8
$\begingroup$

In linear regression, we postulate a model for estimating $E(Y|X)= \beta_0 + \beta_1 x$

We also propose $Y= \beta_0 + \beta_1 x + \mu$ where μ is an error team with expectation equal to zero. We also assume $X$ are fixed, i.e. not random.

If we replace the first equation into the second, it will result: $$Y=E(Y|X) + \mu$$

If we take expectation:

$$E(Y) = EE(Y|X) + E(\mu)$$

The first sumand is E(Y) by the law of iterated expectation. The second is 0 by assumption. So we have:

$$E(Y)=E(Y)$$ which is correct.

But if we directly took expectation in the second equation without replacing, it yields:

$$E(Y) = \beta_0 + \beta_1 x + E(\mu)$$ since $\beta_0 + \beta_1 x$ are not random. So it results:

$$E(Y)= \beta_0 + \beta_1 x \equiv E(Y|X)$$

So we finally arrive to $E(Y) = E(Y|X)$ which is not correct.

What's the explanation about this contradiction? When am I being wrong? Thank you in advance

$\endgroup$
1
  • 3
    $\begingroup$ $E(Y \mid X)$ should be a function of $X$ rather than $x$ (meaning it is a random variable if $X$ is a random variable, and not if not) so perhaps you should start with $E(Y \mid X=x) = \beta_0 + \beta_1 x$ $\endgroup$
    – Henry
    Commented Feb 2, 2023 at 11:06

3 Answers 3

15
$\begingroup$

Disclaimer: I only read your question correctly after i wrote this. If $X$ is non random then yeah $E(Y|X) = E(Y)$.

Your mistake comes down to an abuse of notation. Within usual notation $X$ is a random variable which for simplicity, takes values in the reals. $x$ is one such value $X$ could take. Let's sensibly define $$ Y := \beta_0 + \beta_1X + \mu$$
with $E(\mu|X = x) = 0$ for all $x \in \mathbb{R}$

Now just as $X$ and $x$ are nor the same thing $E(Y|X)$ and $E(Y|X = x)$ are also not the same. The first conditions on the random value of $X$ making it itself a random value while the second conditions on the specific event of $X=x$ meaning it is not random anymore. Just dependent on the chosen value of $x$.

To finish out:

$E(E(Y|X)) = E(\beta_0 + \beta_1X) = \beta_0 + \beta_1E(X) = E(Y)$

$E(E(Y|X =x)) = E(\beta_0 + \beta_1x) = \beta_0 + \beta_1x = E(Y|X=x)$ because that already isn't random anymore and so the second $E$ doesn't do anything.

Speaking of notation abuse: $\mu$ usually refers to a mean while error terms are expressed with something like $\epsilon$ or $e$

$\endgroup$
7
$\begingroup$

If $X$ is not random (i.e. just some constant), then $E[Y|X] = E[Y]$ so it is correct. If $X$ was random then $E[Y] = \beta_0 + \beta_1 E[X] + E[\mu]$ and so, $E[Y|X]$ would not be equal to $E[Y]$ in general

$\endgroup$
4
$\begingroup$

$\bullet$ You must bear in mind $\mathbb E[\boldsymbol \varepsilon |\mathbf X]=\mathbf 0\implies \mathbb E[\boldsymbol \varepsilon]=\mathbf 0$ but not $\mathbb E[\boldsymbol \varepsilon]=\mathbf 0\implies \mathbb E[\boldsymbol\varepsilon |\mathbf X]=\mathbf 0.$

$\bullet$ For non-stochastic regressors, you can straightforward use the unconditional conditions rather than conditional. However, when they are random, be careful with conditional and unconditional aspects.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.