Correct note on Multiple Regression.
This commit is contained in:
Marco Oesting
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---
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---
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editor:
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editor:
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markdown:
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markdown:
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wrap: 72
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wrap: 72
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---
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---
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# Multiple Regression Basics
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# Multiple Regression Basics
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## Motivation
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## Motivation
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### Introductory Example: tree dataset from R
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### Introductory Example: tree dataset from R
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``` julia
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``` julia
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using Statistics
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using Statistics
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using Plots
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using Plots
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using RDatasets
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using RDatasets
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trees = dataset("datasets", "trees")
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trees = dataset("datasets", "trees")
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scatter(trees.Volume, trees.Girth,
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scatter(trees.Volume, trees.Girth,
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legend=false, xlabel="Girth", ylabel="Volume")
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legend=false, xlabel="Girth", ylabel="Volume")
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```
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```
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*Aim:* Find relationship between the *response variable* `volume` and
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*Aim:* Find relationship between the *response variable* `volume` and
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the *explanatory variable/covariate* `girth`? Can we predict the volume
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the *explanatory variable/covariate* `girth`? Can we predict the volume
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of a tree given its girth?
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of a tree given its girth?
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``` julia
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``` julia
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scatter(trees.Girth, trees.Volume,
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scatter(trees.Girth, trees.Volume,
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legend=false, xlabel="Girth", ylabel="Volume")
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legend=false, xlabel="Girth", ylabel="Volume")
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plot!(x -> -37 + 5*x)
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plot!(x -> -37 + 5*x)
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```
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```
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First Guess: There is a linear relation!
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First Guess: There is a linear relation!
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## Simple Linear Regression
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## Simple Linear Regression
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Main assumption: up to some error term, each measurement of the response
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Main assumption: up to some error term, each measurement of the response
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variable $y_i$ depends linearly on the corresponding value $x_i$ of the
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variable $y_i$ depends linearly on the corresponding value $x_i$ of the
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covariate
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covariate
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$\leadsto$ **(Simple) Linear Model:**
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$\leadsto$ **(Simple) Linear Model:**
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$$y_i = \beta_0 + \beta_1 x_i + \varepsilon_i, \qquad i=1,...,n,$$
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$$y_i = \beta_0 + \beta_1 x_i + \varepsilon_i, \qquad i=1,...,n,$$
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where $\varepsilon_i \sim \mathcal{N}(0,\sigma^2)$ are independent
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where $\varepsilon_i \sim \mathcal{N}(0,\sigma^2)$ are independent
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normally distributed errors with unknown variance $\sigma^2$.
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normally distributed errors with unknown variance $\sigma^2$.
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*Task:* Find the straight line that fits best, i.e., find the *optimal*
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*Task:* Find the straight line that fits best, i.e., find the *optimal*
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estimators for $\beta_0$ and $\beta_1$.
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estimators for $\beta_0$ and $\beta_1$.
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*Typical choice*: Least squares estimator (= maximum likelihood
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*Typical choice*: Least squares estimator (= maximum likelihood
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estimator for normal errors)
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estimator for normal errors)
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$$ (\hat \beta_0, \hat \beta_1) = \mathrm{argmin} \ \| \mathbf{y} - \mathbf{1} \beta_0 - \mathbf{x} \beta_1\|^2 $$
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$$ (\hat \beta_0, \hat \beta_1) = \mathrm{argmin} \ \| \mathbf{y} - \mathbf{1} \beta_0 - \mathbf{x} \beta_1\|^2 $$
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where $\mathbf{y}$ is the vector of responses, $\mathbf{x}$ is the
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where $\mathbf{y}$ is the vector of responses, $\mathbf{x}$ is the
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vector of covariates and $\mathbf{1}$ is a vector of ones.
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vector of covariates and $\mathbf{1}$ is a vector of ones.
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Written in matrix style:
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Written in matrix style:
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$$
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$$
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(\hat \beta_0, \hat \beta_1) = \mathrm{argmin} \ \left\| \mathbf{y} - (\mathbf{1},\mathbf{x}) \left( \begin{array}{c} \beta_0\\ \beta_1\end{array}\right) \right\|^2
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(\hat \beta_0, \hat \beta_1) = \mathrm{argmin} \ \left\| \mathbf{y} - (\mathbf{1},\mathbf{x}) \left( \begin{array}{c} \beta_0\\ \beta_1\end{array}\right) \right\|^2
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$$
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$$
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Note: There is a closed-form expression for
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Note: There is a closed-form expression for
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$(\hat \beta_0, \hat \beta_1)$. We will not make use of it here, but
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$(\hat \beta_0, \hat \beta_1)$. We will not make use of it here, but
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rather use Julia to solve the problem.
