Classical test theory: Covariance of two variables I'm stuck with the definition of the Covariance in the classical test theory (CTT) framework:
In CTT it is assumed that for two items:
$X = \tau_x + \epsilon_x$
$E(X) = \tau_x$  

$Y = \tau_y + \epsilon_y$
$E(Y) = \tau_y$  

From these assumptions it follows that:
$Cov(X,Y) = E[(\tau_x+\epsilon_x)*(\tau_y+\epsilon_y)] - E(\tau_x+\epsilon_x)*E(\tau_y+\epsilon_y)$
$= E(\tau_x\tau_y) - E(\tau_x)E(\tau_y) $
$= Cov(\tau_x,\tau_y)$ 

I tried to come to the same conclusion using a slightly different approach:
$Cov(X,Y) = E((X-E(X))(Y-E(Y))$
$=E((\tau_x+\epsilon_x-\tau_x)*(\tau_y+\epsilon_y-\tau_y))$
$=E(\epsilon_x*\epsilon_y)$  

As $E(\epsilon_x) = 0$ the following terms should be equivalent:    

$E(\epsilon_x*\epsilon_y) = E[(\epsilon_x - E(\epsilon_x))*(\epsilon_y - E(\epsilon_y))] = Cov(\epsilon_x, \epsilon_y)$  

As the conclusion drawn from my approach contradicts with the first conclusion I assume that there is a flaw in my math. Unfortunately I seem to be unable to identify my mistake. Can someone point me to it?
I do not myself use classical test theory, but from a statistical point of view the answer to this should be straightforward. The answer is that you are right, and that the first solution is wrong. 

An early indication that the first formula is incorrect is that a :
$$
Cov(\tau_{x},\tau_{y}) = 0
$$
because $\tau_{x}$ and $\tau_{y}$ are constants (I assume $\tau_{x}$ and $\tau_{y}$ are constant in your question because your equations state that they are expected values, which are always constant). The formula above holds because constants can't covary with one another. 

To show you are the one who is correct, we need to dig a little bit deeper. One basic property of covariance is, for $X$, $Y$ random variables and $a$, $b$ constants:
$$
Cov(X+a, Y+b) = Cov(X,Y)
$$
Intuitively, adding a constant to a random variable doesn't affect its linear relationship with another random variable. 

In your particular example, if I understand correctly, $\epsilon_{x}$, $\epsilon_{y}$ are random variables, and $\tau_{x}$, $\tau_{y}$ are fixed constants. Plugging these into the above formula, 
$$
Cov(\epsilon_{x} + \tau_{x}, \epsilon_{y} + \tau_{y}) = Cov(\epsilon_{x}, \epsilon_{y})
$$
And of course, using the notation in your question, $X$ = $\tau_{x}$ + $\epsilon_{x}$= $\epsilon_{x}$ + $\tau_{x}$ and $Y$ = $\tau_{y}$ + $\epsilon_{y}$= $\epsilon_{y}$ + $\tau_{y}$. Therefore, using the variables from your question,
$$
Cov(X,Y) = Cov(\epsilon_{x}, \epsilon_{y})
$$