Updated Text & Code on Regression.
This commit is contained in:
Marco Oesting
52
material/3_wed/regression/Code_Snippets.jl
Normal file
52
material/3_wed/regression/Code_Snippets.jl
Normal file
@@ -0,0 +1,52 @@
|
|||||||
|
############################################################################
|
||||||
|
#### Execute code chunks separately in VSCODE by pressing 'Alt + Enter' ####
|
||||||
|
############################################################################
|
||||||
|
|
||||||
|
using Statistics
|
||||||
|
using Plots
|
||||||
|
using RDatasets
|
||||||
|
using GLM
|
||||||
|
|
||||||
|
##
|
||||||
|
|
||||||
|
trees = dataset("datasets", "trees")
|
||||||
|
|
||||||
|
scatter(trees.Girth, trees.Volume,
|
||||||
|
legend=false, xlabel="Girth", ylabel="Volume")
|
||||||
|
|
||||||
|
##
|
||||||
|
|
||||||
|
scatter(trees.Girth, trees.Volume,
|
||||||
|
legend=false, xlabel="Girth", ylabel="Volume")
|
||||||
|
plot!(x -> -37 + 5*x)
|
||||||
|
|
||||||
|
##
|
||||||
|
|
||||||
|
linmod1 = lm(@formula(Volume ~ Girth), trees)
|
||||||
|
|
||||||
|
##
|
||||||
|
|
||||||
|
linmod2 = lm(@formula(Volume ~ Girth + Height), trees)
|
||||||
|
|
||||||
|
##
|
||||||
|
|
||||||
|
r2(linmod1)
|
||||||
|
r2(linmod2)
|
||||||
|
|
||||||
|
linmod3 = lm(@formula(Volume ~ Girth + Height + Girth*Height), trees)
|
||||||
|
|
||||||
|
r2(linmod3)
|
||||||
|
|
||||||
|
##
|
||||||
|
|
||||||
|
using CSV
|
||||||
|
using HTTP
|
||||||
|
|
||||||
|
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")
|
||||||
|
|
||||||
|
##
|
||||||
|
|
||||||
|
model = glm(@formula(participation ~ age), SwissLabor, Binomial(), ProbitLink())
|
||||||
@@ -1,253 +1,292 @@
|
|||||||
---
|
---
|
||||||
editor:
|
editor:
|
||||||
markdown:
|
markdown:
|
||||||
wrap: 72
|
wrap: 72
|
||||||
---
|
---
|
||||||
|
|
||||||
# Multiple Regression Basics
|
# Multiple Regression Basics
|
||||||
|
|
||||||
## Motivation
|
## Motivation
|
||||||
|
|
||||||
### Introductory Example: tree dataset from R
|
### Introductory Example: tree dataset from R
|
||||||
|
|
||||||
\[figure of raw data\]
|
```{julia}
|
||||||
|
using Statistics
|
||||||
*Aim:* Find relationship between the *response variable* `volume` and
|
using Plots
|
||||||
the *explanatory variable/covariate* `girth`? Can we predict the volume
|
using RDatasets
|
||||||
of a tree given its girth?
|
|
||||||
|
trees = dataset("datasets", "trees")
|
||||||
\[figure including a straight line\]
|
|
||||||
|
scatter(trees.Volume, trees.Girth,
|
||||||
First Guess: There is a linear relation!
|
legend=false, xlabel="Girth", ylabel="Volume")
|
||||||
|
```
|
||||||
## Simple Linear Regression
|
|
||||||
|
*Aim:* Find relationship between the *response variable* `volume` and
|
||||||
Main assumption: up to some error term, each measurement of the response
|
the *explanatory variable/covariate* `girth`? Can we predict the volume
|
||||||
variable $y_i$ depends linearly on the corresponding value $x_i$ of the
|
of a tree given its girth?
