Small changes in admonitions

Author: VaclavMacha

docs/src/lecture_11/glm.md

@@ -19,16 +19,12 @@ function plot_histogram(xs, f; kwargs...)
 end
 ```
 
-
 # [Linear regression revisited](@id statistics)
 
 This section revisits the linear regression. The classical statistical approach uses derives the same formulation for linear regression as the optimization approach. Besides point estimates for parameters, it also computes their confidence intervals and can test whether some parameters can be omitted from the model. We will start with hypothesis testing and then continue with regression.
 
 Julia provides lots of statistical packages. They are summarized at the [JuliaStats](https://juliastats.org/) webpage. This section will give a brief introduction to many of them.
 
-
-
-
 ## Theory of hypothesis testing
 
 Hypothesis testing verifies whether data satisfy a given null hypothesis ``H_0``. Most of the tests need some assumptions about the data, such as normality. Under the validity of the null hypothesis, the test derives that a transformation of the data follows some distribution. Then it constructs a confidence interval of this distribution and checks whether the transformed variable lies in this confidence interval. If it lies outside of it, the test rejects the null hypothesis. In the opposite case, it fails to reject the null hypothesis. The latter is different from confirming the null hypothesis. Hypothesis testing is like a grumpy professor during exams. He never acknowledges that a student knows the topic sufficiently, but he is often clear that the student does not know it.
@@ -47,13 +43,8 @@ p = 2\min\{\mathbb P(T\le t \mid H_0), \mathbb P(T\ge t\mid H_0)\}
 
 If the ``p``-value is smaller than a given threshold, usually ``5\%``, the null hypothesis is rejected. In the opposite case, it is not rejected. The ``p``-value is a measure of the probability that an observed difference could have occurred just by random chance.
 
-
-
-
 ## Hypothesis testing
 
-
-
 We first randomly generate data from the normal distribution with zero mean.
 
 ```@example glm
@@ -72,20 +63,18 @@ nothing # hide
 
 The following exercise performs the ``t``-test to check whether the data come from a distribution with zero mean.
 
-
-
-
-
 ```@raw html
 <div class="admonition is-category-exercise">
 <header class="admonition-header">Exercise:</header>
 <div class="admonition-body">
 ```
+
 Use the ``t``-test to verify whether the samples were generated from a distribution with zero mean.
 
-**Hint**: the Student's distribution is invoked by `TDist()`.
+**Hints:**
+- The Student's distribution is invoked by `TDist()`.
+- The probability ``\mathbb P(T\le t)`` equals to the [distribution function](https://en.wikipedia.org/wiki/Cumulative_distribution_function) ``F(t)``, which can be called by `cdf`.
 
-**Hint**: the probability ``\mathbb P(T\le t)`` equals to the [distribution function](https://en.wikipedia.org/wiki/Cumulative_distribution_function) ``F(t)``, which can be called by `cdf`.
 ```@raw html
 </div></div>
 <details class = "solution-body">
@@ -109,14 +98,6 @@ The ``p``-value is significantly larger than ``5\%``. Therefore, we cannot rejec
 </p></details>
 ```
 
-
-
-
-
-
-
-
-
 Even though the computation of the ``p``-value is simple, we can use the [HypothesisTests](https://juliastats.org/HypothesisTests.jl/stable/) package. When we run the test, it gives us the same results as we computed.
 
 ```@example glm
@@ -125,12 +106,6 @@ using HypothesisTests
 OneSampleTTest(xs)
 ```
 
-
-
-
-
-
-
 ## Theory of generalized linear models
 
 The statistical approach to linear regression is different from the one from machine learning. It also assumes a linear prediction function:
@@ -153,11 +128,6 @@ Since the density is the derivative of the distribution function, the term ``f(y
 
 is often maximized. Since the logarithm is an increasing function, these two formulas are equivalent.
 
-
-
-
-
-
 #### Case 1: Linear regression
 
 The first case considers ``g(z)=z`` to be the identity function and ``y\mid x`` with the [normal distribution](https://en.wikipedia.org/wiki/Normal_distribution) ``N(\mu_i, \sigma^2)``. Then
@@ -174,15 +144,12 @@ and, therefore, we need the solve the following optimization problem:
 
 Since we maximize with respect to ``w``, most terms behave like constants, and this optimization problem is equivalent to
 
-
 ```math
 \operatorname{minimize}_w\qquad \sum_{i=1}^n (y_i - w^\top x_i)^2.
 ```
 
 This is precisely linear regression as derived in the previous lectures.
 
