In a question in which I am being asked to find the Variance of a sum of random variables $X_1....X_{100}$
With $Cov(X_i,X_j)$
The solution says the following: When considering the all of the covariances for each pair $X_i,X_j$, the narrative says that
A. $2\sum_{i>j}^{100}Cov(X_i,X_j)$, and that this expression, in particular the lower limit of summation: "i>j", means that we are only pairing each pair once.
B. The narration continues on to say that if one rearranges what was found in A to be $\sum_{i \neq j}^{100}Cov(X_i,X_j)$, that this means the same thing but is mathematically simpler.
My questions are as follows.
- In part A, is the coefficient 2 a result of the formula $Var[aX+bY+c]=a^2Var[X]+b^2Var[Y]+2abCov[X,Y)]$?
- In part A also, I more or less understand why the the lower limit of summation: "i>j" makes sure we don't double count but I wanted to check my understanding. If we are trying to figure out the sum of all of these $X_i's$, let's say that since, among others we are adding up $X_3$ and $X_{45}$, for the covariance of these particular $X_i's$, we only need to account for $(Cov(X_3,X_{45}))$, and as long as we do that, it doesn't matter whether it is in that form or $(Cov(X_{45},X_{3}))$, so long as we don't add in the covariance twice. Here, that the left-hand side is always greater than the righthand side, still accounts for for all of the pairs, just once. Is this correct?
- In part B, in removing the 2 and rewriting it as $\sum_{i \neq j}^{100}Cov(X_i,X_j)$, I see that it is mathematically the same, but isn't this double counting the different pairs of $X_i's$?