Question

Random Variable function prediction please see pic... HELP PLEASE

Answer 1

where are you stuck?

Answer 2

I honestly don't think I understand the question or what is it asking me to do with it

Answer 3

start with ...

\[E[Y-g(X)]^2=E[(Y-E[Y|X])+(E[Y|X]-g(X))]^2\]

Answer 4

\[=E[(Y-E[Y|X])^2+2(Y-E[Y|X])(E[Y|X]-g(X))+(E[Y|X]-g(X))^2]\]

Answer 5

show

\[E[2(Y-E[Y|X])(E[Y|X]-g(X))]=0\]

Answer 6

on your second line what is after + (EX[Y|X] - g(x.... I can't see what is next

Answer 7

\[=E[(Y-E[Y|X])^2\]\[+2(Y-E[Y|X])(E[Y|X]-g(X))\]\[+(E[Y|X]-g(X))^2]\]

Answer 8

\[E((W+Z)^2=E(W^2+2WZ+Z^2)\]

Answer 9

you understand?

Answer 10

I think I do conceptually because I have been learning theory but I am horrible working through this.. like I said I am helping a friend try and work some examples and I am a bit out of my league.

Answer 11

thought there would be smart ppl out there that could help... thanks btw

Answer 12

so you need to show ...

\[E[2(Y-E[Y|X])(E[Y|X]-g(X))]=0\]


to do this you need to know that

\[E(Y)=E(E(Y|X))\]


and for any measurable function \(f\)

\[E(F(X)|X)=f(X)\]

Answer 13

typo..both f's need to be the same
\[E(f(X)|X)=f(X)\]

Answer 14

ok and what about the minimizing in part 2... i think I may have the concept of part A

Answer 15

@Zarkon

Answer 16

part 2 is trivial

Answer 17

but i wopuld like to know about it if you have a minute to spare on the details

Answer 18

we want to minimize

\[E[Y-g(X)]^2\]


but by part 1 it is equal to

\[E[Y-E[Y|X]]^2+E[E[Y|X]-g(X)]^2\]

we only have control of the second part of the sum


and the smallest it can be is zero since it is being squared

so let \(g(X)=E[Y|X]\)

Answer 19

Thanks!!! and would you revisit http://openstudy.com/updates/509c04a6e4b0077374586887

Answer 20

maybe tomorrow...I'm going to be now

Answer 21

ok just when you can thanks!!! really