# Everything You Cared to Know about Simple Linear Regression

Views and opinions expressed are solely my own.

Users of linear regression often assume when performing linear regression, error terms have to be normally distributed and have constant variance for each fixed $$x$$-value. While these assumptions are true to some extent, not every aspect of linear regression relies on these two assumptions.

The aim of this post is to give a (mostly) complete exposition of simple linear regression, carefully only making assumptions where necessary, and providing a summary at the end.

We show that the justification for fitting a line of best fit through two-dimensional points only requires the assumption that error terms, given a fixed $$x$$-value, have a mean of $$0$$, with no other assumptions required. With no additional assumptions, we demonstrate that the least-squares estimates of the slope and intercept are unbiased for the corresponding linear model parameters. Lastly, we demonstrate that it is only once we begin discussing variances and hypothesis tests with the least-squares estimates of the slope and intercept that we need heteroskedasticity (constant variance of the error terms) and normality of the error terms.

## The Underlying Model

Suppose we have $$n$$ two-dimensional random vectors $$(X_1, Y_1), \dots, (X_n, Y_n)$$. The assumption underlying linear regression is that the true process,

$Y_i = \beta_0 + \beta_1X_i + \epsilon_i$

for $$i = 1, \dots, n$$. In other words, we are saying that

\begin{align} Y_i &= \beta_0 + \beta_1 X_i + \epsilon_i = \text{intercept } + \text{slope }\cdot X_i + \unicode{x201C}\text{noise/error}\unicode{x201D}\text{.}\tag{1} \end{align}

Notice that no assumption of normality of residuals is necessary at this point.

It is also worth noting that we are also assuming that $$Y_i$$ is a linear equation with respect to $$X_i$$ for each data point, with deviations from the linear equation explainable by “noise.” We also assume that for any particular value of $$X_i$$, $$\epsilon_i$$ (our noise/error) has a mean of $$0$$ (that is, $$\mathbb{E}[\epsilon_i \mid X_i] = 0$$).

Note also that most of the time in reality, it is impossible to know if the mechanism described by $$(1)$$ is true for the mechanism which describes how $$Y_i$$ corresponds with $$X_i$$.

## Estimation

The response variable $$Y_i$$, is equal to the equation $Y_i = \beta_0 + \beta_1X_i + \epsilon_i = f(X_i)$ which the notation $$f(X_i)$$ indicating that $$Y_i$$ is a function of $$X_i$$. We wish to estimate $$f(X_i) = \beta_0 + \beta_1X_i + \epsilon_i$$. How do we do it?

### Finding an Optimal Estimator

Let $$\widehat{f}(X_i)$$ be our estimate of $$f(X_i)$$. One way to estimate $$\widehat{f}(X_i)$$ would be to find the equation for $$\widehat{f}(X_i)$$ that would minimize the average squared difference between $$Y_i$$ and $$\widehat{f}(X_i)$$.

That is, we wish to minimize $\mathbb{E}\left\{\left[Y_i - \widehat{f}(X_i)\right]^2\right\}\text{.}$ Observe that this quantity is really difficult to work with: we have a function of two random variables here that we need to find the expectation of. Ideally, instead of working with a two-dimensional distribution, we should try to simplify the above quantity by fixing one of the random variables. Indeed, this is possible through double expectation: $\mathbb{E}\left\{\left[Y_i - \widehat{f}(X_i)\right]^2\right\} = \mathbb{E}\left\{\mathbb{E}\left\{\left[Y_i - \widehat{f}(X_i)\right]^2 \mid X_i\right\} \right\}\text{.}$ Let’s deal with the inner quantity $$\mathbb{E}\left\{\left[Y_i - \widehat{f}(X_i)\right]^2 \mid X_i\right\}$$ for now.

