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Suppose we have a variance of $c_i$ as $Var(c_i)$ for $i = 1, \dots ,k$.

We also have $k$ scalars $a_1, \dots, a_k$.

Then what is the variance of the linear combination $C = \sum_i^k a_i \cdot c_i$??

Let's denote it by $Var(C)$, then is it possible to define $Var(C)$ by using $Var(c_i)$ and $a_i$?? If so, how do we do it?

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  • $\begingroup$ No, it is not. You can have a look at the Wikipedia page to see that you have to square the scalars, assuming you have independence. Otherwise terms called "covariances" appear. $\endgroup$
    – asdf
    Commented Jul 8, 2019 at 8:42
  • $\begingroup$ $Var(C) = (\sum_{i=1}^k |a_i|^2 \cdot Var(c_i))^{\frac{1}{2}}$ is this true? (Assuming independence) $\endgroup$
    – mallea
    Commented Jul 8, 2019 at 8:46

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Assuming independence (as you did in the comment) it holds in fact that $$Var(\sum_i a_i c_i) = \sum_i a_i^2 Var(c_i). $$ I am not quite sure, where your power $0.5$ comes from but it is not correct, anyway. In general, you have for any sum of type $\sum_i a_i c_i$, such that the variance of all $c_i$ exists, the formula $$Var(\sum_i a_i c_i) = \sum_i a_i^2 Var(c_i) + 2\sum_{i\neq j} a_i a_j Cov(c_i,c_j)$$, which follows directly from the binomial formula for $\left(\sum_i a_i c_i\right)^2$ (and can also be found on wikipedia)

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