A simple conditional expectation question

[michonamona]

Let v be a random variable distributed according to F(.). Let X be a set containing the objects x1 and x2. Suppose

E(v|x1) = b AND E(v|x2) = b (The expected value of v conditional on x1 is b, etc)

where b is some constant.

Does it follow that E(v|x1,x2) = b? If so, why?



Additionally, does the converse hold?

i.e. does E(v|x1,x2) = b imply E(v|x1) = b and E(v|x2) = b?


Thank you!
M

 

[jambaugh]

Your converse is not true. Since the "If" part is always true for some b, it does not follow that every subdivision gives equal values. Find a random two cases E(v|x1) not equal to E(v|x2) any example of this would give you a counter example when you set b equal to E(v|x1,x2).

The original holds (I'm pretty sure) and you can verify it by simply breaking down the (conditional) expectation values into their definitions, sums over values times the corresponding (conditional) probabilities. You then use the P(A|B) = P(A and B)/P(B) formula and some algebra...at least that's how I'd start. There might be a lot of regrouping terms in a sum of sums but you should be able to manipulate definition of one side to a definition of the other.

 

[haruspex]

Let v be a random variable distributed according to F(.). Let X be a set containing the objects x1 and x2. Suppose

E(v|x1) = b AND E(v|x2) = b (The expected value of v conditional on x1 is b, etc)

where b is some constant.

Does it follow that E(v|x1,x2) = b?

Roll two dice. v is total, x1 is first die showing 1, x2 is second showing 1.

 

[michonamona]

Is this a counter example?

 

[haruspex]

Is this a counter example?

That's what I'm suggesting.