|
||||||
covariate
|
|
||||||
|
```{julia}
|
||||||
$\leadsto$ **(Simple) Linear Model:**
|
scatter(trees.Girth, trees.Volume,
|
||||||
$$y_i = \beta_0 + \beta_1 x_i + \varepsilon_i, \qquad i=1,...,n,$$
|
legend=false, xlabel="Girth", ylabel="Volume")
|
||||||
where $\varepsilon_i \sim \mathcal{N}(0,\sigma^2)$ are independent
|
plot!(x -> -37 + 5*x)
|
||||||
normally distributed errors with unknown variance $\sigma^2$.
|
```
|
||||||
|
|
||||||
*Task:* Find the straight line that fits best, i.e., find the *optimal*
|
First Guess: There is a linear relation!
|
||||||
estimators for $\beta_0$ and $\beta_1$.
|
|
||||||
|
## Simple Linear Regression
|
||||||
*Typical choice*: Least squares estimator (= maximum likelihood
|
|
||||||
estimator for normal errors)
|
Main assumption: up to some error term, each measurement of the response
|
||||||
|
variable $y_i$ depends linearly on the corresponding value $x_i$ of the
|
||||||
$$ (\hat \beta_0, \hat \beta_1) = \mathrm{argmin} \ \| \mathbf{y} - \mathbf{1} \beta_0 - \mathbf{x} \beta_1\|^2 $$
|
covariate
|
||||||
|
|
||||||
where $\mathbf{y}$ is the vector of responses, $\mathbf{x}$ is the
|
$\leadsto$ **(Simple) Linear Model:**
|
||||||
vector of covariates and $\mathbf{1}$ is a vector of ones.
|
$$y_i = \beta_0 + \beta_1 x_i + \varepsilon_i, \qquad i=1,...,n,$$
|
||||||
|
where $\varepsilon_i \sim \mathcal{N}(0,\sigma^2)$ are independent
|
||||||
Written in matrix style:
|
normally distributed errors with unknown variance $\sigma^2$.
|
||||||
|
|
||||||
$$
|
*Task:* Find the straight line that fits best, i.e., find the *optimal*
|
||||||
(\hat \beta_0, \hat \beta_1) = \mathrm{argmin} \ \left\| \mathbf{y} - (\mathbf{1},\mathbf{x}) \left( \begin{array}{c} \beta_0\\ \beta_1\end{array}\right) \right\|^2
|
estimators for $\beta_0$ and $\beta_1$.
|
||||||
$$
|
|
||||||
|
*Typical choice*: Least squares estimator (= maximum likelihood
|
||||||
Note: There is a closed-form expression for
|
estimator for normal errors)
|
||||||
$(\hat \beta_0, \hat \beta_1)$. We will not make use of it here, but
|
|
||||||
rather use Julia to solve the problem.
|
$$ (\hat \beta_0, \hat \beta_1) = \mathrm{argmin} \ \| \mathbf{y} - \mathbf{1} \beta_0 - \mathbf{x} \beta_1\|^2 $$
|
||||||
|
|
||||||
\[use Julia code (existing package) to perform linear regression for
|
where $\mathbf{y}$ is the vector of responses, $\mathbf{x}$ is the
|
||||||
`volume ~ girth`\]
|
vector of covariates and $\mathbf{1}$ is a vector of ones.
|
||||||
|
|
||||||
*Interpretation of the Julia output:*
|
Written in matrix style:
|
||||||
|
|
||||||
- column `estimate` : least square estimates for $\hat \beta_0$ and
|
$$
|
||||||
$\hat \beta_1$
|
(\hat \beta_0, \hat \beta_1) = \mathrm{argmin} \ \left\| \mathbf{y} - (\mathbf{1},\mathbf{x}) \left( \begin{array}{c} \beta_0\\ \beta_1\end{array}\right) \right\|^2
|
||||||
|
$$
|
||||||
- column `Std. Error` : estimated standard deviation
|
|
||||||
$\hat s_{\hat \beta_i}$ of the estimator $\hat \beta_i$
|
Note: There is a closed-form expression for
|
||||||
|
$(\hat \beta_0, \hat \beta_1)$. We will not make use of it here, but
|
||||||
- column `t value` : value of the $t$-statistics
|
rather use Julia to solve the problem.
