a**m
发帖数: 102 | 1 Let (Omega, P, F) be a probability space, and G, H two sigma subfields.
Suppose X is a random variable which is independent of H.
Then I am wondering if the following is true:
E[X|G] = E[X|sigma{G,H}], where sigma{G,H} is the sigma subfield generated
by G and H.
I think the eqaulity above is true because X and H are independent, and
hence H provides no information when estimating X. But could anyone provide
a formal proof or disproof? Thanks. | f****e
发帖数: 590 | 2 应该是不对的
E[X|G]和H未必independent
回头有空想反例
provide
【在 a**m 的大作中提到】
: Let (Omega, P, F) be a probability space, and G, H two sigma subfields.
: Suppose X is a random variable which is independent of H.
: Then I am wondering if the following is true:
: E[X|G] = E[X|sigma{G,H}], where sigma{G,H} is the sigma subfield generated
: by G and H.
: I think the eqaulity above is true because X and H are independent, and
: hence H provides no information when estimating X. But could anyone provide
: a formal proof or disproof? Thanks.
| n******t
发帖数: 4406 | 3 E[X | G,H] may not even be measurable w.r.t. G.
provide
【在 a**m 的大作中提到】
: Let (Omega, P, F) be a probability space, and G, H two sigma subfields.
: Suppose X is a random variable which is independent of H.
: Then I am wondering if the following is true:
: E[X|G] = E[X|sigma{G,H}], where sigma{G,H} is the sigma subfield generated
: by G and H.
: I think the eqaulity above is true because X and H are independent, and
: hence H provides no information when estimating X. But could anyone provide
: a formal proof or disproof? Thanks.
| a**m
发帖数: 102 | 4 That's the only thing we need prove, since the averaging property is obviously true. But never mind: I know counterexample exists...
【在 n******t 的大作中提到】
: E[X | G,H] may not even be measurable w.r.t. G.
:
: provide
| Q***5
发帖数: 994 | 5 Here is a simple counter example (hope I got it right):
Omega = {1,2,3,4}
P(i) = 1/4, i = 1,2,3,4
G = {{2},{1,3,4},Omega, EmptySet}
H = {{1,2},{3,4},Omega, EmptySet}
X(i) = 1 for i = 2,3; 0 otherwise
X is independent of H
E[X|G](i) = 1 for i = 2 and 1/3 otherwise
E(X|sigma(G,H))(i) = 1 for i = 2, 0 for i = 1 and 1/2 otherwise
provide
【在 a**m 的大作中提到】
: Let (Omega, P, F) be a probability space, and G, H two sigma subfields.
: Suppose X is a random variable which is independent of H.
: Then I am wondering if the following is true:
: E[X|G] = E[X|sigma{G,H}], where sigma{G,H} is the sigma subfield generated
: by G and H.
: I think the eqaulity above is true because X and H are independent, and
: hence H provides no information when estimating X. But could anyone provide
: a formal proof or disproof? Thanks.
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