-
-
 #### Case 2: Logistic regression
 
 The second case considers ``g(z)=\log z`` to be the logarithm function and ``y\mid x`` with the [Poisson distribution](https://en.wikipedia.org/wiki/Poisson_distribution) ``Po(\lambda)``. The inverse function to ``g`` is ``g^{-1}(z)=e^z``. Since the Poisson distribution has non-negative discrete values with probabilities ``\mathbb P(Y=k) = \frac{1}{k!}\lambda^ke^{-\lambda}``, labels ``y_i`` must also be non-negative integers. The same formula for the conditional expectation as before yields:
@@ -205,7 +172,6 @@ By using the formula for ``\lambda_i`` and getting rid of constants, we transfor
 
 This function is similar to the one derived for logistic regression.
 
-
 ## Linear models
 
 We will use the [Employment and Wages in Spain](https://vincentarelbundock.github.io/Rdatasets/doc/plm/Snmesp.html) dataset because it is slightly larger than the iris dataset. It contains 5904 observations of wages from 738 companies in Spain from 1983 to 1990. We will estimate the dependence of wages on other factors such as employment or cash flow. We first load the dataset and transform the original log-wages into non-normalized wages. We use base ``2`` to obtain relatively small numbers.
@@ -241,22 +207,18 @@ model = lm(@formula(W ~ 1 + N + Y + I + K + F), wages)
 
 The table shows the parameter values and their confidence intervals. Besides that, it also tests the null hypothesis ``H_0: w_j = 0`` whether some of the regression coefficients can be omitted. The ``t`` statistics is in column `t`, while its ``p``-value in column `Pr(>|t|)`. The next exercise checks whether we can achieve the same results with fewer features.
 
-
-
-
-
-
-
 ```@raw html
 <div class="admonition is-category-exercise">
 <header class="admonition-header">Exercise:</header>
 <div class="admonition-body">
 ```
+
 Check that the solution computed by hand and by `lm` are the same.
 
 Then remove the feature with the highest ``p``-value and observe whether there was any performance drop. The performance is usually evaluated by the [coeffient of determination](https://en.wikipedia.org/wiki/Coefficient_of_determination) denoted by ``R^2\in[0,1]``. Its higher values indicate a better model.
 
 **Hint**: Use functions `coef` and `r2`.
+
 ```@raw html
 </div></div>
 <details class = "solution-body">
@@ -287,13 +249,6 @@ Since we observe only a small performance drop, we could omit this feature witho
 </p></details>
 ```
 
-
-
-
-
-
-
-
 The core assumption of this approach is that ``y`` follows the normal distribution. We use the `predict` function for predictions and then use the `plot_histogram` function written earlier to plot the histogram and a density of the normal distribution. For the normal distribution, we need to specify the correct mean and variance.
 
 ```@example glm
@@ -316,8 +271,6 @@ test_normality = ExactOneSampleKSTest(y_hat, Normal(mean(y_hat), std(y_hat)))
 
 The result is expected. The ``p``-value is close to ``1\%``, which means that we reject the null hypothesis that the data follow the normal distribution even though it is not entirely far away.
 
-
-
 ## Generalized linear models
 
 While the linear models do not transform the labels, the generalized models transform them by the link function. Moreover, they allow choosing other than the normal distribution for labels. Therefore, we need to specify the link function ``g`` and the distribution of ``y \mid x``.
@@ -328,21 +281,16 @@ We repeat the same example with the link function ``g(z) = \sqrt{z}`` and the [i
 model = glm(@formula(W ~ 1 + N + Y + I + K + F), wages, InverseGaussian(), SqrtLink())
 ```
 
-
-
-
-
-
-
-
 The following exercise plots the predictions for the generalized linear model.
 
 ```@raw html
 <div class="admonition is-category-exercise">
 <header class="admonition-header">Exercise:</header>
 <div class="admonition-body">
 ```
+
 Create the scatter plot of predictions and labels. Do not use the `predict` function.
+
 ```@raw html
 </div></div>
 <details class = "solution-body">
@@ -371,7 +319,6 @@ scatter(y, y_hat;
 savefig("glm_predict.svg")
 ```
 
-
 ```@raw html
 </p></details>
 ```