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Let $X$ be an $n\times p$ full-rank matrix of predictors with $n > p$, and $y$ be the vector of responses. Assume $y \sim N(X\beta, \sigma^2I)$, and let $\hat{\beta}$ be the least-squares estimator of $\beta$. Let $e = y - X\hat{\beta}$ the vector of residuals.

Since $y$ is normal, then $\hat{\beta} = (X^TX)^{-1}X^Ty \sim N(\beta, \sigma^2(X^TX)^{-1})$. And $E(\hat{\beta}) = \beta$.

What does it mean to take the conditional expectation given the residuals? i.e., $E(\hat{\beta}|e = e_0),$ where "=" is interpreted component-wise.

I have two questions:

  1. Intuitively, what exactly does it mean to condition on the residuals $e = e_0$? Since residuals $e = y - X\hat{\beta}$, does this mean that we're assuming that $y = y_0$ and $\hat{\beta} = \hat{\beta}^*$ such that $e_0 = y_0 - X\hat{\beta}^*$? That is, we are conditioning on a specific vector of realizations for the responses $y_0$ and a specific vector of OLS coefficients $\hat{\beta}^*$?
  2. If the above were true, I thought the probability of a continuous random variable is 0. Therefore, is the probability of $e = e_0$ also 0 since $e$ are continuous?
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Question 1

If $y\sim\mathcal{N}(X\beta, \sigma^2I)$, you can obtain $e$ via a linear transformation of $y$, $e=(I-H)y$, this means $E$ is normally distributed with mean $(I-H)X\beta$ and covariance $\sigma^2(I-H)(I-H)^T$. Now you have the pdf $f_E(e)$ and its value at some $e_0$.

You need to estimate the joint distribution $f(\hat{\beta}, E)$. This can be done through a linear transformation $A$ with blocks $(I-H)$ and $(X^TX)^{-1}X^T$

Question 2

assuming all variables have pdfs, you obviously have $P(V=v) = 0$ for any random variable $V$. However as long as you have that the pdf is non-zero $f_V(v) > 0$ you can define conditional pdfs as $$ f(w\mid v) = \dfrac{f_{W,V}(w,v)}{f_V(v)}$$ and compute expectations using $$E[\phi(W)\mid V=v] = \int\phi(w)f(w\mid v)\mathrm{d}w $$ For example think of $f(W\mid V=v)$ as being $W\sim\mathcal{N}(v,\sigma^2)$. You have zero probability, but the pdf changes as the variable $v$ changes.

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  • $\begingroup$ Thanks. So what you're saying is, since $e \sim N(0, \sigma^2(I-H))$, where $H = X(X^TX)^{-1}X^T$, then $P(e = e_0) = 0$ because $e$ is a continuous random variable, correct? $\endgroup$
    – Adrian
    Jan 26 at 13:58
  • $\begingroup$ $P(e=e_0)$ is indeed 0 but thats not a problem, since the pdf $f_E$ is normal and therefore never 0 $\endgroup$
    – ArnoV
    Jan 26 at 14:22
  • $\begingroup$ Could you explain what does it intuitively mean to condition on $e = e_0$? i.e., for question 1: "does this mean that we're assuming that $y = y_0$ and $\hat{\beta} = \hat{\beta}^*$ such that $e_0 = y_0 - X\hat{\beta}^*$? That is, we are conditioning on a specific vector of realizations for the responses $y_0$ and a specific vector of OLS coefficients $\hat{\beta}^*$?" $\endgroup$
    – Adrian
    Jan 26 at 14:26

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