#### Motivating a Solution

To motivate how to deal with this, assuming $$Y_i \mid X_i$$ is a continuous random variable with density $$f_{Y_i \mid X_i}$$, we have \begin{align} \mathbb{E}\left\{\left[Y_i - \widehat{f}(X_i)\right]^2 \mid X_i\right\} = \int_{-\infty}^{\infty}[y_i - \widehat{f}(X_i)]^2f_{Y_i \mid X_i}(y_i \mid X_i)\text{ d}y_i \end{align} Assuming we can interchange the order of differentiation and integration, we can obtain that \begin{align} \dfrac{\partial}{\partial \widehat f(X_i)}\mathbb{E}\left\{\left[Y_i - \widehat{f}(X_i)\right]^2 \mid X_i\right\} &= \int_{-\infty}^{\infty}\dfrac{\partial}{\partial \widehat f(X_i)}[y_i - \widehat{f}(X_i)]^2f_{Y_i \mid X_i}(y_i \mid X_i)\text{ d}y_i \\ &= \int_{-\infty}^{\infty}-2[y_i - \widehat{f}(X_i)]f_{Y_i \mid X_i}(y_i \mid X_i)\text{ d}y_i \end{align} which is equal to $$0$$ if (factoring out the $$-2$$) $\int_{-\infty}^{\infty}y_if_{Y_i \mid X_i}(y_i \mid X_i)\text{ d}y_i -\widehat{f}(X_i)\int_{-\infty}^{\infty}f_{Y_i \mid X_i}(y_i \mid X_i)\text{ d}y_i = 0$ or $\widehat{f}(X_i) = \mathbb{E}[Y_i \mid X_i]$ using the fact that $$\int_{-\infty}^{\infty}f_{Y_i \mid X_i}(y_i \mid X_i)\text{ d}y_i = 1$$. We have \begin{align} \dfrac{\partial^2}{\partial [\widehat f(X_i)]^2}\mathbb{E}\left\{\left[Y_i - \widehat{f}(X_i)\right]^2 \mid X_i\right\} &= \int_{-\infty}^{\infty}2f_{Y_i \mid X_i}(y_i \mid X_i)\text{ d}y_i = 2 > 0 \end{align} regardless of what $$\widehat f(X_i)$$ is, hence $$\widehat f(X_i) = \mathbb{E}[Y_i \mid X_i]$$ minimizes $$\mathbb{E}\left\{\left[Y_i - \widehat{f}(X_i)\right]^2 \mid X_i\right\}$$.

#### A Solution with Less Restrictive Conditions

While the solution above motivates the solution well, it is restrictive in that it requires a continuous random variable with density, and requires allowing the interchange of integration and differentiation. To avoid this problem, let’s start off with the quantity $$\mathbb{E}\left\{\left[Y_i - \widehat{f}(X_i)\right]^2 \mid X_i\right\}$$ again. Subtract and add $$\mathbb{E}[Y_i \mid X_i]$$: \begin{align} \mathbb{E}\left\{\left[Y_i - \widehat{f}(X_i)\right]^2 \mid X_i\right\} &= \mathbb{E}\left\{\left[Y_i - \mathbb{E}[Y_i \mid X_i] + \mathbb{E}[Y_i \mid X_i] - \widehat{f}(X_i)\right]^2 \mid X_i\right\}\text{.} \end{align} Let us focus on the part that is squared and expand it: \begin{align} \left[Y_i - \mathbb{E}[Y_i \mid X_i] + \mathbb{E}[Y_i \mid X_i] - \widehat{f}(X_i)\right]^2 = &\left\{Y_i - \mathbb{E}[Y_i \mid X_i]\right\}^2 + \left\{\mathbb{E}[Y_i \mid X_i] - \widehat{f}(X_i)\right\}^2 \\ &+ 2\left\{Y_i - \mathbb{E}[Y_i \mid X_i]\right\}\left\{\mathbb{E}[Y_i \mid X_i] - \widehat{f}(X_i)\right\}\text{.} \end{align}

Now, focusing on the third term on the right, observe $$2\left\{\mathbb{E}[Y_i \mid X_i] - \widehat{f}(X_i)\right\}$$ depends on $$X_i$$ (and thus would be pulled out of the inner expectation), so we focus on

$\mathbb{E}\left\{Y_i - \mathbb{E}[Y_i \mid X_i] \mid X_i\right\} = \mathbb{E}[Y_i \mid X_i] - \mathbb{E}[Y_i \mid X_i] = 0$ thus the third term on the right is $$0$$ when we take its expectation. Thus $\mathbb{E}\left\{\left[Y_i - \widehat{f}(X_i)\right]^2 \mid X_i\right\} = \mathbb{E}\left[\left\{Y_i - \mathbb{E}[Y_i \mid X_i]\right\}^2\mid X_i\right] + \mathbb{E}\left[\left\{\mathbb{E}[Y_i \mid X_i] - \widehat{f}(X_i)\right\}^2\mid X_i\right]$

and now, taking the expectation with respect to $$X_i$$, by double expectation, we obtain