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rather use Julia to solve the problem.
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``` julia
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``` julia
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lm(@formula(Volume ~ Girth), trees)
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lm(@formula(Volume ~ Girth), trees)
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```
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```
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*Interpretation of the Julia output:*
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*Interpretation of the Julia output:*
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- column `estimate` : least square estimates for $\hat \beta_0$ and
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- column `estimate` : least square estimates for $\hat \beta_0$ and
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$\hat \beta_1$
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$\hat \beta_1$
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- column `Std. Error` : estimated standard deviation
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- column `Std. Error` : estimated standard deviation
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$\hat s_{\hat \beta_i}$ of the estimator $\hat \beta_i$
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$\hat s_{\hat \beta_i}$ of the estimator $\hat \beta_i$
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- column `t value` : value of the $t$-statistics
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- column `t value` : value of the $t$-statistics
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$$ t_i = {\hat \beta_i \over \hat s_{\hat \beta_i}}, \quad i=0,1, $$
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$$ t_i = {\hat \beta_i \over \hat s_{\hat \beta_i}}, \quad i=0,1, $$
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Under the hypothesis $\beta_i=0$, the test statistics $t_i$ would
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Under the hypothesis $\beta_i=0$, the test statistics $t_i$ would
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follow a $t$-distribution.
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follow a $t$-distribution.
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- column `Pr(>|t|)`: $p$-values for the hypotheses $\beta_i=0$ for
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- column `Pr(>|t|)`: $p$-values for the hypotheses $\beta_i=0$ for
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$i=0,1$
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$i=0,1$
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::: callout-tip
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::: callout-tip
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The command `rand(n)` generates a sample of `n` "random" (i.e.,
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The command `rand(n)` generates a sample of `n` "random" (i.e.,
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uniformly distributed) random numbers. If you want to sample from
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uniformly distributed) random numbers. If you want to sample from
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another distribution, use the `Distributions` package, define an object
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another distribution, use the `Distributions` package, define an object
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being the distribution of interest, e.g. `d = Normal(0.0, 2.0)` for a
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being the distribution of interest, e.g. `d = Normal(0.0, 2.0)` for a
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normal distribution with mean 0.0 and standard deviation 2.0, and sample
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normal distribution with mean 0.0 and standard deviation 2.0, and sample
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`n` times from this distribution by `rand(d, n)`.
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`n` times from this distribution by `rand(d, n)`.
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:::
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:::
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::: {.callout-caution collapse="false"}
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::: {.callout-caution collapse="false"}
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## Task 1
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## Task 1
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1. Generate $n=20$ covariates $\mathbf{x}$ randomly.
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1. Generate $n=20$ covariates $\mathbf{x}$ randomly.
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2. Given these covariates and the true parameters $\beta_0=-3$,
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2. Given these covariates and the true parameters $\beta_0=-3$,
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$\beta_1=2$ and $\sigma=0.5$, simulate responses from a linear model
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$\beta_1=2$ and $\sigma=0.5$, simulate responses from a linear model
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(with normally distributed errors) and estimate the coefficients
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(with normally distributed errors) and estimate the coefficients
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$\beta_0$ and $\beta_1$.
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$\beta_0$ and $\beta_1$.
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3. Play with different choices of the parameters above (including the
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3. Play with different choices of the parameters above (including the
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sample size $n$) to see the effects on the parameter estimates and
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sample size $n$) to see the effects on the parameter estimates and
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the $p$-values.
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the $p$-values.
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:::
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::: {callout-note}
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The more covariates you add the more variance can be explained by the linear model - $R^2$ increases. In order to balance goodness-of-tit of a model and its complexity, information criteria such as `aic` are considered.
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## Multiple Regression Model
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:::
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*Idea*: Generalize the simple linear regression model to multiple
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## Multiple Regression Model
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covariates, w.g., predict `volume` using `girth` and \`height\`\`.