|
||||||
|
|
||||||
$$ t_i = {\hat \beta_i \over \hat s_{\hat \beta_i}}, \quad i=0,1, $$
|
\[use Julia code (existing package) to perform linear regression for
|
||||||
|
`volume ~ girth`\]
|
||||||
Under the hypothesis $\beta_i=0$, the test statistics $t_i$ would
|
|
||||||
follow a $t$-distribution.
|
```{julia}
|
||||||
|
lm(@formula(Volume ~ Girth), trees)
|
||||||
- column `Pr(>|t|)`: $p$-values for the hyptheses $\beta_i=0$ for
|
```
|
||||||
$i=0,1$
|
|
||||||
|
*Interpretation of the Julia output:*
|
||||||
**Task 1**: Generate a random set of covariates $\mathbf{x}$. Given
|
|
||||||
these covariates and true parameters $\beta_0$, $\beta_1$ and $\sigma^2$
|
- column `estimate` : least square estimates for $\hat \beta_0$ and
|
||||||
(you can choose them)), simulate responses from a linear model and
|
$\hat \beta_1$
|
||||||
estimate the coefficients $\beta_0$ and $\beta_1$. Play with different
|
|
||||||
choices of the parameters to see the effects on the parameter estimates
|
- column `Std. Error` : estimated standard deviation
|
||||||
and the $p$-values.
|
$\hat s_{\hat \beta_i}$ of the estimator $\hat \beta_i$
|
||||||
|
|
||||||
## Multiple Regression Model
|
- column `t value` : value of the $t$-statistics
|
||||||
|
|
||||||
*Idea*: Generalize the simple linear regression model to multiple
|
$$ t_i = {\hat \beta_i \over \hat s_{\hat \beta_i}}, \quad i=0,1, $$
|
||||||
covariates, w.g., predict `volume` using `girth` and \`height\`\`.
|
|
||||||
|
Under the hypothesis $\beta_i=0$, the test statistics $t_i$ would
|
||||||
$\leadsto$ **Linear Model:**
|
follow a $t$-distribution.
|
||||||
$$y_i = \beta_0 + \beta_1 x_{i1} + \ldots + \beta_p x_{ip} + \varepsilon_i, \qquad i=1,...,n,$$where
|
|
||||||
|
- column `Pr(>|t|)`: $p$-values for the hyptheses $\beta_i=0$ for
|
||||||
- $y_i$: $i$-th measurement of the response,
|
$i=0,1$
|
||||||
|
|
||||||
- $x_{i1}$: $i$ th value of first covariate,
|
**Task 1**: Generate a random set of covariates $\mathbf{x}$. Given
|
||||||
|
these covariates and true parameters $\beta_0$, $\beta_1$ and $\sigma^2$
|
||||||
- ...
|
(you can choose them)), simulate responses from a linear model and
|
||||||
|
estimate the coefficients $\beta_0$ and $\beta_1$. Play with different
|
||||||
- $x_{ip}$: $i$-th value of $p$-th covariate,
|
choices of the parameters to see the effects on the parameter estimates
|
||||||
|
and the $p$-values.
|
||||||
- $\varepsilon_i \sim \mathcal{N}(0,\sigma^2)$: independent normally
|
|
||||||
distributed errors with unknown variance $\sigma^2$.
|
## Multiple Regression Model
|
||||||
|
|
||||||
*Task:* Find the *optimal* estimators for
|
*Idea*: Generalize the simple linear regression model to multiple
|
||||||
$\mathbf{\beta} = (\beta_0, \beta_1, \ldots, \beta_p)$.
|
covariates, w.g., predict `volume` using `girth` and \`height\`\`.
