# If $e_0$ are the OLS residuals, what is random in $\hat{\beta}_{OLS}|f(e_0) < \hat{\beta} < f^*(e_0)$?

This is a follow up question to the question I've posted here.

Suppose $$Y \sim N(X\beta, \sigma^2I)$$, where $$y \in \mathbb{R}^n$$. Let $$X \in \mathbb{R}^{n \times p}$$ denote a full rank design matrix. By ordinary least squares, the residuals are $$\hat{e} = (I - X(X^TX)^{-1}X^T)Y$$ Let $$u_1 \in \mathbb{R}^p$$ denote a unit vector in $$\mathbb{R}^p$$. Let $$\hat{\beta}_1 = u_1^T (X^TX)^{-1}X^TY$$ denote the OLS estimate for the first covariate in $$X$$. Suppose I'm interested in the conditional distribution of $$\hat{\beta}_1 | f(\hat{e}) < \hat{\beta}_1 < f^*(\hat{e}), \hat{e} = e_0$$

where $$f(e_0)$$ and $$f^*(e_0)$$ are some functions of $$e_0.$$ From the answer to my previous post, $$\hat{\beta}_1$$ and residuals $$\hat{e}$$ are independent, so I can drop $$\hat{e} = e_0$$ from the conditioning to obtain

$$\hat{\beta}_1 | f(e_0) < \hat{\beta}_1 < f^*(e_0)$$

Since $$\hat{\beta}_1$$ is normal, then truncating $$\hat{\beta}_1$$ between some fixed quantities $$f(e_0)$$ and $$f^*(e_0)$$ leads to a truncated normal. Therefore, $$\hat{\beta}_1|f(e_0) < \hat{\beta}_1 < f^*(e_0)$$ is a truncated normal with lower truncation limit $$f(e_0)$$ and upper truncation limit $$f^*(e_0)$$

My question is, once I condition on $$\hat{e} = e_0,$$ is there still randomness left in $$\hat{\beta}_1$$? Based on my previous post, I learned that $$[\hat{\beta}_1|\hat{e} = e_0] \overset{d}{=} \hat{\beta}_1$$

Is the same true in this case? i.e.,

$$[\hat{\beta}_1 | f(e_0) < \hat{\beta}_1 < f^*(e_0)]\overset{d?}{=} \hat{\beta}_1$$

The reason I feel that this is not true is because changing the value of $$e_0$$ would lead to a different set of truncation limits on $$\hat{\beta}_1$$. On the other hand, if $$\hat{\beta}_1$$ is independent of $$\hat{e}$$, then it's independent from any functions of $$\hat{e}$$. This might suggest that $$[\hat{\beta}_1 | f(e_0) < \hat{\beta}_1 < f^*(e_0)]\overset{d}{=} \hat{\beta}_1$$. I am quite confused and would greatly appreciate any insight on this.

• I'm a little confused by your notation. How does your notation $y\sim N(\mu,\sigma^2 I)$ relate to the rest of the problem. Based on the rest of the problem, I would have guessed you need something like $Y|X\sim N(X\beta,\sigma^2 I)$? Also, what is $f$? Is it assumed known but arbitrary? Does it need to be invertible? And it sounds like it maps from $\mathbb{R}^n$ to $\mathbb{R}$? Apr 1, 2021 at 20:33
• Yes, technically $y|X \sim N(X\beta, \sigma^2I)$. $f$ is arbitrary and maps from $\mathbb{R}^n$ to $\mathbb{R}$. Apr 2, 2021 at 3:47
• Thanks. another question. I see now that my first comment may have misled you, notationally speaking, because you are using $\beta$ in a different sense than I assumed. I had interpreted $\beta_1$ as the regression coefficient corresponding to the first predictor, but you are writing $\beta_1$ as the fitted value of the outcome given $X=u_1$ for some covariate pattern $u_1$. In other words, what you are calling $\hat\beta_1$, I would write as $\hat Y|X=u_1$, or $\hat Y(u_1)$. Is my understanding correct? Apr 2, 2021 at 18:42
• $\beta_1$ is the coefficient corresponding to the first covariate. $u_1$ is a unit vector with 1 in the first position and 0 everywhere else. So $\beta_1 = u_1^T\beta$, where $\beta_1$ is a scalar, and $\beta$ is the entire coefficient vector of length $p$. Apr 2, 2021 at 20:21
• Ok. I understand a unit vector to be any vector with length 1, and I was confused by your stating that $u_1 \in \mathbb{R}^p$ instead of just saying $u_1=\{1, 0, \ldots, 0\}$. Apr 2, 2021 at 20:43

Knowing that $$\hat\beta_1$$ is statistically independent from the residual vector $$\hat e$$ is not relevant to the question. Your question comes down to: is it true that $$[\hat{\beta}_1 | c_l < \hat{\beta}_1 < c_u]\overset{d}{=}\hat{\beta}_1$$ for some arbitrary constants $$c_l$$ and $$c_u$$ where $$c_l? The answer is 'no: they are not equal in distribution'. As a counterexample, assuming that $$\hat\beta_1$$ is normally distributed, $$\Pr(\hat\beta_1 < c_l)>0$$ but $$\Pr(\hat\beta_1 < c_l | c_l < \hat{\beta}_1 < c_u)=0$$.
Addendum To your comment, On the other hand, if $$\hat\beta_1$$ is independent of $$\hat e$$ , then it's independent from any functions of $$\hat e$$, that only applies to functions of $$\hat e$$ alone. You are proposing to condition on a random variable that is a function of both $$\hat e$$ and $$\hat \beta_1$$, which is an entirely different kind of dependence. Altogether.