# Difference between E(Y) and E(Y|X) in regression

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.

• $E(Y \mid X)$ 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 $E(Y \mid X=x) = \beta_0 + \beta_1 x$ Commented Feb 2, 2023 at 11:06

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 $$e$$

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

$$\bullet$$ 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.