Suppose in econometrics, $$ y = \beta_{0} + \beta_{1}x_{1} + \beta_{2}x_{2} + ... + \beta_{k}x_{k} + u$$ In Gujarati's book, it says that the following equation (1) $$ E[u | x_{1}, x_{2},..., x_{k}] = E[u] = 0 \tag{1}$$ given that $$ x_{1}, x_{2},..., x_{k}$$are independent from each other implies (with bilinearity property of covariance) $$ Cov(u, x_{1} + x_{2} + ... +x_{k}) = cov(u, x_{1}) = cov(u, x_{2}) = ... = cov(u,x_{k}) = 0$$ How can we derive this result?
Book's content:
Thank you very much.
Argue as follows: For the $i$th variable $x_i$ we have: $$E(ux_i)\stackrel{(a)}=E[E(ux_i\mid x_1,x_2,\ldots,x_k)]\stackrel{(b)}=E[x_i E(u\mid x_1,x_2,\ldots,x_k)]\stackrel{(1)}=E[x_i\cdot 0]=0 \stackrel{(1)}= E(u)E(x_i)$$ In step (a) we use the tower property of conditional expectation; in (b) we use the fact that $x_i$ is measurable with respect to $\sigma(x_1,x_2,\ldots,x_k)$, so can be pulled out of the conditional expectation (i.e., we can take out what is known).
Conclude that $E(ux_i)=E(u)E(x_i)$, which implies zero covariance between $u$ and $x_i$.