# 15.4. Fitting the Multiple Linear Model¶

For a $$n \times (p + 1)$$ design matrix $$X$$, a $$n$$-dimensional column vector of outcomes $$y$$, and a $$(p + 1)$$-dimensional column vector of model parameters $$\theta$$, we assume that:

\begin{aligned} y = X \theta + \epsilon \end{aligned}

Here, $$\epsilon$$ is a $$n$$-dimensional column vector that represents the sampling error. We define the multiple linear model as:

\begin{aligned} f_{\theta}(X) = X \theta \end{aligned}

Similar to the simple linear model, we’ll fit $$f_{\theta}(X)$$ using the squared loss function. We want to find the model parameters $$\hat{\theta}$$ that minimize the mean squared loss:

\begin{aligned} L(\theta, X, y) &= \frac{1}{n} \left | y - f_{\theta}(X) \right|^2 \end{aligned}

Here, we’re using the notation $$|v|^2$$ for a vector $$v$$ as a shorthand for the sum of each vector element squared 1: $$|v|^2 = \sum_i v_i^2$$ .

In this section, we’ll fit our model by figuring out what the minimizing $$\hat{\theta}$$ is. One idea is to use calculus as we did for the simple linear model. However, this approach needs knowledge of vector calculus that we won’t cover in this book. Instead, we’ll use a geometric argument.

## 15.4.1. A Geometric Problem¶

Our goal is the find the $$\hat{\theta}$$ that minimizes our loss function—we want to make $$L(\theta, X, y)$$ as small as possible for a given $$X$$ and $$y$$. The key insight is that we can restate this goal in a geometric way. Remember: the model predictions $$f_{\theta}(X)$$ and the true outcomes $$y$$ are both vectors. We can treat vectors as points—for example, we can plot the vector $$[ 2, 3 ]$$ at $$x = 2, y = 3$$ in 2D space. Then, minimizing $$L(\theta, X, y)$$ is equivalent to finding $$\hat{\theta}$$ that makes $$f_{\theta}(X)$$ as close as possible to $$y$$ when we plot them as points. As depicted in Figure 15.2, different values of $$\theta$$ give different predictions $$f_{\theta}(X)$$ (hollow points). Then, $$\hat{\theta}$$ is the vector of parameters that put $$f_{\theta}(X)$$ as close to $$y$$ (filled point) as possible.

Fig. 15.2 A plot showing different values of $$f_{\theta}(X)$$ (hollow points) and the outcome vector $$y$$ (filled point).

Next, we’ll look at the possible values of $$f_{\theta}(X)$$. In Figure 15.2, we showed a few possible $$f_{\theta}(X)$$. Instead of just plotting a few possible points, we can plot all possible values of $$f_{\theta}(X)$$ by varying $$\theta$$. This results in a subspace of possible $$f_{\theta}(X)$$ values, as shown in Figure 15.3.

Fig. 15.3 A plot showing all possible values of $$f_{\theta}(X)$$ as a line.

In the above Figure 15.3, we drew the set of possible $$f_{\theta}(X)$$ values as a line. Since our model is $$f_{\theta}(X) = X \theta$$, from a property of matrix-vector multiplication we know that $$f_{\theta}(X)$$ is a linear combination of the columns of $$X$$, which we also call $$\text{span}(X)$$. Now, we need to figure out which point within $$\text{span}(X)$$ lies the closest to $$y$$.

As Figure 15.3 suggests, the closest point to $$y$$ is the point where the error $$\epsilon = y - f_{\theta}(X)$$ is perpendicular to $$\text{span}(X)$$. We’ll leave the complete proof as an exercise. With this final fact, we can solve for $$\hat{\theta}$$:

\begin{split} \begin{aligned} f_\hat{\theta}(X) + \epsilon &= y \\ X \hat{\theta} + \epsilon &= y \\ X^\top X \hat{\theta} + X^\top \epsilon &= X^\top y & (\text{left-multiply both sides by } X^\top) \\ X^\top X \hat{\theta} &= X^\top y & (X^\top \epsilon = 0 \text{ since } \epsilon \perp \text{span}(X)) \\ \hat{\theta} &= (X^\top X)^{-1} X^\top y \end{aligned} \end{split}

With this derivation done, we can now write a short function to fit the multiple linear model using $$X$$ and $$y$$.

def fit(X, y):
return np.linalg.inv(X.T @ X) @ X.T @ y


Notice that deriving $$\hat{\theta}$$ for the multiple linear model also gives us $$\hat{\theta}$$ for the simple linear model too. If we set $$X$$ to contain the intercept column and one column of features, using the formula for $$\hat{\theta}$$ gives us the intercept and slope of the best-fit line.

## 15.4.2. Predicting AUM Using Multiple Variables¶

Now, let’s fit a multiple linear model. Recall that we’re using nine variables to predict AUM, the upward mobility for a commuting zone. We have our design matrix in a dataframe X and a vector of outcomes in y.