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*Idea*: Generalize the simple linear regression model to multiple
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$\leadsto$ **Linear Model:**
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covariates, w.g., predict `volume` using `girth` and \`height\`\`.
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$$y_i = \beta_0 + \beta_1 x_{i1} + \ldots + \beta_p x_{ip} + \varepsilon_i, \qquad i=1,...,n,$$where
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$\leadsto$ **Linear Model:**
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- $y_i$: $i$-th measurement of the response,
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$$y_i = \beta_0 + \beta_1 x_{i1} + \ldots + \beta_p x_{ip} + \varepsilon_i, \qquad i=1,...,n,$$where
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- $x_{i1}$: $i$ th value of first covariate,
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- $y_i$: $i$-th measurement of the response,
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- ...
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- $x_{i1}$: $i$ th value of first covariate,
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- $x_{ip}$: $i$-th value of $p$-th covariate,
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- ...
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- $\varepsilon_i \sim \mathcal{N}(0,\sigma^2)$: independent normally
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- $x_{ip}$: $i$-th value of $p$-th covariate,
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distributed errors with unknown variance $\sigma^2$.
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- $\varepsilon_i \sim \mathcal{N}(0,\sigma^2)$: independent normally
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*Task:* Find the *optimal* estimators for
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distributed errors with unknown variance $\sigma^2$.
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$\mathbf{\beta} = (\beta_0, \beta_1, \ldots, \beta_p)$.
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*Task:* Find the *optimal* estimators for
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*Our choice again:* Least squares estimator (= maximum likelihood
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$\mathbf{\beta} = (\beta_0, \beta_1, \ldots, \beta_p)$.
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estimator for normal errors)
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*Our choice again:* Least squares estimator (= maximum likelihood
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$$
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estimator for normal errors)
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\hat \beta = \mathrm{argmin} \ \| \mathbf{y} - \mathbf{1} \beta_0 - \mathbf{x}_1 \beta_1 - \ldots - \mathbf{x}_p \beta_p\|^2
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$$
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$$
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\hat \beta = \mathrm{argmin} \ \| \mathbf{y} - \mathbf{1} \beta_0 - \mathbf{x}_1 \beta_1 - \ldots - \mathbf{x}_p \beta_p\|^2
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where $\mathbf{y}$ is the vector of responses, $\mathbf{x}$\_j is the
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$$
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vector of the $j$ th covariate and $\mathbf{1}$ is a vector of ones.
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where $\mathbf{y}$ is the vector of responses, $\mathbf{x}$\_j is the
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Written in matrix style:
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vector of the $j$ th covariate and $\mathbf{1}$ is a vector of ones.
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$$
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Written in matrix style:
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\mathbf{\hat \beta} = \mathrm{argmin} \ \left\| \mathbf{y} - (\mathbf{1},\mathbf{x}_1,\ldots,\mathbf{x}_p) \left( \begin{array}{c} \beta_0 \\ \beta_1 \\ \vdots \\ \beta_p\end{array} \right) \right\|^2
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$$
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$$
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\mathbf{\hat \beta} = \mathrm{argmin} \ \left\| \mathbf{y} - (\mathbf{1},\mathbf{x}_1,\ldots,\mathbf{x}_p) \left( \begin{array}{c} \beta_0 \\ \beta_1 \\ \vdots \\ \beta_p\end{array} \right) \right\|^2
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Defining the *design matrix*
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$$
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$$ \mathbf{X} = \left( \begin{array}{cccc}
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Defining the *design matrix*
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1 & x_{11} & \ldots & x_{1p} \\
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\vdots & \vdots & \ddots & \vdots \\
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$$ \mathbf{X} = \left( \begin{array}{cccc}
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1 & x_{11} & \ldots & x_{1p}
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1 & x_{11} & \ldots & x_{1p} \\
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\end{array}\right) \qquad
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\vdots & \vdots & \ddots & \vdots \\
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(\text{size } n \times (p+1)), $$
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1 & x_{11} & \ldots & x_{1p}
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\end{array}\right) \qquad
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we get the short form
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(\text{size } n \times (p+1)), $$
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$$
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we get the short form
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\mathbf{\hat \beta} = \mathrm{argmin} \ \| \mathbf{y} - \mathbf{X} \mathbf{\beta} \|^2 = (\mathbf{X}^\top \mathbf{X})^{-1} \mathbf{X}^\top \mathbf{y}
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$$
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$$
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\mathbf{\hat \beta} = \mathrm{argmin} \ \| \mathbf{y} - \mathbf{X} \mathbf{\beta} \|^2 = (\mathbf{X}^\top \mathbf{X})^{-1} \mathbf{X}^\top \mathbf{y}
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\[use Julia code (existing package) to perform linear regression for
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$$
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`volume ~ girth + height`\]
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\[use Julia code (existing package) to perform linear regression for
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The interpretation of the Julia output is similar to the simple linear
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`volume ~ girth + height`\]
