Conditional expectation and variance

[covariance64]

Let X, Y be independent exponential random variables with means 1 and 2 respectively.

Let
Z = 1, if X < Y
Z = 0, otherwise

Find E(X|Z) and V(X|Z).

We should first find E(X|Z=z)
E(X|Z=z) = integral (from 0 to inf) of xf(x|z).
However, how do we find f(x|z) ?

 

[EnumaElish]

Z is discrete. Since E[.|Z] and V[.|Z] are functions of Z, they too are discrete.

E[X|Z=1] = E[X|X<Y].
E[X|Z=0] = E[X|X>Y].

Similarly for V[X|Z].

 

[covariance64]

I have found that

E[X|Z=1] = E[X|X<Y] = 1/9
E[X|Z=0] = E[X|X>Y] = 8/9
by integrating, and conditioning on the random variable Y.

So E(X) = E(E(X|Z)) = (1/9)(1/3) + (8/9)(2/3) = 17/27,

which contradicts the fact that E(X) = 1, for X exponential with mean 1.

I am wondering where is the error.

 

[EnumaElish]

How are you computing Pr{Z=0} and Pr{Z=1}?