Interpretations of Linear Regression

There are three ways to interpret Linear Regression:

1. Linear Conditional Expectation

2. Best Linear Approximation

3. Causal Model

Now, we talk about each way in detail.

Linear Conditional Expectation

Suppose that:

$$E[Y \mid X]=X^{\prime} \beta$$

and define: $U=Y-E[Y \mid X]$

Recall the Conditional Expectation Function (CEF): $\mathbb{E}[Y \mid X=x]=m(x)$, which is the best estimator of $Y$.

For $\mathbb{E}[Y \mid X]$, we have $\mathbb{E}[Y \mid X] = \mathbb{E}\left[Y \mid X_1, X_2, \ldots, X_k\right] = m(X_1, X_2, \ldots, X_k)$.But in practice, the form of this function $m(\cdot)$ is unknown to us.

This Linear Conditional Expectation has several implications:

1. $\mathbb{E}(U)=0$.

   $\mathbb{E}[U]=\mathbb{E}_{X}[\mathbb{E}[U \mid X]]$

   $=\mathbb{E}_{X}[\mathbb{E}[Y - X^{\prime}\beta \mid X]] =\mathbb{E}_X[\mathbb{E}[Y| X]-X^{\prime} \beta]=\mathbb{E}_X[X^{\prime} \beta - X^{\prime} \beta] = 0$

2. $\mathbb{E}[X U]=0$

   $\mathbb{E}[X U]=\mathbb{E}\left[X \cdot\left(Y-X^{\prime} \beta\right)\right]=\mathbb{E}[X Y]-\mathbb{E}\left[X X^{\prime} \beta\right]$

   $=\mathbb{E}[\mathbb{E}[X Y \mid X]]-\mathbb{E}\left[XX^{\prime} \beta\right]$

   take $E[Y \mid X]=X^{\prime} \beta$ into it we have:

   $= \mathbb{E}[X \mathbb{E}[Y \mid X]]-\mathbb{E}\left[XX^{\prime} \beta\right]=0$

3. $\operatorname{Cov}(X, U)=0$

For $\beta$ we got by this define method, it does not necessarily have a Causal Effect. Since one $X_i$ change, others might also change.

Therefore, we also cannot interpret $\beta_j$ as the ceteris paribus (i.e., holding $X_{-j}$ and $U$ constant) effect of a one unit change in $X_i$ on $Y$. We need more information to check if we can do that.

• We are only holding $X_j, j \neq i$ constant, but we didn't hold $U$ constant in this structure.

• As $U=Y-\mathbb{E}[Y \mid X]$, we have $Y=\mathbb{E}[Y \mid X]+U = m(X) + U$

• So, $\frac{\partial Y}{\partial X_i}=\frac{\partial m(X)}{\partial X_i}+\frac{\partial U}{\partial X_i}$

• $\frac{\partial m(X)}{\partial X_i} = \beta_i$ only if $m(X)$ is a linear function, e.g. $m(X)=\beta_0+\beta_1 X_1+\cdots+\beta_k X_k$

"Best" Linear Approximation

In general, the conditional expectation is probably NOT linear. We are just using a linear function to approximate this.

Suppose that: $E\left[Y^2\right]<\infty$ and $E\left[X X^{\prime}\right]<\infty$ (or, $E\left[X_j^2\right]<\infty$ for $\left.1 \leq j \leq k\right)$

Under these assumptions, one may consider what is the "best" linear approximation (i.e., function of the form $X^{\prime} b$ for some choice of $b \in \mathbf{R}^{k+1}$, $b$ as the best linear predictor for $\beta$) to the conditional expectation.

To this end, consider the minimization problem for approximation/prediction error:

$$\min _{b \in \mathbf{R}^{k+1}} E\left[\left(E[Y \mid X]-X^{\prime} b\right)^2\right]$$

Minimize over $b$ and denote $\beta$ as the solution to this minimization problem. Then,

• $\beta$ is called the best linear predictor, and it is the linear projection coefficient.

• $b$ is the generic coefficient vector.

But we still cannot interpret $\beta_j$ as the ceteris paribus (i.e., holding $X_{-j}$ and $U$ constant) effect of a one unit change in $X_j$ on $Y$. Because we still do not know the information about the error term. Partial derivative only hold for other $X_i$s.

Best Linear Predictor (BLP)

Best linear predictor (BLP) can also be defined as:

$$\beta \in \underset{b \in \mathbf{R}^{k+1}}{\operatorname{argmin}} E\left[\left(Y-X^{\prime} b\right)^2\right]$$

This $\beta$ is also a convenient way of summarizing the “best” linear predictor of $Y$ given $X$.