$\mathbb{E}\left\{\left[Y_i - \widehat{f}(X_i)\right]^2\right\} = \mathbb{E}\left[\left\{Y_i - \mathbb{E}[Y_i \mid X_i]\right\}^2\right] + \mathbb{E}\left[\left\{\mathbb{E}[Y_i \mid X_i] - \widehat{f}(X_i)\right\}^2\right]$

Observe in the above that in the right-hand side, the only term that depends on $$\widehat{f}(X_i)$$ is $\mathbb{E}\left[\left\{\mathbb{E}[Y_i \mid X_i] - \widehat{f}(X_i)\right\}^2\right]\text{.}$ What we have here is the expectation of a squared quantity, which must be non-negative and thus $$\geq 0$$. Thus, the expectation is minimized when $\left\{\mathbb{E}[Y_i \mid X_i] - \widehat{f}(X_i)\right\}^2 = 0$ or $\mathbb{E}[Y_i \mid X_i] = \widehat{f}(X_i)\text{.}$

### Context in Linear Regression

What does the fact that $$\mathbb{E}[Y_i \mid X_i] = \widehat{f}(X_i)$$ mean? This means that since $$Y_i = \beta_0 + \beta_1 X_i + \epsilon_i$$, the best estimator for $$f(X_i)$$ is $\widehat{f}(X_i) = \mathbb{E}[Y_i \mid X_i] = \mathbb{E}[\beta_0 + \beta_1 X_i + \epsilon_i \mid X_i] = \beta_0 + \beta_1 X_i$ using the fact that $$\mathbb{E}[\epsilon_i \mid X_i] = 0$$.

This means, then, that even though the process $$Y_i$$ is given by a linear equation plus an error/noise term, the best estimate we can come up with for $$Y_i$$ is the underlying linear equation. We don’t need to worry about trying to incorporate the error/noise term, given the assumptions we’ve made.

Unfortunately, in reality, we don’t know the values for $$\beta_0$$ and $$\beta_1$$ in our estimator $$\widehat f(X_i) = \beta_0 + \beta_1 X_i$$. We need to estimate $$\beta_0$$ and $$\beta_1$$ somehow.

