254 lines
8.8 KiB
Plaintext
254 lines
8.8 KiB
Plaintext
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---
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editor:
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markdown:
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wrap: 72
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---
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# Multiple Regression Basics
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## Motivation
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### Introductory Example: tree dataset from R
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\[figure of raw data\]
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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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of a tree given its girth?
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\[figure including a straight line\]
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First Guess: There is a linear relation!
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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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variable $y_i$ depends linearly on the corresponding value $x_i$ of the
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covariate
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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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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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*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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*Typical choice*: Least squares estimator (= maximum likelihood
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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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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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Written in matrix style:
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$$
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(\hat \beta_0, \hat \beta_1) = \mathrm{argmin} \ \left\| \mathbf{y} - (\mathbf{1},\mathbf{x}) \left( \begin{array}{c} \beta_0\\ \beta_1\end{array}\right) \right\|^2
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$$
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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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rather use Julia to solve the problem.
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\[use Julia code (existing package) to perform linear regression for
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`volume ~ girth`\]
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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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$\hat \beta_1$
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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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- 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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Under the hypothesis $\beta_i=0$, the test statistics $t_i$ would
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follow a $t$-distribution.
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- column `Pr(>|t|)`: $p$-values for the hyptheses $\beta_i=0$ for
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$i=0,1$
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**Task 1**: Generate a random set of covariates $\mathbf{x}$. Given
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these covariates and true parameters $\beta_0$, $\beta_1$ and $\sigma^2$
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(you can choose them)), simulate responses from a linear model and
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estimate the coefficients $\beta_0$ and $\beta_1$. Play with different
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choices of the parameters to see the effects on the parameter estimates
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and the $p$-values.
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## Multiple Regression Model
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*Idea*: Generalize the simple linear regression model to multiple
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covariates, w.g., predict `volume` using `girth` and \`height\`\`.
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$\leadsto$ **Linear Model:**
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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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- $y_i$: $i$-th measurement of the response,
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- $x_{i1}$: $i$ th value of first covariate,
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- ...
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- $x_{ip}$: $i$-th value of $p$-th covariate,
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- $\varepsilon_i \sim \mathcal{N}(0,\sigma^2)$: independent normally
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distributed errors with unknown variance $\sigma^2$.
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*Task:* Find the *optimal* estimators for
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$\mathbf{\beta} = (\beta_0, \beta_1, \ldots, \beta_p)$.
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*Our choice again:* Least squares estimator (= maximum likelihood
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estimator for normal errors)
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$$
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\hat \beta = \mathrm{argmin} \ \| \mathbf{y} - \mathbf{1} \beta_0 - \mathbf{x}_1 \beta_1 - \ldots - \mathbf{x}_p \beta_p\|^2
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$$
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where $\mathbf{y}$ is the vector of responses, $\mathbf{x}$\_j is the
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vector of the $j$ th covariate and $\mathbf{1}$ is a vector of ones.
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Written in matrix style:
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$$
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\mathbf{\hat \beta} = \mathrm{argmin} \ \left\| \mathbf{y} - (\mathbf{1},\mathbf{x}_1,\ldots,\mathbf{x}_p) \left( \begin{array}{c} \beta_0 \\ \beta_1 \\ \vdots \\ \beta_p\end{array} \right) \right\|^2
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$$
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Defining the *design matrix*
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$$ \mathbf{X} = \left( \begin{array}{cccc}
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1 & x_{11} & \ldots & x_{1p} \\
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\vdots & \vdots & \ddots & \vdots \\
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1 & x_{11} & \ldots & x_{1p}
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\end{array}\right) \qquad
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(\text{size } n \times (p+1)), $$
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we get the short form
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$$
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\mathbf{\hat \beta} = \mathrm{argmin} \ \| \mathbf{y} - \mathbf{X} \mathbf{\beta} \|^2 = (\mathbf{X}^\top \mathbf{X})^{-1} \mathbf{X}^\top \mathbf{y}
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$$
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\[use Julia code (existing package) to perform linear regression for
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`volume ~ girth + height`\]
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The interpretation of the Julia output is similar to the simple linear
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regression model, but we provide explicit formulas now:
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- parameter estimates:
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$$
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(\mathbf{X}^\top \mathbf{X})^{-1} \mathbf{X}^\top \mathbf{y}
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$$
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- estimated standard errors:
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$$
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\hat s_{\beta_i} = \sqrt{([\mathbf{X}^\top \mathbf{X}]^{-1})_{ii} \frac 1 {n-p} \|\mathbf{y} - \mathbf{X} \beta\|^2}
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$$
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- $t$-statistics:
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$$ t_i = \frac{\hat \beta_i}{\hat s_{\hat \beta_i}}, \qquad i=0,\ldots,p. $$
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- $p$-values:
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$$
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p\text{-value} = \mathbb{P}(|T| > t_i), \quad \text{where } T \sim t_{n-p}
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$$
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**Task 2**: Implement functions that estimate the $\beta$-parameters,
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the corresponding standard errors and the $t$-statistics. Test your
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functions with the \`\`\`tree''' data set and try to reproduce the
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output above.
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## Generalized Linear Models
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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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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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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: $$
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\mathbf{y} \mid \mathbf{X} \sim \mathrm{Poisson}(\exp(\mathbf{X}\mathbf{\beta})) \qquad \leadsto \mathbb{E}(\mathbf{y} \mid \mathbf{X}) = \exp(\mathbf{X} \beta)
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$$
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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: $$
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\mathbf{y} \mid \mathbf{X} \sim \mathrm{Bernoulli}\left( \frac{\exp(\mathbf{X}\mathbf{\beta})}{1 + \exp(\mathbf{X}\mathbf{\beta})} \right) \qquad \leadsto \mathbb{E}(\mathbf{y} \mid \mathbf{X}) = \frac{\exp(\mathbf{X}\mathbf{\beta})}{1 + \exp(\mathbf{X}\mathbf{\beta})}
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$$
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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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$$
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\mathbb{E}(\mathbf{y} \mid \mathbf{X}) = g^{-1}(\mathbf{X} \beta).
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$$
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For the models above, these are:
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+----------------------+---------------------+----------------------+
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| Type of Data | Distribution Family | Link Function |
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+======================+=====================+======================+
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| continuous | Normal | identity: |
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| | | $$ |
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| | | g(x)=x |
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+----------------------+---------------------+----------------------+
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| count | Poisson | log: |
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| | | $$ |
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| | | g(x) = \log(x) |
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+----------------------+---------------------+----------------------+
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| binary | Bernoulli | logit: |
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| | | $$ |
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| | | g(x) = \log\left |
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| | | ( |
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| | | \frac{x}{1-x}\right) |
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+----------------------+---------------------+----------------------+
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In general, the parameter vector $\beta$ is estimated via maximizing the
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likelihood, i.e.,
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$$
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\hat \beta = \mathrm{argmax} \prod_{i=1}^n f(y_i \mid \mathbf{X}_{\cdot i}),
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$$
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which is equivalent to the maximization of the log-likelihood, i.e.,
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$$
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\hat \beta = \mathrm{argmax} \sum_{i=1}^n \log f(y_i \mid \mathbf{X}_{\cdot i}),
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$$
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In the Gaussian case, the maximum likelihood estimator is identical to
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the least squares estimator considered above.
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\[\[ Example in Julia: maybe `SwissLabor` \]\]
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**Task 3:** Reproduce the results of our data analysis of the `tree`
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data set using a generalized linear model with normal distribution
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family.
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