Proof:

$\mathbb{E}\left[\left(Y-X^{\prime} b\right)^2\right] =\mathbb{E}\left[\left(Y-\mathbb{E}[Y \mid X]+\mathbb{E}[Y \mid X]-X^{\prime} b\right)^2\right]$

$= \mathbb{E}\left[ (Y - \mathbb{E}[Y|X])^2 \right] + \mathbb{E}\left[ (\mathbb{E}[Y \mid X] - X^{\prime} b)^2 \right]+2 \mathbb{E}\left[(Y-\mathbb{E}[Y \mid X])\left(\mathbb{E}[Y \mid X]-X^{\prime} b\right)\right]$

As for $\mathbb{E}\left[(Y-\mathbb{E}[Y \mid X])\left(\mathbb{E}[Y \mid X]-X^{\prime} b\right)\right]$, we have that it is equal to,

$\mathbb{E}\left[\mathbb{E}\left[(Y-\mathbb{E}[Y \mid X])\left(\mathbb{E}[Y \mid X]-X^{\prime} b\right) \mid X\right]\right)$, and $\mathbb{E}[Y \mid X]-X^{\prime} b$ only depend on $X$, we have:

$\mathbb{E}\left[(Y-\mathbb{E}[Y \mid X])\left(\mathbb{E}[Y \mid X]-X^{\prime} b\right)\right] = \mathbb{E}[(\mathbb{E}[Y \mid X]-X^{\prime} b)(\mathbb{E}[Y-\mathbb{E}[Y \mid X]\mid X])]$

$=\mathbb{E}\left[\left(\mathbb{E}[Y \mid X]-X^{\prime} b\right)(\mathbb{E}[\mathbb{E}[Y \mid X] -\mathbb{E}[Y \mid X]])\right] = 0$

So, we got:

$\mathbb{E}\left[\left(Y-X^{\prime} b\right)^2\right]=\mathbb{E}\left[(Y-\mathbb{E}[Y \mid X])^2\right]+\mathbb{E}\left[\left(\mathbb{E}[Y \mid X]-X^{\prime} b\right)^2\right]$

However, only the part $\mathbb{E}\left[\left(\mathbb{E}[Y \mid X]-X^{\prime} b\right)^2\right]$ is depended on $b$, so optimize the whole function over $b$ is same to optimize over this part. Therefore, this BLP definition is same with the previous one.

Summary: Two in One

Two interpretations from equivalent optimization problems:

$$\beta \in \underset{b \in \mathbf{R}^{k+1}}{\operatorname{argmin}} E\left[\left(E[Y \mid X]-X^{\prime} b\right)^2\right] \text { and } \beta \in \underset{b \in \mathbb{R}^{k+1}}{\operatorname{argmin}} E\left[\left(Y-X^{\prime} b\right)^2\right]$$

Note $E\left[\left(Y-X^{\prime} b\right)^2\right]$ is a convex (as a function of $b$ ) and this has the following implications.

• We can take the derivative on it and use FOC to solve $b$ and get $\beta$

• However, in order to do this, we need to make more assumptions. And I will introduce them in detail later in the estimation of $\beta$

Therefore, we can have that:

$\begin{aligned} \frac{\partial \mathbb{E}\left[\left(Y-X^{\prime} b\right)^2\right]}{\partial b} & =\mathbb{E}\left[\frac{\partial\left(Y-X^{\prime} b\right)^2}{\partial b}\right]=\mathbb{E}\left[\left(Y-X^{\prime} b\right) \cdot X\right] \\ & \left.=\mathbb{E}[Y X]-\mathbb{E}\left[X^{\prime} b X\right]=\mathbb{E}[Y X]-\mathbb{E}\left[X^{\prime} X b\right]\right]=0\end{aligned}$

We can solve $b \rightarrow \beta$. This is the First Order Condition (FOC), note that in above equations:

1. We can change the order between the expectation and partial derivative because of the dominated convergence theorem.

   Lemma:

   Let $X \in \mathcal{X}$ be a random variable $g: \mathbb{R} \times \mathcal{X} \rightarrow \mathbb{R}$ a function such that $g(t, X)$ is integrable for all $t$ and $g$ is continuously differentiable w.r.t. $t$. Assume that there is a random variable $Z$ such that $\left|\frac{\partial}{\partial t} g(t, X)\right| \leq Z$ a.s. for all $t$ and $\mathbb{E}(Z)<\infty$. Then

   $$\frac{\partial}{\partial t} \mathbb{E}(g(t, X))=\mathbb{E}\left(\frac{\partial}{\partial t} g(t, X)\right) .$$