The most popular way to estimate $$\beta_0$$ and $$\beta_1$$ is through least-squares estimation (which can be justified by means of the Gauss-Markov Theorem). We wish to find $$\widehat\beta_0$$ and $$\widehat\beta_1$$ which minimizes the sum of squared differences between $$Y_i$$ and our estimated linear equation $$\widehat\beta_0 + \widehat\beta_1X_i$$; i.e., we aim to minimize $L(\widehat\beta_0, \widehat\beta_1) = \sum_{i=1}^{n}[Y_i - (\widehat\beta_0 + \widehat\beta_1X_i)]^2 = \sum_{i=1}^{n}(Y_i - \widehat\beta_0 - \widehat\beta_1X_i)^2$ Taking partial derivatives, we obtain \begin{align} \dfrac{\partial L}{\partial \widehat\beta_0} &= -2\sum_{i=1}^{n}(Y_i - \widehat\beta_0 - \widehat\beta_1X_i)\\ \dfrac{\partial L}{\partial \widehat\beta_1} &= -2\sum_{i=1}^{n}X_i(Y_i - \widehat\beta_0 - \widehat\beta_1X_i) \\ \dfrac{\partial^2 L}{\partial \widehat\beta_0^2} &= -2\sum_{i=1}^{n}(-1) = 2n\\ \dfrac{\partial^2 L}{\partial \widehat\beta_1^2} &= -2\sum_{i=1}^{n}-X_i^2 = 2\sum_{i=1}^{n}X_i^2 \\ \dfrac{\partial^2 L}{\partial \widehat\beta_0 \widehat\beta_1} &= 0 = \dfrac{\partial^2 L}{\partial \widehat\beta_1 \widehat\beta_0}\\ \end{align} The Hessian matrix is given by $\mathbf{H} = \begin{bmatrix} 2n & 0 \\ 0 & 2\sum_{i=1}^{n}X_i^2 \end{bmatrix}$ and $\det \mathbf{H} = 2n\left(2\sum_{i=1}^{n}X_i^2\right) = 4n\sum_{i=1}^{n}X_i^2 > 0$ thus $$\mathbf{H}$$ is a positive-definite matrix, and whatever solution we get for $$(\widehat\beta_0, \widehat\beta_1)$$ will minimize $$L$$. We start by setting the first partial derivatives equal to $$0$$: $\dfrac{\partial L}{\partial \widehat\beta_0} = 0 \implies \sum_{i=1}^{n}(Y_i - \widehat\beta_0 - \widehat\beta_1X_i) = \sum_{i=1}^{n}Y_i - n\widehat\beta_0 - \widehat\beta_1\sum_{i=1}^{n}X_i = 0$ thus yielding $\widehat\beta_0 = \dfrac{\sum_{i=1}^{n}Y_i - \widehat\beta_1\sum_{i=1}^{n}X_i }{n}\tag{2}$ and $\dfrac{\partial L}{\partial \widehat\beta_1} = 0 \implies \sum_{i=1}^{n}X_i(Y_i - \widehat\beta_0 - \widehat\beta_1X_i) = \sum_{i=1}^{n}X_iY_i-\widehat\beta_0\sum_{i=1}^{n}X_i-\widehat\beta_1\sum_{i=1}^{n}X_i^2$ thus yielding $\widehat\beta_1 = \dfrac{\sum_{i=1}^{n}X_iY_i-\widehat\beta_0\sum_{i=1}^{n}X_i}{\sum_{i=1}^{n}X_i^2}\text{.}\tag{3}$ However, we’ve written $$\widehat\beta_0$$ in terms of $$\widehat\beta_1$$, and vice versa. We thus substitute $$(3)$$ into $$(2)$$ and attempt to combine like terms: \begin{align} &\widehat\beta_0 = \dfrac{\sum_{i=1}^{n}Y_i - \left(\dfrac{\sum_{i=1}^{n}X_iY_i-\widehat\beta_0\sum_{i=1}^{n}X_i}{\sum_{i=1}^{n}X_i^2}\right)\sum_{i=1}^{n}X_i }{n}\\ &\implies \widehat\beta_0 = \dfrac{\sum_{i=1}^{n}Y_i}{n} - \dfrac{\sum_{i=1}^{n}X_iY_i \cdot \sum_{i=1}^{n}X_i}{n\sum_{i=1}^{n}X_i^2}+\widehat\beta_0\cdot \dfrac{\left(\sum_{i=1}^{n}X_i\right)^2}{n\sum_{i=1}^{n}X_i^2} \\ &\implies \left[1 - \dfrac{\left(\sum_{i=1}^{n}X_i\right)^2}{n\sum_{i=1}^{n}X_i^2}\right]\widehat\beta_0 = \dfrac{\sum_{i=1}^{n}Y_i}{n} - \dfrac{\sum_{i=1}^{n}X_iY_i \cdot \sum_{i=1}^{n}X_i}{n\sum_{i=1}^{n}X_i^2} \\ &\implies \left[\dfrac{n\sum_{i=1}^{n}X_i^2 - \left(\sum_{i=1}^{n}X_i\right)^2}{n\sum_{i=1}^{n}X_i^2}\right]\widehat\beta_0 = \dfrac{\sum_{i=1}^{n}Y_i}{n} - \dfrac{\sum_{i=1}^{n}X_iY_i \cdot \sum_{i=1}^{n}X_i}{n\sum_{i=1}^{n}X_i^2} \\ &\implies \left[n\sum_{i=1}^{n}X_i^2 - \left(\sum_{i=1}^{n}X_i\right)^2\right]\widehat\beta_0 = \sum_{i=1}^{n}Y_i \cdot \sum_{i=1}^{n}X_i^2 - \sum_{i=1}^{n}X_iY_i \cdot \sum_{i=1}^{n}X_i \\ &\implies \widehat\beta_0 = \dfrac{\sum_{i=1}^{n}Y_i \cdot \sum_{i=1}^{n}X_i^2 - \sum_{i=1}^{n}X_iY_i \cdot \sum_{i=1}^{n}X_i}{n\sum_{i=1}^{n}X_i^2 - \left(\sum_{i=1}^{n}X_i\right)^2} \end{align} and using $$(2)$$ to obtain $$\widehat\beta_1$$: $\widehat\beta_0 = \dfrac{\sum_{i=1}^{n}Y_i}{n} - \widehat\beta_1 \cdot \dfrac{\sum_{i=1}^{n}X_i}{n} = \overline{Y}_n - \widehat\beta_1\overline{X}_n \implies \widehat\beta_1 = \dfrac{\overline{Y}_n - \widehat\beta_0}{\overline{X}_n}\text{.}\tag{3}$

