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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: enter image description here

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Thank you very much.

SayMyNameHeisenberg
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1 Answers1

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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 $x_i$.

grand_chat
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  • Thank you grand_chat, but I didn't get step (b) that x can be measured by $$\sigma(x_{1},x_{2},...,x_{k}$$ so that we can pull it out? Could you please elaborate this a liitle bit? I appreciate it! – SayMyNameHeisenberg Jan 08 '19 at 22:51
  • @commentallez-vous Saying that "$x_i$ is measurable with respect to $\sigma(x_1,\ldots,x_k)$" means, roughly speaking, that $x_i$ is a function of $x_1,\ldots, x_k$, and therefore we can 'take out what is known'. See my edit. – grand_chat Jan 08 '19 at 22:54
  • Oh...so it is basically like for a given observation, this particular x is known as a constant, therefore the expected value of a constant is the constant itself, is this understanding right? – SayMyNameHeisenberg Jan 08 '19 at 22:57
  • @commentallez-vous Exactly. If you condition on $x_1,\ldots, x_k$ then the value of $x_i$ is known, so $x_i$ can be treated as a constant and "factored out" of $E(ux_i\mid x_1,\ldots,x_k)$. – grand_chat Jan 08 '19 at 22:59
  • Thank you so much! This is a really nice proof and you saved my life! Much obliged! – SayMyNameHeisenberg Jan 08 '19 at 23:13