|
||||||
|
|
||||||
*Our choice again:* Least squares estimator (= maximum likelihood
|
$\leadsto$ **Linear Model:**
|
||||||
estimator for normal errors)
|
$$y_i = \beta_0 + \beta_1 x_{i1} + \ldots + \beta_p x_{ip} + \varepsilon_i, \qquad i=1,...,n,$$where
|
||||||
|
|
||||||
$$
|
- $y_i$: $i$-th measurement of the response,
|
||||||
\hat \beta = \mathrm{argmin} \ \| \mathbf{y} - \mathbf{1} \beta_0 - \mathbf{x}_1 \beta_1 - \ldots - \mathbf{x}_p \beta_p\|^2
|
|
||||||
$$
|
- $x_{i1}$: $i$ th value of first covariate,
|
||||||
|
|
||||||
where $\mathbf{y}$ is the vector of responses, $\mathbf{x}$\_j is the
|
- ...
|
||||||
vector of the $j$ th covariate and $\mathbf{1}$ is a vector of ones.
|
|
||||||
|
- $x_{ip}$: $i$-th value of $p$-th covariate,
|
||||||
Written in matrix style:
|
|
||||||
|
- $\varepsilon_i \sim \mathcal{N}(0,\sigma^2)$: independent normally
|
||||||
$$
|
distributed errors with unknown variance $\sigma^2$.
|
||||||
\mathbf{\hat \beta} = \mathrm{argmin} \ \left\| \mathbf{y} - (\mathbf{1},\mathbf{x}_1,\ldots,\mathbf{x}_p) \left( \begin{array}{c} \beta_0 \\ \beta_1 \\ \vdots \\ \beta_p\end{array} \right) \right\|^2
|
|
||||||
$$
|
*Task:* Find the *optimal* estimators for
|
||||||
|
$\mathbf{\beta} = (\beta_0, \beta_1, \ldots, \beta_p)$.
|
||||||
Defining the *design matrix*
|
|
||||||
|
*Our choice again:* Least squares estimator (= maximum likelihood
|
||||||
$$ \mathbf{X} = \left( \begin{array}{cccc}
|
estimator for normal errors)
|
||||||
1 & x_{11} & \ldots & x_{1p} \\
|
|
||||||
\vdots & \vdots & \ddots & \vdots \\
|
$$
|
||||||
1 & x_{11} & \ldots & x_{1p}
|
\hat \beta = \mathrm{argmin} \ \| \mathbf{y} - \mathbf{1} \beta_0 - \mathbf{x}_1 \beta_1 - \ldots - \mathbf{x}_p \beta_p\|^2
|
||||||
\end{array}\right) \qquad
|
$$
|
||||||
(\text{size } n \times (p+1)), $$
|
|
||||||
|
where $\mathbf{y}$ is the vector of responses, $\mathbf{x}$\_j is the
|
||||||
we get the short form
|
vector of the $j$ th covariate and $\mathbf{1}$ is a vector of ones.
