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How do I show that $$\text{Var}\bigg(\sum_{i=1}^m X_i\bigg) = \sum_{i=1}^m \text{Var}(X_i) + 2\sum_{i\lt j} \text{Cov}(X_i,X_j)$$

when I know that $$ \left(\sum_{i=1}^{n}a_i\right)^2= \sum_{i=1}^{n}\sum_{j=1}^{n}a_ia_j $$ and so $$ {\rm var} \left( \sum_{i=1}^{n} X_i \right) = \sum_{i=1}^{n} \sum_{j=1}^{n} \big( E(X_i X_j)-E(X_i) E(X_j) \big) = \sum_{i=1}^{n} \sum_{j=1}^{n} {\rm cov}(X_i, X_j)$$ where do I go from here to get the result?

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  • $\begingroup$ Separate the last sum depending on whether $i<j$, $i>j$ or $i=j$. It will give the terms you are seeking. $\endgroup$
    – Suzet
    Jul 27, 2019 at 19:16
  • $\begingroup$ $cov(X_i,X_i)=var(X_i)$. For $i\ne j$, there are two equal terms for each $(i,j)$ combination. $\endgroup$ Jul 27, 2019 at 20:03

2 Answers 2

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Please think about the RHS term $\displaystyle \sum_{i=1}^{n} \sum_{j=1}^{n} Cov(X_i, X_j)$

What happens when $i = j$?

$Cov(X_i, X_j)$ becomes $Var(X_i)$

When $i \neq j$, there are two terms viz., $Cov(X_i, X_j)$ and $Cov(X_j, X_i)$ in the summation. They are equal. Hence it is enough to replace the sum of these two terms by $2 Cov(X_i, X_j)$ when $i < j$. That is why you have a factor of $2$ before the single summation.

Hence

$\displaystyle \sum_{i=1}^{n} \sum_{j=1}^{n} Cov(X_i, X_j)$

$ = \displaystyle \sum_{i=1}^{n} Var (X_i) + 2 \displaystyle \sum_{i < j} Cov(X_i, X_j)$

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You can simplify the proof by introducing the variable $Y_i=X_i-EX_i$. Using the fact that variance of $X$ is same as variance of $X+c$ for any constant $c$ the given statement is equivalent to: $var ( \sum\limits_{i=1}^{n} Y_i) =\sum\limits_{i=1}^{n} var(Y_i) + \sum\limits_{i \neq j} EY_iY_j$ or $E ( \sum\limits_{i=1}^{n} Y_i)^{2} =\sum\limits_{i=1}^{n} EY_i^{2} + \sum\limits_{i\neq j} EY_iY_j$. This follows from the fact that $( \sum\limits_{i=1}^{n} Y_i)^{2} =\sum\limits_{i=1}^{n} Y_i^{2} + \sum\limits_{i\neq j} Y_iY_j$.

Note that $2 \sum\limits_{i < j} cov(X_i,X_j) = \sum\limits_{i \neq j} cov(X_i,X_j)$

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