## Bias of Estimators

With $$\mathbf{X} = (X_1, \dots, X_n)$$ fixed (or in the language of probability, “conditioned on” $$X_1, \dots, X_n$$), observe that \begin{align} \mathbb{E}\left[\widehat\beta_0 \mid \mathbf{X}\right] &= \dfrac{\sum_{i=1}^{n}X_i^2 \cdot \sum_{i=1}^{n}\mathbb{E}\left[Y_i \mid \mathbf{X}\right] - \sum_{i=1}^{n}X_i\mathbb{E}\left[Y_i \mid \mathbf{X}\right] \cdot \sum_{i=1}^{n}X_i}{n\sum_{i=1}^{n}X_i^2 - \left(\sum_{i=1}^{n}X_i\right)^2} \\ &= \dfrac{\sum_{i=1}^{n}X_i^2 \cdot \sum_{i=1}^{n}\mathbb{E}\left[Y_i \mid X_i\right] - \sum_{i=1}^{n}X_i\mathbb{E}\left[Y_i \mid X_i\right] \cdot \sum_{i=1}^{n}X_i}{n\sum_{i=1}^{n}X_i^2 - \left(\sum_{i=1}^{n}X_i\right)^2} \\ &= \dfrac{\sum_{i=1}^{n}X_i^2 \cdot \sum_{i=1}^{n}(\beta_0 + \beta_1X_i) - \sum_{i=1}^{n}X_i(\beta_0 + \beta_1X_i) \cdot \sum_{i=1}^{n}X_i}{n\sum_{i=1}^{n}X_i^2 - \left(\sum_{i=1}^{n}X_i\right)^2} \\ &= \dfrac{\sum_{i=1}^{n}X_i^2 \cdot (n\beta_0 + \beta_1\sum_{i=1}^{n}X_i) - \left(\beta_0\sum_{i=1}^{n}X_i + \beta_1\sum_{i=1}^{n}X_i^2\right) \cdot \sum_{i=1}^{n}X_i}{n\sum_{i=1}^{n}X_i^2 - \left(\sum_{i=1}^{n}X_i\right)^2} \\ &= \dfrac{n\beta_0\sum_{i=1}^{n}X_i^2 + \beta_1\sum_{i=1}^{n}X_i \cdot \sum_{i=1}^{n}X_i^2 - \beta_0\left(\sum_{i=1}^{n}X_i\right)^2 - \beta_1\sum_{i=1}^{n}X_i^2 \cdot \sum_{i=1}^{n}X_i}{n\sum_{i=1}^{n}X_i^2 - \left(\sum_{i=1}^{n}X_i\right)^2} \\ &= \dfrac{n\beta_0\sum_{i=1}^{n}X_i^2 - \beta_0\left(\sum_{i=1}^{n}X_i\right)^2}{n\sum_{i=1}^{n}X_i^2 - \left(\sum_{i=1}^{n}X_i\right)^2} \\ &= \beta_0\left[\dfrac{n\sum_{i=1}^{n}X_i^2 - \left(\sum_{i=1}^{n}X_i\right)^2}{n\sum_{i=1}^{n}X_i^2 - \left(\sum_{i=1}^{n}X_i\right)^2} \right] \\ &= \beta_0\text{.} \end{align} Thus, conditioned on $$\mathbf{X}$$, $$\widehat\beta_0$$ is an unbiased estimator for $$\beta_0$$. We can also observe that by $$(3)$$ that \begin{align} \mathbb{E}\left[\widehat\beta_1 \mid \mathbf{X}\right] = \dfrac{\mathbb{E}\left[\overline{Y}_n \mid \mathbf{X}\right] - \mathbb{E}\left[\widehat\beta_0 \mid \mathbf{X}\right]}{\overline{X}_n}\text{.} \end{align} Now $\mathbb{E}\left[\overline{Y}_n \mid \mathbf{X}\right] = \dfrac{1}{n}\sum_{i=1}^{n}\mathbb{E}\left[Y_i \mid X_i \right] = \dfrac{1}{n}\sum_{i=1}^{n}(\beta_0 + \beta_1 X_i) = \beta_0 + \beta_1\overline{X}_n$ thus \begin{align} \mathbb{E}\left[\widehat\beta_1 \mid \mathbf{X}\right] = \dfrac{\mathbb{E}\left[\overline{Y}_n \mid \mathbf{X}\right] - \mathbb{E}\left[\widehat\beta_0 \mid \mathbf{X}\right]}{\overline{X}_n} = \dfrac{\beta_0 + \beta_1\overline{X}_n - \beta_0}{\overline{X}_n} = \beta_1 \end{align} showing that conditioned on $$\mathbf{X}$$, $$\widehat\beta_1$$ is unbiased for $$\beta_1$$.