|
||||||
|
|
||||||
$$
|
Written in matrix style:
|
||||||
\mathbf{\hat \beta} = \mathrm{argmin} \ \| \mathbf{y} - \mathbf{X} \mathbf{\beta} \|^2 = (\mathbf{X}^\top \mathbf{X})^{-1} \mathbf{X}^\top \mathbf{y}
|
|
||||||
$$
|
$$
|
||||||
|
\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
|
||||||
\[use Julia code (existing package) to perform linear regression for
|
$$
|
||||||
`volume ~ girth + height`\]
|
|
||||||
|
Defining the *design matrix*
|
||||||
The interpretation of the Julia output is similar to the simple linear
|
|
||||||
regression model, but we provide explicit formulas now:
|
$$ \mathbf{X} = \left( \begin{array}{cccc}
|
||||||
|
1 & x_{11} & \ldots & x_{1p} \\
|
||||||
- parameter estimates:
|
\vdots & \vdots & \ddots & \vdots \\
|
||||||
|
1 & x_{11} & \ldots & x_{1p}
|
||||||
$$
|
\end{array}\right) \qquad
|
||||||
(\mathbf{X}^\top \mathbf{X})^{-1} \mathbf{X}^\top \mathbf{y}
|
(\text{size } n \times (p+1)), $$
|
||||||
$$
|
|
||||||
|
we get the short form
|
||||||
- estimated standard errors:
|
|
||||||
|
$$
|
||||||
$$
|
\mathbf{\hat \beta} = \mathrm{argmin} \ \| \mathbf{y} - \mathbf{X} \mathbf{\beta} \|^2 = (\mathbf{X}^\top \mathbf{X})^{-1} \mathbf{X}^\top \mathbf{y}
|
||||||
\hat s_{\beta_i} = \sqrt{([\mathbf{X}^\top \mathbf{X}]^{-1})_{ii} \frac 1 {n-p} \|\mathbf{y} - \mathbf{X} \beta\|^2}
|
$$
|
||||||
$$
|
|
||||||
|
\[use Julia code (existing package) to perform linear regression for
|
||||||
- $t$-statistics:
|
`volume ~ girth + height`\]
|
||||||
|
|
||||||
$$ t_i = \frac{\hat \beta_i}{\hat s_{\hat \beta_i}}, \qquad i=0,\ldots,p. $$
|
The interpretation of the Julia output is similar to the simple linear
|
||||||
|
regression model, but we provide explicit formulas now:
|
||||||
- $p$-values:
|
|
||||||
|
- parameter estimates:
|
||||||
$$
|
|
||||||
p\text{-value} = \mathbb{P}(|T| > t_i), \quad \text{where } T \sim t_{n-p}
|
$$
|
||||||
$$
|
(\mathbf{X}^\top \mathbf{X})^{-1} \mathbf{X}^\top \mathbf{y}
|
||||||
|
$$
|
||||||
**Task 2**: Implement functions that estimate the $\beta$-parameters,
|
|
||||||
the corresponding standard errors and the $t$-statistics. Test your
|
- estimated standard errors:
|
||||||
functions with the \`\`\`tree''' data set and try to reproduce the
|
|
||||||
output above.
|
$$
|
||||||
|
\hat s_{\beta_i} = \sqrt{([\mathbf{X}^\top \mathbf{X}]^{-1})_{ii} \frac 1 {n-p} \|\mathbf{y} - \mathbf{X} \beta\|^2}
|
||||||
## Generalized Linear Models
|
$$
|
||||||
|
|
||||||
Classical linear model
|
- $t$-statistics:
|
||||||
|
|
||||||
$$
|
$$ t_i = \frac{\hat \beta_i}{\hat s_{\hat \beta_i}}, \qquad i=0,\ldots,p. $$
|
||||||
\mathbf{y} = \mathbf{X} \beta + \varepsilon
|
|
||||||
$$
|
- $p$-values:
|
||||||
|
|
||||||
implies that
|
$$
|
||||||
$$ \mathbf{y} \mid \mathbf{X} \sim \mathcal{N}(\mathbf{X} \mathbf{\beta}, \sigma^2\mathbf{I}).$$
|
p\text{-value} = \mathbb{P}(|T| > t_i), \quad \text{where } T \sim t_{n-p}
|
||||||
|
$$
|
||||||
In particular, the conditional expectation satisfies
|
|
||||||
$\mathbb{E}(\mathbf{y} \mid \mathbf{X}) = \mathbf{X} \beta$.
|
**Task 2**: Implement functions that estimate the $\beta$-parameters,
|
||||||
|
the corresponding standard errors and the $t$-statistics. Test your
|
||||||
However, the assumption of a normal distribution is unrealistic for
|
functions with the \`\`\`tree''' data set and try to reproduce the
|
||||||
non-continuous data. Popular alternatives include:
|
output above.
