Relative part of Variance The following problem I have $n$ students who are taking a test in which two items of information $X_1$ and $X_2$ are collected. Now I form another variable $X_3=X_1+X_2$ and want to find out how large the relative proportion of the variance of $X_1$ and $X_2$ is in the variance of $X_3$. Would it be enough to simply look at the quotient $\operatorname{Var}(X_1)\big/\operatorname{Var}(X_3)$? Somehow that doesn't seem right to me, since the proportion of the covariance $\operatorname{Cov}(X_1,X_2)$ contained in $\operatorname{Var}(X_3)$ is not taken into account, is it?

As you noticed
$$
\DeclareMathOperator{\Var}{\mathrm{Var}}
\Var(X_1 + X_2) = \Var(X_1) + \Var(X_2) + 2\,\mathrm{Cov}(X_1, X_2) 
$$
So if you look only at the ratio of the variances, you are missing the covariance. This would work only if the $X_1$ and $X_2$ are uncorrelated (so the covariance is zero).
Consider two cases of variables:

*

*$X_1 = X_2$, in such a case
$$
\frac{\Var(X_1)}{\Var(X_1 + X_2)} = \frac{\Var(X_1)}{\Var(X_1) + \Var(X_1) + 2\,\Var(X_1)} = \frac{\sigma^2}{4 \sigma^2} = \frac{1}{4}
$$


*$X_1$ and $X_2$ are independent and identically distributed (so have same variance)
$$
\frac{\Var(X_1)}{\Var(X_1 + X_2)} = \frac{\Var(X_1)}{\Var(X_1) + \Var(X_2) + 0} = \frac{\sigma^2}{2 \sigma^2} = \frac{1}{2}
$$
In both cases, you summed things with the same variances, but didn't consider the covariance.
The problem is that you cannot consider "relative proportion of the variance" without taking into account covariance. It will influence the variance of the sum unless the variables are independent.
Also notice that your idea of dividing $\Var(X_1)$ by $\Var(X_1 + X_2)$ would in the best case (independent variables) tell you only how much bigger, or smaller, $\Var(X_1)$ is from $\Var(X_2)$, and it doesn't sound as a very interesting result.
You seem to be asking about covariance between $X_1$ and $X_2$.
