U<u, V<vX<Y) + Pr(U<u, V<vX>Y)
The correction is:
F(u,v)= Pr(U<u, V<v)= Pr(U<u, V<vX<Y) Pr(X<Y) + P(U<u, V<vX>Y) Pr(X>Y)
Now redo it.


This post was updated on .
Oh yes!! how could I miss that. (thank you so much Sir)
EDIT
F(u,v)= Pr(U<u, V<v)= Pr(U<u, V<vX<Y) Pr(X<Y) + P(U<u, V<vX>Y) Pr(X>Y) (as pointed by Amit Sir and now it seems fairly obvious).
Pr(U<u, V<v AND X<Y) * Pr(X<Y) + P(U<u, V<v AND X>Y) * Pr(X>Y)
=  
Pr(X<Y) Pr(X>Y)
= P ( X<u,Y<v) + P (Y<u,X<v)
/EDIT
Rest of the solution is same:
=P( X<u)*P(Y<v) + P(Y<u)*P(X<v) since X and Y are independent.
= Integral (x=0 to u) 2x^2 dx * Integral(y=0 to v) 3y^3 dy + Integral(y=0 to u) 3y^3 dy * Integral (x=0 to v) 2x^2 dx
=u^2*v^3 + v^2*u^3
f(u,v)= ∂²F(u,v)/∂u∂v
= 6uv(u+v) 0<u<v<1

"You don't have to believe in God, but you should believe in The Book." Paul Erdős

Administrator

There is still a mistake. Because U<u, V<v AND X<Y implies that X< u and Y < v but is not implied by it. Hence {U<u, V<v AND X<Y} is not the same event as {X<u,Y<v} but is a subset of it and you can't replace {U<u, V<v AND X<Y} by {X<u,Y<v}


This post was updated on .
Ques: X and Y are independent rv
f(x) = 2x 0<x<1
= 0 otherwise
f(y)= 3y^2 ; 0<y<1
= 0 otherwise
U=Min{X,Y} & V=Max{X,Y}. Find joint pdf of U and V.

f(x,y)= 6xy^2 ; 0<x<1 & 0<y<1 (since X and Y are independent)
= 0 ; elsewhere.
F(u,v)= Pr(U<u, V<v)= Pr(U<u, V<vX<=Y) Pr(X<=Y) + Pr(U<u, V<vX>Y) Pr(X>Y)
Pr(U<u, V<v AND X<Y) * Pr(X<=Y) + Pr(U<u, V<v AND X>Y) * Pr(X>Y)
=  
Pr(X<=Y) Pr(X>Y)
= Pr(U<u, V<v AND X<=Y) + Pr(U<u, V<v AND X>Y)
= Pr(Min{X,Y}<u, Max{X,Y}<v AND X<=Y) + Pr(Min{X,Y}<u, Max{X,Y}<v AND X>Y)
Pr(Min{X,Y}<u, Max{X,Y}<v AND X<=Y) :</b>
Pr(Min{X,Y}<u, Max{X,Y}<v AND X>Y):
Adding the above 4 integrals would give Pr(U<u, V<v)=F(u,v); [Assuming U<=V since Min(X,Y) <= Max(X,Y)].

"You don't have to believe in God, but you should believe in The Book." Paul Erdős

Administrator

Now this is correct. Good.

