# How to fit a multivariate linear model

## Description

Let’s say we have several independent variables, $x_1, x_2, \ldots, x_k$, and a dependent variable $y$. How can I fit a linear model that uses these independent variables to best predict the dependent variable?

In other words, what are the model coefficients $\beta_0, \beta_1, \beta_2, \ldots, \beta_k$ that give me the best linear model $\hat{y}=\beta_0 + \beta_1x + \beta_2x + \cdots + \beta_kx$ based on my data?

## Using statsmodels, in Python

View this solution alone.

We’re going to use fake data here for illustrative purposes. You can replace our fake data with your real data in the code below.

We’ll put the data into a dataframe and then make a variable with a list of the independent variables and a variable with the outcome variable.

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import pandas as pd

# Replace this fake data with your real data
df = pd.DataFrame( {
'x1':[2, 7, 4, 3, 11, 18, 6, 15, 9, 12],
'x2':[4, 6, 10, 1, 18, 11, 8, 20, 16, 13],
'x3':[11, 16, 20, 6, 14, 8, 5, 23, 13, 10],
'y':[24, 60, 32, 29, 90, 45, 130, 76, 100, 120]
} )

xs = df[['x1', 'x2', 'x3']]  # list of independent variables
y = df['y']                  # dependent variable


We can use StatsModels’ OLS to build our multivariate linear model. We’ll print out the coefficients and the intercept, and the coefficients will be in the form of an array when we print them.

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import statsmodels.api as sm

# Add a constant to the dependent variables first

# Build the model
model = sm.OLS(y, xs).fit()

# Show the model summary to get the coefficients and the intercept
model.summary()

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/opt/conda/lib/python3.10/site-packages/scipy/stats/_stats_py.py:1736: UserWarning: kurtosistest only valid for n>=20 ... continuing anyway, n=10
warnings.warn("kurtosistest only valid for n>=20 ... continuing "

Dep. Variable: R-squared: y 0.594 OLS 0.390 Least Squares 2.921 Mon, 24 Jul 2023 0.122 20:46:50 -45.689 10 99.38 6 100.6 3 nonrobust
coef std err t P>|t| [0.025 0.975] 77.2443 27.366 2.823 0.030 10.282 144.206 -2.7009 2.855 -0.946 0.381 -9.686 4.284 7.2989 2.875 2.539 0.044 0.265 14.333 -4.8607 2.187 -2.223 0.068 -10.211 0.490
 Omnibus: Durbin-Watson: 2.691 2.123 0.26 1.251 0.524 0.535 1.62 58.2

Notes:
[1] Standard Errors assume that the covariance matrix of the errors is correctly specified.

The coefficients and intercept appear on the left hand side of the output, about half way down, under the heading “coef.”

Thus the multivariate linear model from the example data is $\hat y = 77.2443 - 2.7009x_1 + 7.2989x_2 - 4.8607x_3$.

See a problem? Tell us or edit the source.

## Solution, in R

View this solution alone.

We’re going to use fake data here for illustrative purposes. You can replace our fake data with your real data in the code below.

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# Replace this fake data with your real data
x1 <- c(2, 7, 4, 3, 11, 18, 6, 15, 9, 12)
x2 <- c(4, 6, 10, 1, 18, 11, 8, 20, 16, 13)
x3 <- c(11, 16, 20, 6, 14, 8, 5, 23, 13, 10)
y <- c(24, 60, 32, 29, 90, 45, 130, 76, 100, 120)

# If you'll need the model coefficients later, store them as variables like this:
model <- lm(y ~ x1 + x2 + x3)
beta0 <- model$coefficients[1] beta1 <- model$coefficients[2]
beta2 <- model$coefficients[3] beta3 <- model$coefficients[4]

# To see the model summary, which includes the coefficients and much more, do this:
summary(model)

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Call:
lm(formula = y ~ x1 + x2 + x3)

Residuals:
Min      1Q  Median      3Q     Max
-25.031 -20.218  -8.373  22.937  35.640

Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept)   77.244     27.366   2.823   0.0302 *
x1            -2.701      2.855  -0.946   0.3806
x2             7.299      2.875   2.539   0.0441 *
x3            -4.861      2.187  -2.223   0.0679 .
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Residual standard error: 30.13 on 6 degrees of freedom
Multiple R-squared:  0.5936,	Adjusted R-squared:  0.3904
F-statistic: 2.921 on 3 and 6 DF,  p-value: 0.1222


The coefficients and intercept appear on the left hand side of the output, about half way down, under the heading “Estimate.”

Thus the multivariate linear model from the example data is $\hat y = 77.244 - 2.701x_1 + 7.299x_2 - 4.861x_3$.