Thread: conditional expectation

if Z1 and Z1 are r.v with distribution N(0,1) and X1,X2 are r.v

and X1 = a + bZ1 and X2 = c + dZ1 + eZ2

then to calculate E[X2|X1=x1] can we just rearange and substitute so

E[X2|X1=x1] = E[c + d((x1 -a)/b) + eZ2] = c + (d/b)E[X1] - a/b
= c+ a/b(d-1)
thanks

I think you have the right idea but im not sure your algebra is quite right


To do this problem evaluate teh expectation on the right hand side, one term at a time.


edit: (ive conditioned on Z1 in the last term for convenience later. This is ok as Z1 is the only random part of X1)





To get the last line you must assume that Z2 and Z1 are independant.



In your derivation, you used E(X1|X1=x) = E(X1)=0. however, E(X1|X1=x) E(X1)
E(X1|X1=x1) = x1