|
||||||
|
|
||||||
- for counting data: $$
|
```{julia}
|
||||||
\mathbf{y} \mid \mathbf{X} \sim \mathrm{Poisson}(\exp(\mathbf{X}\mathbf{\beta})) \qquad \leadsto \mathbb{E}(\mathbf{y} \mid \mathbf{X}) = \exp(\mathbf{X} \beta)
|
r2(linmod1)
|
||||||
$$
|
r2(linmod2)
|
||||||
|
|
||||||
Here, the components are considered to be independent and the
|
linmod3 = lm(@formula(Volume ~ Girth + Height + Girth*Height), trees)
|
||||||
exponential function is applied componentwise.
|
|
||||||
|
r2(linmod3)
|
||||||
- for binary data: $$
|
```
|
||||||
\mathbf{y} \mid \mathbf{X} \sim \mathrm{Bernoulli}\left( \frac{\exp(\mathbf{X}\mathbf{\beta})}{1 + \exp(\mathbf{X}\mathbf{\beta})} \right) \qquad \leadsto \mathbb{E}(\mathbf{y} \mid \mathbf{X}) = \frac{\exp(\mathbf{X}\mathbf{\beta})}{1 + \exp(\mathbf{X}\mathbf{\beta})}
|
|
||||||
$$
|
## Generalized Linear Models
|
||||||
|
|
||||||
Again, the components are considered to be independent and all the
|
Classical linear model
|
||||||
operations are applied componentwise.
|
|
||||||
|
$$
|
||||||
All these models are defined by the choice of a family of distributions
|
\mathbf{y} = \mathbf{X} \beta + \varepsilon
|
||||||
and a function $g$ (the so-called *link function*) such that
|
$$
|
||||||
|
|
||||||
$$
|
implies that
|
||||||
\mathbb{E}(\mathbf{y} \mid \mathbf{X}) = g^{-1}(\mathbf{X} \beta).
|
$$ \mathbf{y} \mid \mathbf{X} \sim \mathcal{N}(\mathbf{X} \mathbf{\beta}, \sigma^2\mathbf{I}).$$
|
||||||
$$
|
|
||||||
|
In particular, the conditional expectation satisfies
|
||||||
For the models above, these are:
|
$\mathbb{E}(\mathbf{y} \mid \mathbf{X}) = \mathbf{X} \beta$.
|
||||||
|
|
||||||
+----------------------+---------------------+----------------------+
|
However, the assumption of a normal distribution is unrealistic for
|
||||||
| Type of Data | Distribution Family | Link Function |
|
non-continuous data. Popular alternatives include:
|
||||||
+======================+=====================+======================+
|
|
||||||
| continuous | Normal | identity: |
|
- for counting data: $$
|
||||||
| | | |
|
\mathbf{y} \mid \mathbf{X} \sim \mathrm{Poisson}(\exp(\mathbf{X}\mathbf{\beta})) \qquad \leadsto \mathbb{E}(\mathbf{y} \mid \mathbf{X}) = \exp(\mathbf{X} \beta)
|
||||||
| | | $$ |
|
$$
|
||||||
| | | g(x)=x |
|
|
||||||
| | | $$ |
|
Here, the components are considered to be independent and the
|
||||||
+----------------------+---------------------+----------------------+
|
exponential function is applied componentwise.
|
||||||
| count | Poisson | log: |
|
|
||||||
| | | |
|
- for binary data: $$
|
||||||
| | | $$ |
|
\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})}
|
||||||
| | | g(x) = \log(x) |
|
$$
|
||||||
| | | $$ |
|
|
||||||
+----------------------+---------------------+----------------------+
|
Again, the components are considered to be independent and all the
|
||||||
| binary | Bernoulli | logit: |
|
operations are applied componentwise.
