# 15.2. Fitting the Simple Linear Model¶

After picking a model, the predictor variable(s), and the outcome variable, we want to use our data to find the model parameters $$\theta_0$$ and $$\theta_1$$. This process is called fitting a model to data.

To fit our model, we’ll use loss minimization (first introduced in Chapter 4). We’ll pick a loss function, then find out which values of $$\theta_0$$ and $$\theta_1$$ give the smallest loss for our data. For linear models, we typically choose the squared loss. For a given data point $$(x, y)$$ and model parameters $$\theta$$, the squared loss is:

\begin{aligned} {\cal l}(\theta, x, y) = (y - f_{\theta}(x))^2 \end{aligned}

This loss function says: find the difference between the actual observed $$y$$ and the prediction $$f_{\theta}(x)$$, then square it.

We say that the dataset has $$n$$ points: $$(x_1, y_1), \ldots, (x_n, y_n)$$. To simplify the notation, we say that the column vector $$x = [x_1, \ldots, x_n]$$ and $$y = [y_1, \ldots, y_n]$$. The mean squared loss (MSE) is the average loss for the entire dataset:

\begin{split} \begin{aligned} L(\theta, x, y) &= \frac{1}{n} \sum_{i} {\cal l}(\theta, x_i, y_i) \\ &= \frac{1}{n} \sum_{i}(y_i - f_{\theta}(x_i))^2 \end{aligned} \end{split}

With our loss function defined, we can now use calculus to find the $$\hat{\theta}_0$$ and $$\hat{\theta}_1$$ that minimize the loss.

## 15.2.1. Minimizing the Loss¶

Let’s derive the best fit model parameter $$\hat{\theta}_0$$. We start by plugging in the model equation $$f_{\theta}(x) = \theta_0 + \theta_1 x$$ into the loss equation:

\begin{split} \begin{aligned} L(\theta, x, y) &= \frac{1}{n} \sum_{i}(y_i - f_{\theta}(x_i))^2 \\ &= \frac{1}{n} \sum_{i}(y_i - \theta_1 x_i - \theta_0)^2 \\ \end{aligned} \end{split}

To find $$\hat{\theta}_0$$, we take the partial derivative of the loss:

\begin{aligned} \frac{\partial}{\partial \theta_0} L(\theta, x, y) &= \frac{1}{n} \sum_{i} 2 (y_i - \theta_1 x_i - \theta_0) (-1) \end{aligned}

Then, we set the partial derivative equal to 0 and solve for $$\hat{\theta}_0$$:

\begin{split} \begin{aligned} \frac{1}{n} \sum_{i} 2 (y_i - \hat{\theta}_1 x_i - \hat{\theta}_0) (-1) &= 0 \\ \sum_{i} (\hat{\theta}_0 + \hat{\theta}_1 x_i - y_i) &= 0 \\ n \hat{\theta}_0 + \hat{\theta}_1 \textstyle \sum x_i - \textstyle \sum y_i &= 0 \\ \hat{\theta}_0 &= \textstyle \frac{1}{n} \sum y_i - \hat{\theta}_1 \frac{1}{n} \textstyle \sum x_i \\ \end{aligned} \end{split}

Using the shorthand $$\bar x = \sum x_i$$, the final expression for $$\hat{\theta}_0$$ can be simplified to:

\begin{aligned} \hat{\theta}_0 &= \bar y - \hat{\theta}_1 \bar x \end{aligned}

Here we run into a complication: $$\hat{\theta}_0$$ depends on $$\hat{\theta}_1$$. To solve this, we can start solving for $$\hat{\theta}_1$$ by taking the derivative of $$L(\theta, x, y)$$ with respect to $$\theta_1$$. If we do this, we’ll find that the formula for $$\hat{\theta}_1$$ depends on $$\hat{\theta}_0$$, so we can plug in $$\hat{\theta}_0 = \bar y - \hat{\theta}_1 \bar x$$ and solve. The algebra is long, so we’ll leave this the derivation as an exercise. After the derivation, the resulting $$\hat{\theta}_1$$ is:

\begin{aligned} \hat{\theta}_1 &= r \frac{\sigma_y}{\sigma_x} \end{aligned}

In the formula above, $$\sigma_x$$ and $$\sigma_y$$ are the standard deviations of $$x$$ and $$y$$. The correlation coefficient $$r$$ is the average of the products of $$x$$ and $$y$$ in standard units (subtracting the mean, then dividing by the standard deviation). Since pandas.Series objects have built-in methods to compute $$\sigma_x$$, $$\sigma_y$$, and $$r$$, we can quickly define functions that fit the model to our data:

def theta_1(x, y):
r = x.corr(y)
return r * y.std() / x.std()

def theta_0(x, y):
return y.mean() - theta_1(x, y) * x.mean()


