j****i
发帖数: 305 | 1 Suppose X,Y ~U(0,1), X,Y independent.
Let U = X + Y, V = X - Y. What is the conditional pdf f(V|U)? Are U, V
independent? | m***s
发帖数: 605 | 2 they are not independent, when u=2,v must be 0.
【在 j****i 的大作中提到】
: Suppose X,Y ~U(0,1), X,Y independent.
: Let U = X + Y, V = X - Y. What is the conditional pdf f(V|U)? Are U, V
: independent?
| j****i
发帖数: 305 | 3 What is E(V|U) then?
【在 m***s 的大作中提到】
: they are not independent, when u=2,v must be 0.
| m***s
发帖数: 605 | 4 E(V|U=u)=E(U-2Y|U=u)=u-2*E(Y|U)
E(Y|U=u)= u/2 if u<=1
E(Y|U=u)= u/2 if u>1
so , E(V|U=u)= 0
【在 j****i 的大作中提到】
: What is E(V|U) then?
| j****i
发帖数: 305 | 5 Thanks. Then U, V are uncorrelated, but dependent.
【在 m***s 的大作中提到】
: E(V|U=u)=E(U-2Y|U=u)=u-2*E(Y|U)
: E(Y|U=u)= u/2 if u<=1
: E(Y|U=u)= u/2 if u>1
: so , E(V|U=u)= 0
| m******g
发帖数: 12 | 6 From either Jacobian or graphic examination, conditioning on any linear
combo of uniform, any other linear combo of uniforms is uniformly
distributed. The trick is usually to figure out the range.
In this case, conditional on U=X+Y = u, V has a range [-u, u] regardless of
the value of u. Thus,
f(V=v| U=u) = 1/(2u) 1_{|v|<=u} | w*****e
发帖数: 197 | 7 Yeah, you provide the best idea here.
Using Jacobian, it is easy to see that
for every possible pair of ( u, v ), its
joint density must be uniform.
So now it is just a matter of determine
the range of v given u. And it takes a bit
patience and is best done through 2D graphic.
So:
if u is [ 0, 1 ], f(v|u) = 1/2u
if u is [ 1, 2 ], f(v|u) = 1/(2(2-u))
of
【在 m******g 的大作中提到】
: From either Jacobian or graphic examination, conditioning on any linear
: combo of uniform, any other linear combo of uniforms is uniformly
: distributed. The trick is usually to figure out the range.
: In this case, conditional on U=X+Y = u, V has a range [-u, u] regardless of
: the value of u. Thus,
: f(V=v| U=u) = 1/(2u) 1_{|v|<=u}
|
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