# Thread: conditional expectation

1. ## 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

2. 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.

$\displaystyle E(X2 | X1) = E(C|X1) + dE(Z1|X1) + eE(Z2|Z1)$
edit: (ive conditioned on Z1 in the last term for convenience later. This is ok as Z1 is the only random part of X1)

$\displaystyle E(X2 | X1) = C + dE(\frac{X1-a}{b}|X1) + eE(Z2|Z1)$
$\displaystyle E(X2 | X1) = C + d\frac{X1-a}{b} + eE(Z2|Z1)$

$\displaystyle E(X2 | X1) = C + d\frac{X1-a}{b} + e*0$
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) $\displaystyle \neq$ E(X1)
E(X1|X1=x1) = x1