## Variance of Estimators

Here, we make the assumption that in $$(1)$$ that conditioned on $$\mathbf{X}$$, $$\epsilon_i$$ are independent with $$\text{Var}\left[\epsilon_i\mid X_i\right] = \sigma^2 > 0$$. It follows easily that assuming $$\beta_0, \beta_1$$ are constants that, conditioned on $$\mathbf{X}$$, $$Y_1, \dots, Y_n$$ are independent with $\text{Var}\left[Y_i \mid X_i\right] = \sigma^2\tag{4}$ as well. Following very similar steps to what we did for computing $$\mathbb{E}\left[\widehat\beta_0 \mid \mathbf{X}\right]$$, we obtain \begin{align} \text{Var}\left[\widehat\beta_0 \mid \mathbf{X}\right] &= \dfrac{\left(\sum_{i=1}^{n}X_i^2\right)^2 \cdot \sum_{i=1}^{n}\text{Var}\left[Y_i \mid \mathbf{X}\right] - \left(\sum_{i=1}^{n}X_i\right)^2\sum_{i=1}^{n}X_i^2\text{Var}\left[Y_i \mid \mathbf{X}\right]}{\left[n\sum_{i=1}^{n}X_i^2 - \left(\sum_{i=1}^{n}X_i\right)^2\right]^2} \\ &= \dfrac{\left(\sum_{i=1}^{n}X_i^2\right)^2 \cdot \sum_{i=1}^{n}\text{Var}\left[Y_i \mid X_i\right] - \left(\sum_{i=1}^{n}X_i\right)^2\sum_{i=1}^{n}X_i^2\text{Var}\left[Y_i \mid X_i\right]}{\left[n\sum_{i=1}^{n}X_i^2 - \left(\sum_{i=1}^{n}X_i\right)^2\right]^2} \\ &= \dfrac{\left(\sum_{i=1}^{n}X_i^2\right)^2 \cdot n\sigma^2 - \left(\sum_{i=1}^{n}X_i\right)^2\sum_{i=1}^{n}X_i^2\sigma^2}{\left[n\sum_{i=1}^{n}X_i^2 - \left(\sum_{i=1}^{n}X_i\right)^2\right]^2} \\ &= \sigma^2 \cdot \dfrac{n\left(\sum_{i=1}^{n}X_i^2\right)^2 - \left(\sum_{i=1}^{n}X_i\right)^2\sum_{i=1}^{n}X_i^2}{\left[n\sum_{i=1}^{n}X_i^2 - \left(\sum_{i=1}^{n}X_i\right)^2\right]^2}\tag{5} \end{align} and \begin{align} \text{Var}\left[\widehat\beta_1 \mid \mathbf{X} \right] &= \dfrac{\text{Var}\left[\overline{Y}_n \mid \mathbf{X} \right] + \text{Var}\left[\widehat\beta_0 \mid \mathbf{X} \right] - 2 \cdot \text{Cov}(\overline{Y}_n, \widehat\beta_0 \mid \mathbf{X})}{\overline{X}_n^2} \\ &= \dfrac{\sigma^2/n + \text{Var}\left[\widehat\beta_0 \mid \mathbf{X} \right] - 2 \cdot \text{Cov}(\overline{Y}_n, \widehat\beta_0 \mid \mathbf{X})}{\overline{X}_n^2}\text{.