<p>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. </p>

<p>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. </p>

<p>To show you are the one who is correct, we need to dig a little bit deeper. One basic <a href="https://en.wikipedia.org/wiki/Covariance" rel="nofollow noreferrer">property of covariance</a> 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. </p>

<p>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,
$$
Cov(X,Y) = Cov(\epsilon_{x}, \epsilon_{y})
$$</p>

CC BY-SA 4.0
2018-05-15T22:52:09.863Z
2018-05-15T22:52:09.863Z
Stat_Programmer
<p>Thank you very much for your insightful comment. I think I know now the source of the problem. On the lecture slides I took the first approach from, it would have been necessary to disentangle the the true value of the population $\tau_{pop}$ (defined as the expected value of the mean of all $X_i$) and the true value of a person $i$ $\tau_{i}$ in that population. If you do that, you come to the right solution:  </p>

<p>$X_i = \tau_{x_i}+\epsilon_{x_i}$ and $E(X_i) = \tau_{x_i}$<br>
$X_{x_{pop}} = \tau_{x_{pop}}+\epsilon_{x_{pop}}$ and $E(X_{x_{pop}}) = \tau_{x_{pop}}$  </p>

<p>the above equations are equivalent for Y.<br>
For computing the Covariance one would be using the individual values $X_i$ and $Y_i$ and the means in the populaiton $X_{pop}$ and $Y_{pop}$. It follows:   </p>

<p>$Cov(X,Y) = E[(X_i-E(X_{pop}))(Y_i-E(Y_{pop}))]$<br>
$ = E[(\tau_{x_i}+\epsilon_{x_i}-\tau_{x_{pop}})*(\tau_{y_i}+\epsilon_{y_i}-\tau_{y_{pop}})]$<br>
$ = E(\tau_{x_i}\tau_{y_i})+E(\tau_{x_i}\epsilon_{y_i})-E(\tau_{x_i}\tau_{y_{pop}}) + E(\epsilon_{x_i}\tau_{y_i})+E(\epsilon_{x_i}\epsilon_{y_i})-E(\epsilon_{x_i}\tau_{y_{pop}})-E(\tau_{x_{pop}}\tau_{y_i})-E(\tau_{x_{pop}}\epsilon_{y_i})+E(\tau_{x_{pop}}\tau_{y_{pop}})$
As the error terms are assumed to be uncorrelated with each other and the true values in CTT:<br>
$= E(\tau_{x_i}\tau_{y_i})-E(\tau_{x_i}\tau_{y_{pop}})-E(\tau_{x_{pop}}\tau_{y_i})+E(\tau_{x_{pop}}\tau_{y_{pop}})$<br>
$= E[(\tau_{x_i}-\tau_{x_{pop}})(\tau_{y_i}-\tau_{y_{pop}})]$  </p>

<p>So the expected covarance of $X$ and $Y$ is only determined by the covariance of the true values $\tau_x$ and $\tau_y$</p>

CC BY-SA 4.0
2018-05-16T07:39:12.720Z
2018-05-16T07:39:12.720Z
JannikO
<p>I'm stuck with the definition of the Covariance in the <a href="https://en.wikipedia.org/wiki/Classical_test_theory" rel="nofollow noreferrer"><strong>classical test theory</strong></a> (CTT) framework:
In CTT it is assumed that for two items:<br>
$X = \tau_x + \epsilon_x$<br>
$E(X) = \tau_x$  </p>

<p>$Y = \tau_y + \epsilon_y$<br>
$E(Y) = \tau_y$  </p>

<p>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) $<br>
$= Cov(\tau_x,\tau_y)$ </p>

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

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

<p>$E(\epsilon_x*\epsilon_y) = E[(\epsilon_x - E(\epsilon_x))*(\epsilon_y - E(\epsilon_y))] = Cov(\epsilon_x, \epsilon_y)$  </p>

<p>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?</p>

<measurement><test>
Classical test theory: Covariance of two variables
JannikO
# Classical test theory: Covariance of two variables

(Asked by: username_0 on 2018-05-15)

I'm stuck with the definition of the Covariance in the [**classical test theory**](https://en.wikipedia.org/wiki/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?

**Tags:** measurement, test


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## 2 answers

### Answer by username_1 on 2018-05-15. Score: 2 | Accepted answer

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](https://en.wikipedia.org/wiki/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,
$$
Cov(X,Y) = Cov(\epsilon\_{x}, \epsilon\_{y})
$$


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### Answer by username_0 on 2018-05-16. Score: 1

Thank you very much for your insightful comment. I think I know now the source of the problem. On the lecture slides I took the first approach from, it would have been necessary to disentangle the the true value of the population $\tau\_{pop}$ (defined as the expected value of the mean of all $X\_i$) and the true value of a person $i$ $\tau\_{i}$ in that population. If you do that, you come to the right solution:

$X\_i = \tau\_{x\_i}+\epsilon\_{x\_i}$ and $E(X\_i) = \tau\_{x\_i}$  
$X\_{x\_{pop}} = \tau\_{x\_{pop}}+\epsilon\_{x\_{pop}}$ and $E(X\_{x\_{pop}}) = \tau\_{x\_{pop}}$

the above equations are equivalent for Y.  
For computing the Covariance one would be using the individual values $X\_i$ and $Y\_i$ and the means in the populaiton $X\_{pop}$ and $Y\_{pop}$. It follows:

$Cov(X,Y) = E[(X\_i-E(X\_{pop}))(Y\_i-E(Y\_{pop}))]$  
$ = E[(\tau\_{x\_i}+\epsilon\_{x\_i}-\tau\_{x\_{pop}})\*(\tau\_{y\_i}+\epsilon\_{y\_i}-\tau\_{y\_{pop}})]$  
$ = E(\tau\_{x\_i}\tau\_{y\_i})+E(\tau\_{x\_i}\epsilon\_{y\_i})-E(\tau\_{x\_i}\tau\_{y\_{pop}}) + E(\epsilon\_{x\_i}\tau\_{y\_i})+E(\epsilon\_{x\_i}\epsilon\_{y\_i})-E(\epsilon\_{x\_i}\tau\_{y\_{pop}})-E(\tau\_{x\_{pop}}\tau\_{y\_i})-E(\tau\_{x\_{pop}}\epsilon\_{y\_i})+E(\tau\_{x\_{pop}}\tau\_{y\_{pop}})$
As the error terms are assumed to be uncorrelated with each other and the true values in CTT:  
$= E(\tau\_{x\_i}\tau\_{y\_i})-E(\tau\_{x\_i}\tau\_{y\_{pop}})-E(\tau\_{x\_{pop}}\tau\_{y\_i})+E(\tau\_{x\_{pop}}\tau\_{y\_{pop}})$  
$= E[(\tau\_{x\_i}-\tau\_{x\_{pop}})(\tau\_{y\_i}-\tau\_{y\_{pop}})]$

So the expected covarance of $X$ and $Y$ is only determined by the covariance of the true values $\tau\_x$ and $\tau\_y$


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