# Expectation conditional on self and others

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.

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

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 $$E[x_1x_2|x_1,x_2]=x_1x_2=E[x_1|x_1,x_2]E[x_2|x_1,x_2]+cov(x_1,x_2|x_1,x_2)=x_1x_2+cov(x_1,x_2|x_1,x_2)$$.