|
||||||
| | | |
|
|
||||||
| | | $$ |
|
All these models are defined by the choice of a family of distributions
|
||||||
| | | g(x) = \log\left |
|
and a function $g$ (the so-called *link function*) such that
|
||||||
| | | ( |
|
|
||||||
| | | \frac{x}{1-x}\right) |
|
$$
|
||||||
| | | $$ |
|
\mathbb{E}(\mathbf{y} \mid \mathbf{X}) = g^{-1}(\mathbf{X} \beta).
|
||||||
+----------------------+---------------------+----------------------+
|
$$
|
||||||
|
|
||||||
In general, the parameter vector $\beta$ is estimated via maximizing the
|
For the models above, these are:
|
||||||
likelihood, i.e.,
|
|
||||||
|
+--------------+---------------------+--------------------+
|
||||||
$$
|
| Type of Data | Distribution Family | Link Function |
|
||||||
\hat \beta = \mathrm{argmax} \prod_{i=1}^n f(y_i \mid \mathbf{X}_{\cdot i}),
|
+==============+=====================+====================+
|
||||||
$$
|
| continuous | Normal | identity: |
|
||||||
|
| | | |
|
||||||
which is equivalent to the maximization of the log-likelihood, i.e.,
|
| | | $$ |
|
||||||
|
| | | g(x)=x |
|
||||||
$$
|
| | | $$ |
|
||||||
\hat \beta = \mathrm{argmax} \sum_{i=1}^n \log f(y_i \mid \mathbf{X}_{\cdot i}),
|
+--------------+---------------------+--------------------+
|
||||||
$$
|
| count | Poisson | log: |
|
||||||
|
| | | |
|
||||||
In the Gaussian case, the maximum likelihood estimator is identical to
|
| | | $$ |
|
||||||
the least squares estimator considered above.
|
| | | g(x) = \log(x) |
|
||||||
|
| | | $$ |
|
||||||
\[\[ Example in Julia: maybe `SwissLabor` \]\]
|
+--------------+---------------------+--------------------+
|
||||||
|
| binary | Bernoulli | logit: |
|
||||||
**Task 3:** Reproduce the results of our data analysis of the `tree`
|
| | | |
|
||||||
data set using a generalized linear model with normal distribution
|
| | | $$ |
|
||||||
family.
|
| | | g(x) = \log\left |
|
||||||
|
| | | ( |
|
||||||
|
| | | \ |
|
||||||
|
| | | f |
|
||||||
|
| | | rac{x}{1-x}\right) |
|
||||||
|
| | | $$ |
|
||||||
|
+--------------+---------------------+--------------------+
|
||||||
|
|
||||||
|
In general, the parameter vector $\beta$ is estimated via maximizing the
|
||||||
|
likelihood, i.e.,
|
||||||
|
|
||||||
|
$$
|
||||||
|
\hat \beta = \mathrm{argmax} \prod_{i=1}^n f(y_i \mid \mathbf{X}_{\cdot i}),
|
||||||
|
$$
|
||||||
|
|
||||||
|
which is equivalent to the maximization of the log-likelihood, i.e.,
|
||||||
|
|
||||||
|
$$
|
||||||
|
\hat \beta = \mathrm{argmax} \sum_{i=1}^n \log f(y_i \mid \mathbf{X}_{\cdot i}),
|
||||||
|
$$
|
||||||
|
|
||||||
|
In the Gaussian case, the maximum likelihood estimator is identical to
|
||||||
|
the least squares estimator considered above.
|
||||||
|
|
||||||
|
```{julia}
|
||||||
|
using CSV
|
||||||
|
using HTTP
|
||||||
|
|
||||||
|
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")
|
||||||
|
|
||||||
|
model = glm(@formula(participation ~ age^2),
|
||||||
|
SwissLabor, Binomial(), ProbitLink())
|
||||||
|
```
|
||||||
|
|
||||||
|
**Task 3:** Reproduce the results of our data analysis of the `tree`
|
||||||
|
data set using a generalized linear model with normal distribution
|
||||||
|
family.
|
||||||
|
|||||||
Reference in New Issue
Block a user