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regression model, but we provide explicit formulas now:
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The interpretation of the Julia output is similar to the simple linear
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- parameter estimates:
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regression model, but we provide explicit formulas now:
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$$
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- parameter estimates:
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(\mathbf{X}^\top \mathbf{X})^{-1} \mathbf{X}^\top \mathbf{y}
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$$
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$$
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(\mathbf{X}^\top \mathbf{X})^{-1} \mathbf{X}^\top \mathbf{y}
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- estimated standard errors:
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$$
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$$
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- estimated standard errors:
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\hat s_{\beta_i} = \sqrt{([\mathbf{X}^\top \mathbf{X}]^{-1})_{ii} \frac 1 {n-p} \|\mathbf{y} - \mathbf{X} \beta\|^2}
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$$
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$$
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\hat s_{\beta_i} = \sqrt{([\mathbf{X}^\top \mathbf{X}]^{-1})_{ii} \frac 1 {n-p} \|\mathbf{y} - \mathbf{X} \beta\|^2}
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- $t$-statistics:
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$$
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$$ t_i = \frac{\hat \beta_i}{\hat s_{\hat \beta_i}}, \qquad i=0,\ldots,p. $$
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- $t$-statistics:
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- $p$-values:
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$$ t_i = \frac{\hat \beta_i}{\hat s_{\hat \beta_i}}, \qquad i=0,\ldots,p. $$
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$$
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- $p$-values:
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p\text{-value} = \mathbb{P}(|T| > t_i), \quad \text{where } T \sim t_{n-p}
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$$
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$$
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p\text{-value} = \mathbb{P}(|T| > t_i), \quad \text{where } T \sim t_{n-p}
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::: {.callout-caution collapse="false"}
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$$
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## Task 2
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::: {.callout-caution collapse="false"}
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1. Implement functions that estimate the $\beta$-parameters,
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## Task 2
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the corresponding standard errors and the $t$-statistics.
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2. Test your functions with the `tree' data set and try to reproduce the
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1. Implement functions that estimate the $\beta$-parameters,
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output above.
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the corresponding standard errors and the $t$-statistics.
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:::
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2. Test your functions with the `tree' data set and try to reproduce the
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output above.
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Which model is the best? For linear models, one often uses the $R^2$ characteristic.
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:::
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Roughly speaking, it gives the percentage (between 0 and 1) of the variance that can be explained by the linear model.
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Which model is the best? For linear models, one often uses the $R^2$ characteristic.
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``` julia
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Roughly speaking, it gives the percentage (between 0 and 1) of the variance that can be explained by the linear model.
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r2(linmod1)
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r2(linmod2)
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``` julia
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r2(linmod1)
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linmod3 = lm(@formula(Volume ~ Girth + Height + Girth*Height), trees)
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r2(linmod2)
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r2(linmod3)
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linmod3 = lm(@formula(Volume ~ Girth + Height + Girth*Height), trees)
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```
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r2(linmod3)
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::: {callout-note}
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```
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The more covariates you add the more variance can be explained by the linear model - $R^2$ increases. In order to balance goodness-of-tit of a model and its complexity, information criteria such as `aic` are considered.
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:::
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## Generalized Linear Models
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## Generalized Linear Models
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Classical linear model
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Classical linear model
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$$
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\mathbf{y} = \mathbf{X} \beta + \varepsilon
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$$
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$$
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\mathbf{y} = \mathbf{X} \beta + \varepsilon
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$$
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implies that
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$$ \mathbf{y} \mid \mathbf{X} \sim \mathcal{N}(\mathbf{X} \mathbf{\beta}, \sigma^2\mathbf{I}).$$
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implies that
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$$ \mathbf{y} \mid \mathbf{X} \sim \mathcal{N}(\mathbf{X} \mathbf{\beta}, \sigma^2\mathbf{I}).$$
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In particular, the conditional expectation satisfies
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$\mathbb{E}(\mathbf{y} \mid \mathbf{X}) = \mathbf{X} \beta$.
