$x_1, x_2, x_3, x_4,\ldots,x_k$ being random signs that are independent |-> $\operatorname{Var}(\sum_{k=1}^n b_k x_k) = \sum_{k=1}^n {b_k}^2$

Question

$x_1, x_2, x_3, x_4,\ldots,x_k$ are random signs that are independent. (I suppose this means that $P(x_i=1)=0.5=P(x_i=−1))$

Does the following hold?

$$\operatorname{Var}\left(\sum_{k=1}^n b_k x_k\right) = \sum_{k=1}^n {b_k}^2$$ for $b_1,b_2,b_3,\ldots,b_k\in\mathbb R$.

So variance is the second derivative of a generating function. And I supposed the generating function for $x_k$ would be $0.5t^1+0.5t^{-1}$? Doesn't seem right when i plug it in there.

Answer 1

Since the $x_k$ are independent, the $b_kx_k$ are independent as well. Remember, that the variance of the sum of independent random variables is the same as the sum of the variances of those variables. Furthermore, for any $k$, it holds that $$Var(b_kx_k) = b_{k}^{2}Var(x_k) = b_k^2 (E(x_k^2)-E(x_k)^2)= b_k^2(1-0) = b_k^2 .$$

Combining all of the above statements, we get: $$ Var(\sum_{k=1}^{n}b_kx_k) = \sum_{k=1}^{n}Var(b_kx_k) = \sum_{k=1}^{n}b_k^2 .$$

So indeed the statement holds.