   Proof:

   We have

   $$\begin{aligned} \frac{\partial}{\partial t} \mathbb{E}(g(t, X)) & =\lim _{h \rightarrow 0} \frac{1}{h}(\mathbb{E}(g(t+h, X))-\mathbb{E}(g(t, X))) \\ & =\lim _{h \rightarrow 0} \mathbb{E}\left(\frac{g(t+h, X)-g(t, X)}{h}\right) \\ & =\lim _{h \rightarrow 0} \mathbb{E}\left(\frac{\partial}{\partial t} g(\tau(h), X)\right), \end{aligned}$$

   where $\tau(h) \in(t, t+h)$ exists by the mean value theorem. By assumption we have

   $$\left|\frac{\partial}{\partial t} g(\tau(h), X)\right| \leq Z$$

   and thus we can use the dominated convergence theorem to conclude

   $$\frac{\partial}{\partial t} \mathbb{E}(g(t, X))=\mathbb{E}\left(\lim _{h \rightarrow 0} \frac{\partial}{\partial t} g(\tau(h), X)\right)=\mathbb{E}\left(\frac{\partial}{\partial t} g(t, X)\right) .$$

   This completes the proof.

2. We can change the order between the $X$ and $b$ because they are scalers.

Casual Model

In order to analysis the Casual Effect using the model, suppose that: $Y=g(X, U)$, where $X$ are the observed determinants of $Y$ and $U$ are the unobserved determinants of $Y$.

Such a relationship is a model of how $Y$ is determined and may come from physics, economics, etc. The effect of $X_j$ on $Y$ holding $X_{-j}$ and $U$ constant (i.e., ceteris paribus) is determined by this function $g$.

If $g$ is differentiable. then it is given by

$$D_{x_j} g(X, U) \text {. }$$

If we assume further that $g$ is linear, such that,

$$g(X, U)=X^{\prime} \beta+U,$$

then the ceteris paribus effect of $X_j$ on $Y$ is simply $\beta_j$. We may normalize $U$ so that $E[U]=0$, we can achieve this by replacing $U$ with $U-E[U]$ and $\beta_0$ with $\beta_0+E[U]$ if $E[U]=0$ is not the original case.

On the other hand, $E[U \mid X], E\left[U \mid X_j\right]$ and $E\left[U X_j\right]$ for $1 \leq j \leq k$ may or may not equal zero. This aspect is crucial in causal inference as it relates to the independence or correlation of the unobserved determinants with the observed variables. So, $\beta_j$ might be biased for the ceteris paribus effect. To get the ceteris paribus effect, we need more assumptions.

Potential Outcomes

Potential outcomes are an easy way to think about causal relationships.

Illustration: randomized controlled experiment where individuals are randomly assigned to a treatment (a drug) that is intended to improve their health status.

Notation: Let $Y$ denote the observed health status and $X \in \{0,1\}$ denote whether the individual takes the drug or not.

The causal relationship between $X$ and $Y$ can be described using the so-called potential outcomes:

• $Y(0)$ potential outcome in the absence of treatment

• $Y(1)$ potential outcome in the presence of treatment

Under health status, we have two health status variables: $(Y(0), Y(1))$, where

• $Y(0)$ is the value of the outcome that would have been observed if (possibly counter-to-fact) $X$ were 0.

• $Y(1)$ is the value of the outcome that would have been observed if (possibly counter-to-fact) $X$ were 1.

Treatment Effects

• The difference $Y(1)-Y(0)$ is called the treatment effect (TE).

• The quantity $E[Y(1)-Y(0)]$ is usually referred to as the average treatment effect (ATE).

Using this notation, we may rewrite the observed outcome as:

$$Y=\beta_0+\beta_1 X+U \text { with } \beta_1=Y(1)-Y(0) .$$

Note that this is not quite "the" linear model: since the ccoefficient $\beta_1$ is random. For $\beta_1$ to be constant. we need to assume that $Y(1)-Y(0)$ is constant across individuals.

Under all these assumptions: we end up with a linear constant effect causal model with $U$ independent with $X$ (from the nature of the randomized experiment), $E[U]=0$, and so $E[X U]=0$.

Without assuming constant treatment effects, it can be shown that a regression of $Y$ on $X$ identifies the average treatment effect,

$$\beta=\frac{\operatorname{Cov}[Y, X]}{\operatorname{Var}[X]}=E[Y(1)-Y(0)]$$

which is often called a causal parameter given that it is an average of causal effects.