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In particular, the conditional expectation satisfies
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$\mathbb{E}(\mathbf{y} \mid \mathbf{X}) = \mathbf{X} \beta$.
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However, the assumption of a normal distribution is unrealistic for
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non-continuous data. Popular alternatives include:
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However, the assumption of a normal distribution is unrealistic for
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non-continuous data. Popular alternatives include:
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- for counting data: $$
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\mathbf{y} \mid \mathbf{X} \sim \mathrm{Poisson}(\exp(\mathbf{X}\mathbf{\beta})) \qquad \leadsto \mathbb{E}(\mathbf{y} \mid \mathbf{X}) = \exp(\mathbf{X} \beta)
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- for counting data: $$
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$$
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\mathbf{y} \mid \mathbf{X} \sim \mathrm{Poisson}(\exp(\mathbf{X}\mathbf{\beta})) \qquad \leadsto \mathbb{E}(\mathbf{y} \mid \mathbf{X}) = \exp(\mathbf{X} \beta)
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$$
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Here, the components are considered to be independent and the
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exponential function is applied componentwise.
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Here, the components are considered to be independent and the
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exponential function is applied componentwise.
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- for binary data: $$
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\mathbf{y} \mid \mathbf{X} \sim \mathrm{Bernoulli}\left( \frac{\exp(\mathbf{X}\mathbf{\beta})}{1 + \exp(\mathbf{X}\mathbf{\beta})} \right) \qquad \leadsto \mathbb{E}(\mathbf{y} \mid \mathbf{X}) = \frac{\exp(\mathbf{X}\mathbf{\beta})}{1 + \exp(\mathbf{X}\mathbf{\beta})}
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- for binary data: $$
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$$
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\mathbf{y} \mid \mathbf{X} \sim \mathrm{Bernoulli}\left( \frac{\exp(\mathbf{X}\mathbf{\beta})}{1 + \exp(\mathbf{X}\mathbf{\beta})} \right) \qquad \leadsto \mathbb{E}(\mathbf{y} \mid \mathbf{X}) = \frac{\exp(\mathbf{X}\mathbf{\beta})}{1 + \exp(\mathbf{X}\mathbf{\beta})}
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$$
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Again, the components are considered to be independent and all the
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operations are applied componentwise.
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Again, the components are considered to be independent and all the
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operations are applied componentwise.
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All these models are defined by the choice of a family of distributions
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and a function $g$ (the so-called *link function*) such that
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All these models are defined by the choice of a family of distributions
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||||||
|
and a function $g$ (the so-called *link function*) such that
|
||||||
$$
|
|
||||||
\mathbb{E}(\mathbf{y} \mid \mathbf{X}) = g^{-1}(\mathbf{X} \beta).
|
$$
|
||||||
$$
|
\mathbb{E}(\mathbf{y} \mid \mathbf{X}) = g^{-1}(\mathbf{X} \beta).
