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In the context of linear regression where there is an assumption of "constant variance" I have read this:

$$\mathbb{V}(\epsilon_i \mid X_i)=\sigma^2$$

But there are two ways I can read this. Either $\mathbb{V}(\epsilon_i \mid X_i)$ is a random variable and the right hand side is a constant, in which case we are saying $\mathbb{V}(\epsilon_i \mid X_i)$ is a random variable with a point mass distribution, or $\mathbb{V}(\epsilon_i \mid X_i=x_i)$ is a function that takes as input a possible value for $X_i$, and returns a value for the variance that depends on the specific value of $x_i$ that was passed in. In other words are we truly assuming the variance is constant regardless of the data (in which case why did we bother conditioning?), or are we saying for any particular fixed data value, the variance of the errors specifically for occurrences of that data value is a constant? If the latter it seems like $\mathbb{V}(\epsilon_i \mid X_i)=\sigma^2(x_i)$ would maybe be clearer.

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  • $\begingroup$ Where exactly have you seen it? How is the notation explained/defined there? $\endgroup$
    – corey979
    Commented May 20, 2020 at 22:28
  • $\begingroup$ @corey979 earlier in the text they say $V(A \mid B)$ is a random variable and $V(A \mid B=b)$ is a function that takes values of $b$ and spits out a specific scalar. But as I say above both interpretations are weird in this context. $\endgroup$ Commented May 22, 2020 at 3:56

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I cannot see anything weird, both interpretation sort of say the same thing in different ways. Your confusion is highlighted by the last sentence, If the latter it seems like $\mathbb{V}(\epsilon_i \mid X_i)=\sigma^2(x_i)$ would maybe be clearer. But the point is that, for constant variance, that function is a constant function, $\sigma^2(x_i)=\sigma^2$. If it is any other, non-constant function, we would not have constant variance.

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