} \end{align} Now \begin{align} \text{Cov}(\overline{Y}_n, \widehat\beta_0 \mid \mathbf{X}) &= \text{Cov}\left(\dfrac{1}{n}\sum_{i=1}^{n}Y_i, \dfrac{\sum_{i=1}^{n}Y_i \cdot \sum_{i=1}^{n}X_i^2 - \sum_{i=1}^{n}X_iY_i \cdot \sum_{i=1}^{n}X_i}{n\sum_{i=1}^{n}X_i^2 - \left(\sum_{i=1}^{n}X_i\right)^2} \mid \mathbf{X}\right) \\ &= \dfrac{\text{Cov}\left(\sum_{i=1}^{n}Y_i, \sum_{i=1}^{n}Y_i \cdot \sum_{i=1}^{n}X_i^2 - \sum_{i=1}^{n}X_iY_i \cdot \sum_{i=1}^{n}X_i \mid \mathbf{X}\right)}{n\left[n\sum_{i=1}^{n}X_i^2 - \left(\sum_{i=1}^{n}X_i\right)^2 \right]} \\ &= \dfrac{\sum_{i=1}^{n}X_i^2 \cdot \text{Cov}\left(\sum_{i=1}^{n}Y_i, \sum_{i=1}^{n}Y_i \mid \mathbf{X}\right) - \sum_{i=1}^{n}X_i \cdot \text{Cov}\left(\sum_{i=1}^{n}Y_i,\sum_{i=1}^{n}X_iY_i \mid \mathbf{X}\right)}{n\left[n\sum_{i=1}^{n}X_i^2 - \left(\sum_{i=1}^{n}X_i\right)^2 \right]} \\ &= \dfrac{\sum_{i=1}^{n}X_i^2 \cdot \text{Cov}\left(\sum_{i=1}^{n}Y_i, \sum_{j=1}^{n}Y_j \mid \mathbf{X}\right) - \sum_{i=1}^{n}X_i \cdot \text{Cov}\left(\sum_{i=1}^{n}Y_i,\sum_{j=1}^{n}X_jY_j \mid \mathbf{X}\right)}{n\left[n\sum_{i=1}^{n}X_i^2 - \left(\sum_{i=1}^{n}X_i\right)^2 \right]} \\ &= \dfrac{\sum_{i=1}^{n}X_i^2 \cdot \sum_{i=1}^{n}\sum_{j=1}^{n}\text{Cov}\left(Y_i, Y_j \mid \mathbf{X}\right) - \sum_{i=1}^{n}X_i \cdot \sum_{i=1}^{n}\sum_{j=1}^{n}\text{Cov}\left(Y_i,X_jY_j \mid \mathbf{X}\right)}{n\left[n\sum_{i=1}^{n}X_i^2 - \left(\sum_{i=1}^{n}X_i\right)^2 \right]} \\ &= \dfrac{\sum_{i=1}^{n}X_i^2 \cdot n\sigma^2 - \sigma^2\left(\sum_{i=1}^{n}X_i\right)^2}{n\left[n\sum_{i=1}^{n}X_i^2 - \left(\sum_{i=1}^{n}X_i\right)^2 \right]} \\ &= \sigma^2 \cdot \dfrac{n\sum_{i=1}^{n}X_i^2 - \left(\sum_{i=1}^{n}X_i\right)^2}{n\left[n\sum_{i=1}^{n}X_i^2 - \left(\sum_{i=1}^{n}X_i\right)^2 \right]} \\ &= \dfrac{\sigma^2}{n} \end{align} thus $\text{Var}\left[\widehat\beta_1 \mid \mathbf{X} \right] = \dfrac{\text{Var}\left[\widehat\beta_0 \mid \mathbf{X} \right] - \sigma^2/n}{\overline{X}_n^2}\tag{6}$ where $$\text{Var}\left[\widehat\beta_0 \mid \mathbf{X} \right]$$ is computed as in $$(5)$$.

