Is proving that $var[X_1+X_2]=var[X_1]+var[X_2]$ sufficient in proving $var(\sum\limits_{i=1}^n X_i $)=$\sum\limits_{i=1}^n var[X_i] $

Question

Assuming $E[X_iX_j]=E[X_i]E[X_j]$ for i and j between 1 and n

Is proving that $var[X_1+X_2]=var[X_1]+var[X_2]$ sufficient in proving var($\sum\limits_{i=1}^n X_i $)=$\sum\limits_{i=1}^n var[X_i] $.

It seems pretty self-evident to me but perhaps do I need some intermediate steps to show this?

Answer 1

Let $X=\sum\limits_{i=1}^nX_i$. Then Var$(X)=E(X^2)-E(X)^2$.

Expand $X^2$ and take the expected value of it and then subtract the square of $E(X)=\sum\limits_{i=1}^nE(X_i)$.

The terms $2E(X_iX_j)-2E(X_i)E(X_j)$ will all cancel out leaving you with the sum of all the required variances. There is therefore no need to do the case of two variables first. Looking back at your proof you will then see why the result holds for any number of variables.

Answer 2

As a rule of thumb: If a theorem/formula seems self-evident to you, but you are not sure if it really holds, you better prove it.
If it really is self-evident, the proof should not be hard. But it could also be that you just thought the wrong way about it and trying the proof you see where the difficulties are.

In your case, starting from $\operatorname{Var} [X+Y]=\operatorname{Var}[X] + \operatorname{Var}[Y], $ you can use induction to show $ \operatorname{Var}[\sum_{i=1}^n X_i]= \sum_{i=1}^n \operatorname{Var}[X_i] .$