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I would simply like to know if:

$E[x_1|x_1,x_2]=E[x_1|x_2]$

or

$E[x_1|x_1,x_2]=E[x_1|x_1]=x_1$

or something completely different and why.

This is not homework. It came up because I'm trying to figure out the bias of omitting an interaction term. When the population model is...

$y=\beta_0 + \beta_1 x_1 + \beta_2 x_2 + x_1 x_2 + \epsilon$

...regressing without the interaction yields:

$E[y|x_1,x_2]=\hat{\beta}_0 + \hat{\beta}_1 x_1 + \hat{\beta}_2 x_2 + E[x_1 x_2|x_1,x_2]$

I expand that conditional expectation at the end to:

$E[x_1|x_1,x_2]E[x_2|x_1,x_2]+cov(x_1,x_2|x_1,x_2)$

But then I'm not sure what to do with $E[x_1|x_1,x_2]$ and $E[x_2|x_1,x_2]$.

Bonus points if you confirm or deny that $cov(x_1,x_2|x_1,x_2)$ is zero, regardless of whether $x_1$ and $x_2$ are independent.

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  • $\begingroup$ $X_1$ being measurable wrt $X_1$, $\mathbb{E}[X_1|X_1]=X_1$. $\endgroup$
    – Xi'an
    Jan 16, 2019 at 17:31
  • $\begingroup$ There is no regression coefficient before interaction in your model? $\endgroup$
    – user158565
    Jan 16, 2019 at 17:39
  • $\begingroup$ @user158565 the coefficient is one, for simplicity $\endgroup$
    – suckrates
    Jan 16, 2019 at 17:39
  • $\begingroup$ @Xi'an so $E[x_1|x_1,x_2]=x_1$? $\endgroup$
    – suckrates
    Jan 17, 2019 at 9:32
  • $\begingroup$ Yes and $\text{cov}(X_1,X_2|X_1)=0$ as well. $\endgroup$
    – Xi'an
    Jan 17, 2019 at 10:31

1 Answer 1

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In $E[y|x_1,x_2]$ it seems you already said that $E[x_i|x_1,x_2]=x_i$'s by pulling $x_i$ out of the expectation. $E[x_1x_2|x_1,x_2]$ is also $x_1x_2$ because this is like saying what is the expectation of $x_1x_2$, if we know $x_1,x_2$. It's going to be $x_1x_2$. This directly means that $cov(x_1,x_2|x_1,x_2)=0$, since $E[x_1x_2|x_1,x_2]=x_1x_2=E[x_1|x_1,x_2]E[x_2|x_1,x_2]+cov(x_1,x_2|x_1,x_2)=x_1x_2+cov(x_1,x_2|x_1,x_2)$.

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