X.head(2)

intr frac_traveltime_lt15 gini dropout_r ... taxrate gradrate_r frac_worked1416 cs_born_foreign
0 1 0.33 0.47 -0.02 ... 0.02 -2.43e-03 3.75e-03 0.01
1 1 0.28 0.43 -0.02 ... 0.02 -1.01e-01 4.78e-03 0.02

2 rows × 10 columns

As with the simple linear model, we’ll split our data into a training set and a test set.

from sklearn.model_selection import train_test_split

X_train, X_test, y_train, y_test = train_test_split(
X, y, test_size=0.2, random_state=42,
)

print(f'Training set size: {len(X_train)} rows')
print(f'    Test set size: {len(X_test)} rows')

Training set size: 383 rows
Test set size: 96 rows


Let’s fit the model on the training set:

theta = fit(X_train, y_train)


We can now find the MSE on the test set:

def predict(theta, X):
return X.values @ theta

def mse(y, pred):
return np.mean((y - pred) ** 2)

mse(y_test, predict(theta, X_test))

5.3667257699216195


To understand this error, we can start by comparing the multiple linear model with the simple linear model from before that only used the fraction with ≤15 min commute time.

theta_simple = fit(X_train[['intr', 'frac_traveltime_lt15']], y_train)
mse(y_test, predict(theta_simple, X_test[['intr', 'frac_traveltime_lt15']]))

12.925251956432646


We see that the multiple linear model has about 60% lower error than the simple linear model—a big improvement! Let’s make residual plots for both multiple and simple linear models.

pred = predict(theta, X_train)
resid = y_train - pred

pred_simple = predict(theta_simple, X_train[['intr', 'frac_traveltime_lt15']])
resid_simple = y_train - pred_simple

all_resid = pd.concat([
pd.DataFrame({'model': 'multi', 'pred': pred, 'resid': resid}),
pd.DataFrame({'model': 'simple', 'pred': pred_simple, 'resid': resid_simple}),
], ignore_index=True)

fig = px.scatter(all_resid, x='pred', y='resid', facet_col='model',
width=550, height=300)
margin(fig, t=20)


We can see that the multiple linear model’s residuals are much closer to 0 in general, which helps to explain why the MSE is also lower.

## 15.4.3. Interpreting Linear Models¶

Now that we have a linear model, we want to know what it says about our data. After fitting our simple linear model, we have:

\begin{aligned} f_{\theta}(x) = 31.1 + 29.3 x \end{aligned}

Here, $$x$$ is the fraction of people with ≤15 min commute times. We interpret this model as saying: if we take two commuting zones, and one of them has 10% more people with a ≤15 min commute time, then the model predicts that the AUM will be 2.93 points higher on average. Simply put, commuting zones that have short commute times generally have more economic mobility or opportunity.

We need to be careful about this interpretation. It’s tempting to say that increasing $$x$$ by 0.1 causes or is associated with an AUM that’s 2.93 points higher. But, our model doesn’t let us make this conclusion. Our model really only lets us draw conclusions about average differences between commuting zones. It doesn’t let us draw conclusions about what would happen if we took a single commuting zone and increased its $$x$$.

We can try interpreting the the intercept term too. We might say that the model predicts that commuting zones where $$x = 0$$ have an average AUM of 31.1. However, there aren’t actually any data points in our data that have $$x = 0$$, so this interpretation isn’t useful. In general, we have to check the meaning of the intercept on a case-by-case basis.

Now, let’s look at our multiple linear model. Since our model has 9 predictor variables, we’ll show the model coefficients in a table rather than writing out the long model equation:

theta.index = ['intr', *predictors]
display_df(theta.rename('coeff').to_frame(), 10)

coeff
intr 40.32
frac_traveltime_lt15 14.73
gini -1.23
dropout_r -36.60
rel_tot 5.86
cs_fam_wkidsinglemom -43.89
taxrate 56.97
frac_worked1416 440.29
cs_born_foreign 6.80

Let’s look at the coefficient for gradrate_r, the college graduation rate. For a multiple linear model, we interpret this coefficient as saying: if we took two commuting zones that are the same in all predictors except gradrate_r, the model predicts that the commuting zone with a 0.1 larger gradrate_r also has a 0.051 higher AUM. The key difference for multiple linear models is that we only have a useful interpretation for a coefficient when we can hold the other predictors constant.

Unfortunately, this is not always the case. As an extreme example, let’s say our model happened to have an extra predictor 2 * gradrate_r which doubles each value from gradrate_r. If our model had this predictor, it doesn’t make sense to talk about gradrate_r’s coefficient since we can’t hold 2 * gradrate_r constant. In general, we need to watch out for collinearity, where two predictor variables are highly correlated with each other. If our model has variables with high collinearity, we can’t make useful interpretations for the model coefficients.

To check for collinearity, we can look at the correlations between each pair of predictors, as shown in the heatmap below.

plt.figure(figsize=(5, 4))
sns.heatmap(df[predictors].corr().round(1), center=0, annot=True);


We see that cs_fam_wkidsinglemom, the fraction of families with a single mom, has a high correlation with several other predictors, like gini and frac_worked1416. Here, we face a central tension in modeling. If we want the most interpretable model, we should try to avoid collinearity. For instance, we could remove variables that are highly correlated with others. However, doing this often makes our model less accurate. Having an interpretable model can mean having a less accurate model. Is is better to have a highly interpretable model or a highly accurate one? It depends. Each modeling situation needs a different tradeoff, and it’s the job of the data scientist to understand and make this tradeoff.

This section covered how to fit a multiple linear model using a geometric perspective. We then used our derivation to fit a multiple linear model on the Opportunity data and found that it has a much better fit than the simple linear model. Finally, we interpreted our model coefficients and talked about how it can be tricky to interpret multiple linear models. In the next section, we’ll introduce feature engineering techniques that make linear models useful for different kinds of data.

1

$$|v|$$ is also called the $$\ell_2$$ norm of a vector $$v$$.