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becuase

$$\left(\sum_{i=1}^n a_n\right)^2=\sum_{i=1}^n a_i^2 + 2\sum_{i=1}^n\sum_{j=1}^{i-1}a_ja_i$$

therefore

$$\text{Var}\left(\sum_{i=1}^n{X}\right) = E\left\{ \sum_{i=1}^n [X_i-E(X_i)]^2 + 2\sum_{j=1}^n\sum_{k=1}^{j-1} [X_j-E(X_j)][X_k-E(X_k)] \right\}$$

but I don't understand there is no $2$ in the following proof.

enter image description here

Can someone give me a hint?

Thanks

Ben
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  • Please do not use pictures for critical portions of your post. Pictures may not be legible, cannot be searched and are not view-able to some, such as those who use screen readers. – Another User Sep 30 '22 at 18:32
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    I believe the "double sum" in the included picture is incorrect, it was only intended to apply for $k\neq j$, but that was not stated in the summation. – user3257842 Sep 30 '22 at 18:40
  • @BrianMoehring in the range of $[1, j-1]$ – Ben Sep 30 '22 at 18:46
  • @BrianMoehring yep, thanks. – Ben Oct 01 '22 at 05:02

1 Answers1

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The proof is applying your identity with $a_i:=X_i-E(X_i)$ but has an error. The identity is derived as follows: $$\left(\sum_{i=1}^na_i\right)^2 =\left(\sum_{j=1}^na_j\right)\left(\sum_{k=1}^na_k\right)= \sum_{j=1}^n\sum_{k=1}^na_ja_k\tag1 $$ At this point the idea is to separate out the $j=k$ case: $$ =\sum_{i=1}^na_i^2 + \sum_{j\ne k} a_ja_k\tag2 $$ The proof writes $\sum_{j\ne k} a_ja_k$ as $$\sum_{j=1}^n\sum_{k=1}^n a_ja_k,$$ which is incorrect. Instead the sum over $k$ should exclude $k=j$. Another way to write the RHS of (2) is:

$$ \sum_{j\ne k} a_ja_k \stackrel{(a)}= 2\sum_{1\le j<k\le n}a_ja_k\stackrel{(b)}=2\sum_{j=1}^n\sum_{k=1}^{j-1} a_ja_k $$ Version (a), which is proved here, leads directly to version (b), which is the form you're familiar with.

grand_chat
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