## Statistical Inference on Estimators

Let us assume now that $$\epsilon_i$$ is normally distributed for each $$i = 1, 2, \dots, n$$. Then, \begin{align*} \widehat\beta_0 &= \dfrac{\sum_{j=1}^{n}Y_j \cdot \sum_{i=1}^{n}X_i^2 - \sum_{j=1}^{n}X_jY_j \cdot \sum_{i=1}^{n}X_i}{n\sum_{i=1}^{n}X_i^2 - \left(\sum_{i=1}^{n}X_i\right)^2} \\ &= \dfrac{\left(n\beta_0 + \beta_1\sum_{j=1}^{n}X_i + \sum_{j=1}^{n}\epsilon_j\right) \cdot \sum_{i=1}^{n}X_i^2 - \sum_{j=1}^{n}X_j\left(\beta_0 + \beta_1X_j + \epsilon_j\right) \cdot \sum_{i=1}^{n}X_i}{n\sum_{i=1}^{n}X_i^2 - \left(\sum_{i=1}^{n}X_i\right)^2} \\ &= C\left[\text{a bunch of stuff} + \left(\sum_{i=1}^{n}X_i^2 \right)\sum_{j=1}^{n}\epsilon_j - \sum_{i=1}^{n}X_i\left(\sum_{j=1}^{n}X_j\epsilon_j \right)\right] \\ &= C\left[\text{a bunch of stuff} + \sum_{j=1}^{n}\left(\sum_{i=1}^{n}X_i^2 - X_j\sum_{k=1}^{n}X_k \right)\epsilon_j \right] \end{align*} where $$C$$ is a placeholder constant (assuming $$\mathbf{X}$$ fixed). Thus, we have demonstrated that conditioned on $$\mathbf{X}$$, $$\widehat\beta_0$$ is normally distributed since it is a linear combination of $$\epsilon_i$$ plus a constant, hence $\widehat\beta_0 \mid \mathbf{X} \sim \mathcal{N}\left(\beta_0, \sigma^2 \cdot \dfrac{n\left(\sum_{i=1}^{n}X_i^2\right)^2 - \left(\sum_{i=1}^{n}X_i\right)^2\sum_{i=1}^{n}X_i^2}{\left[n\sum_{i=1}^{n}X_i^2 - \left(\sum_{i=1}^{n}X_i\right)^2\right]^2} \right)\text{.}$ Next, using the fact that $\widehat{\beta}_1 = \dfrac{\overline{Y}_n - \widehat\beta_0}{\overline{X}_n}$ we obtain \begin{align} \widehat{\beta}_1 &= \dfrac{\beta_0 + \beta_1\overline{X}_n + \frac{1}{n}\sum_{j=1}^{n}\epsilon_j - C\left[\text{a bunch of stuff} + \sum_{j=1}^{n}\left(\sum_{i=1}^{n}X_i^2 - X_j\sum_{k=1}^{n}X_k \right)\epsilon_j \right]}{\overline{X}_n} \\ &= C^{\prime}\left[\text{a bunch of other stuff} + \sum_{j=1}^{n}\left(\dfrac{1}{n} - C\sum_{i=1}^{n}X_i^2 + CX_j\sum_{k=1}^{n}X_k\right)\epsilon_j \right] \end{align} which is another linear combination of $$\epsilon_i$$ plus a constant, hence $\widehat{\beta}_1 \mid \mathbf{X} \sim \mathcal{N}\left(\beta_1, \dfrac{\text{Var}\left[\widehat\beta_0 \mid \mathbf{X} \right] - \sigma^2/n}{\overline{X}_n^2}\right)\text{.}$

With these results in hand, we may construct confidence intervals and perform hypothesis tests for $$\beta_0 \mid \mathbf{X}$$ and $$\beta_1 \mid \mathbf{X}$$.

## Theoretical Conclusions

• The true process that is assumed is that $$Y_i = \beta_0 + \beta_1 X_i + \epsilon_i$$, with $$\mathbb{E}\left[\epsilon_i \mid X_i\right] = 0$$.
• Based on expected squared differences between $$Y_i$$ and a given function we can use to estimate $$Y_i$$ dependent only on $$X_i$$, the optimal choice is $$\mathbb{E}[Y_i \mid X_i] = \beta_0 + \beta_1 X_i$$.
• Finding corresponding coefficients for a linear equation that best fits a set of points is our attempt to estimate $$\mathbb{E}[Y_i \mid X_i]$$.
• The Gauss-Markov Theorem provides rationale as to why least-squares estimators for the coefficients are the “best” estimators to use.
• The least-squares estimators of $$\beta_0$$ and $$\beta_1$$ are unbiased, conditioned on $$X_1, \dots, X_n$$ fixed.
• The heteroskedasticity assumption that $$\text{Var}[\epsilon_i \mid X_i]= \sigma^2 > 0$$ and independence of $$\epsilon_1, \dots, \epsilon_n$$ given $$\mathbf{X}$$ is only necessary for computation of the variances of the least-squares estimators of $$\beta_0$$ and $$\beta_1$$.
• Independence of $$\epsilon_1, \dots, \epsilon_n$$ given $$\mathbf{X}$$ implies independence of $$Y_1, \dots, Y_n$$ given $$\mathbf{X}$$.
• Hypothesis tests for the least-squares estimators of $$\beta_0$$ and $$\beta_1$$ rely on the assumption that $$\epsilon_1, \dots, \epsilon_n$$ are normally distributed.

## Next Steps

It is my philosophy that theory needs to drive good practice in statistics. In a future blog post, I will be detailing how these theoretical results have implications for practical analytical issues.

##### Yeng Miller-Chang

I am a Senior Data Scientist - Global Knowledge Solutions with General Mills, Inc. Views and opinions expressed are my own.