## 15.2.2. Example: What’s the Relationship Between Commute Time and Upward Mobility?¶

Now, let’s fit a simple linear model on the economic mobility data. Once we do this, we can look at the relationship between commute time and upward mobility.

When modeling, we first split our dataset into a training set and a test set. We fit our model using only the data in the training set. Then, we report our model’s accuracy using only the test set.

Note

Why do we need to separate the training set and the test set? There are deep statistical reasons for splitting up the data, but rigorously showing this requires more space than we have. Instead, we’ll try to give intuition using an analogy. Let’s say your friend Jan is studying for a math exam. To “train”, he reviews the examples from the class’s lecture slides. To check how well Jan knows the material, we wouldn’t want to reuse the lecture examples for the exam—since he specifically reviewed those examples, he’d do unfairly well. Instead, we’d write new “test” problems that Jan hasn’t seen yet.

Similarly, if we reused the training set to evaluate our model, our model would do unfairly well on those examples. To see how well our model performs, we need to set aside data that our model didn’t use for training.

First, we’ll subset out the predictor $$x$$ and outcome $$y$$ from the data.

x = df['frac_traveltime_lt15']
y = df['aum']


Then, we’ll randomly off 20% of the data into the test set. We’ll use the train_test_split split function from the scikit-learn Python package.

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: 564 rows
Test set size: 141 rows


Then, we can fit the model by computing $$\hat{\theta}_0$$ and $$\hat{\theta}_1$$ on the training set.

t1 = theta_1(x_train, y_train)
t0 = theta_0(x_train, y_train)
print(f'Model: f(x) = {t0:.2f} + {t1:.2f}x')

Model: f(x) = 31.10 + 29.29x


At last, we have a model! This model is a simple linear model that predicts AUM for a commuting zone using the fraction of people with ≤15 min commute times. After fitting a model, we want to check how well our model does on the data.

## 15.2.3. Examining the Model¶

One way to look at the model’s performance is to look at the the model’s MSE on the test set.

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

pred = t0 + t1 * x_test
print(f'Test set MSE: {mse(y_test, pred):.2f}')

Test set MSE: 17.70


However, this number isn’t very easy to interpret on its own—we usually use the MSE to compare two different models, so we’ll return to the MSE later in the chapter. Instead, we’ll plot the model’s predictions on top of the scatter plot of the data.

fig = px.scatter(x=x_train, y=y_train,
width=450, height=250)

pred_x = np.array([x.min(), x.max()])
pred_y = t0 + t1 * pred_x

fig


We can see that the line has a positive slope, which captures the positive association between AUM and the fraction of people with ≤15 min commute.

To check the model fit, we use a residual plot, which puts the predicted $$f_{\theta}(x)$$ values on the x-axis and the residuals $$y - f_{\theta}(x)$$ on the y-axis:

predicted = t0 + t1 * x_train
resid = pd.DataFrame({'predicted': predicted,
'residuals': y_train - predicted})
fig = px.scatter(resid, x='predicted', y='residuals',
width=350, height=250)


In this section, we derived the model parameters $$\hat{\theta}_0$$ and $$\hat{\theta}_1$$ using calculus. Then, we fit a simple linear model that predicts AUM for a commuting zone using the fraction of people with a ≤15 min commute. A residual plot shows that this linear model doesn’t fit the data well. But we didn’t expect a great fit to begin with—economic mobility is a complicated metric that probably can’t be predicted using a single variable. To improve our model, we can use multiple predictor variables rather than just one. This model is called a multiple linear model, and we’ll introduce this model in the next section.