|
||||||
|
$$
|
||||||
For the models above, these are:
|
|
||||||
|
For the models above, these are:
|
||||||
+----------------+------------------+--------------------------+
|
|
||||||
| Type of Data | Distribution | Link Function |
|
+----------------+------------------+--------------------------+
|
||||||
| | Family | |
|
| Type of Data | Distribution | Link Function |
|
||||||
+================+==================+==========================+
|
| | Family | |
|
||||||
| continuous | Normal | identity: |
|
+================+==================+==========================+
|
||||||
| | | |
|
| continuous | Normal | identity: |
|
||||||
| | | $$ |
|
| | | |
|
||||||
| | | g(x)=x |
|
| | | $$ |
|
||||||
| | | $$ |
|
| | | g(x)=x |
|
||||||
+----------------+------------------+--------------------------+
|
| | | $$ |
|
||||||
| count | Poisson | log: |
|
+----------------+------------------+--------------------------+
|
||||||
| | | |
|
| count | Poisson | log: |
|
||||||
| | | $$ |
|
| | | |
|
||||||
| | | g(x) = \log(x) |
|
| | | $$ |
|
||||||
| | | $$ |
|
| | | g(x) = \log(x) |
|
||||||
+----------------+------------------+--------------------------+
|
| | | $$ |
|
||||||
| binary | Bernoulli | logit: |
|
+----------------+------------------+--------------------------+
|
||||||
| | | |
|
| binary | Bernoulli | logit: |
|
||||||
| | | $$ |
|
| | | |
|
||||||
| | | g(x) = \log\left( |
|
| | | $$ |
|
||||||
| | | \frac{x}{1-x} |
|
| | | g(x) = \log\left( |
|
||||||
| | | \right) |
|
| | | \frac{x}{1-x} |
|
||||||
| | | $$ |
|
| | | \right) |
|
||||||
+----------------+------------------+--------------------------+
|
| | | $$ |
|
||||||
|
+----------------+------------------+--------------------------+
|
||||||
In general, the parameter vector $\beta$ is estimated via maximizing the
|
|
||||||
likelihood, i.e.,
|
In general, the parameter vector $\beta$ is estimated via maximizing the
|
||||||
|
likelihood, i.e.,
|
||||||
$$
|
|
||||||
\hat \beta = \mathrm{argmax} \prod_{i=1}^n f(y_i \mid \mathbf{X}_{\cdot i}),
|
$$
|
||||||
$$
|
\hat \beta = \mathrm{argmax} \prod_{i=1}^n f(y_i \mid \mathbf{X}_{\cdot i}),
|
||||||
|
$$
|
||||||
which is equivalent to the maximization of the log-likelihood, i.e.,
|
|
||||||
|
which is equivalent to the maximization of the log-likelihood, i.e.,
|
||||||
$$
|
|
||||||
\hat \beta = \mathrm{argmax} \sum_{i=1}^n \log f(y_i \mid \mathbf{X}_{\cdot i}),
|
$$
|
||||||
$$
|
\hat \beta = \mathrm{argmax} \sum_{i=1}^n \log f(y_i \mid \mathbf{X}_{\cdot i}),
|
||||||
|
$$
|
||||||
In the Gaussian case, the maximum likelihood estimator is identical to
|
|
||||||
the least squares estimator considered above.
|
In the Gaussian case, the maximum likelihood estimator is identical to
|
||||||
|
the least squares estimator considered above.
|
||||||
``` julia
|
|
||||||
using CSV
|
``` julia
|
||||||
using HTTP
|
using CSV
|
||||||
|
using HTTP
|
||||||
http_response = HTTP.get("https://vincentarelbundock.github.io/Rdatasets/csv/AER/SwissLabor.csv")
|
|
||||||
SwissLabor = DataFrame(CSV.File(http_response.body))
|
http_response = HTTP.get("https://vincentarelbundock.github.io/Rdatasets/csv/AER/SwissLabor.csv")
|
||||||
|
SwissLabor = DataFrame(CSV.File(http_response.body))
|
||||||
SwissLabor[!,"participation"] .= (SwissLabor.participation .== "yes")
|
|
||||||
|
SwissLabor[!,"participation"] .= (SwissLabor.participation .== "yes")
|
||||||
model = glm(@formula(participation ~ age^2),
|
|
||||||
SwissLabor, Binomial(), ProbitLink())
|
model = glm(@formula(participation ~ age^2),
|
||||||
```
|
SwissLabor, Binomial(), ProbitLink())
|
||||||
|
```
|
||||||
::: {.callout-caution collapse="false"}
|
|
||||||
|
::: {.callout-caution collapse="false"}
|
||||||
## Task 3:
|
|
||||||
|
## Task 3:
|
||||||
1. Reproduce the results of our data analysis of the `tree` data set
|
|
||||||
using a generalized linear model with normal distribution family.
|
1. Reproduce the results of our data analysis of the `tree` data set
|
||||||
2. Generate $n=20$ random covariates $\mathbf{x}$ and Poisson-distributed counting data with parameters $\beta_0 + \beta_1 x_i$. Re-estimate the parameters by a generalized linear model.
|
using a generalized linear model with normal distribution family.
|
||||||
:::
|
2. Generate $n=20$ random covariates $\mathbf{x}$ and Poisson-distributed counting data with parameters $\beta_0 + \beta_1 x_i$. Re-estimate the parameters by a generalized linear model.
|
||||||
|
:::
|
||||||
|
|||||||
Reference in New Issue
Block a user