update of notes

This commit is contained in:
Morten Hjorth-Jensen
2023-09-19 09:55:08 +02:00
parent 752365eafb
commit 3ded8d6726
11 changed files with 187 additions and 162 deletions
+18 -8
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@@ -55,7 +55,7 @@ const thebe_selector_output = ".output, .cell_output"
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@@ -284,7 +284,17 @@ const thebe_selector_output = ".output, .cell_output"
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Week 37: Statitsitcal interpretations and Resampling Methods
Week 37: Statistical interpretations and Resampling Methods
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Exercises week 38
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Week 38: Logistic Regression and Optimization
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@@ -649,7 +659,7 @@ which polynomial fits the data best.</p>
</div>
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<div class="cell_output docutils container">
<div class="output stderr highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>/var/folders/td/3yk470mj5p931p9dtkk0y6jw0000gn/T/ipykernel_7740/39730396.py:11: MatplotlibDeprecationWarning: Calling gca() with keyword arguments was deprecated in Matplotlib 3.4. Starting two minor releases later, gca() will take no keyword arguments. The gca() function should only be used to get the current axes, or if no axes exist, create new axes with default keyword arguments. To create a new axes with non-default arguments, use plt.axes() or plt.subplot().
<div class="output stderr highlight-myst-ansi notranslate"><div class="highlight"><pre><span></span>/var/folders/td/3yk470mj5p931p9dtkk0y6jw0000gn/T/ipykernel_20604/39730396.py:11: MatplotlibDeprecationWarning: Calling gca() with keyword arguments was deprecated in Matplotlib 3.4. Starting two minor releases later, gca() will take no keyword arguments. The gca() function should only be used to get the current axes, or if no axes exist, create new axes with default keyword arguments. To create a new axes with non-default arguments, use plt.axes() or plt.subplot().
ax = fig.gca(projection=&#39;3d&#39;)
</pre></div>
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@@ -805,17 +815,17 @@ term which measures the deviation from the true data and the mean value of the m
That is, show that</p>
<div class="math notranslate nohighlight">
\[
\mathbb{E}\left[(\boldsymbol{y}-\boldsymbol{\tilde{y}})^2\right]=(\mathrm{Bias}[\tilde{y}])^2+\mathrm{var}[\tilde{f}]+\sigma^2,
\mathbb{E}\left[(\boldsymbol{y}-\boldsymbol{\tilde{y}})^2\right]=\mathrm{Bias}[y]+\mathrm{var}[\tilde{y}]+\sigma^2,
\]</div>
<p>with</p>
<div class="math notranslate nohighlight">
\[
(\mathrm{Bias}[\tilde{y}])^2=\left(\boldsymbol{y}-\mathbb{E}\left[\boldsymbol{\tilde{y}}\right]\right)^2,
\mathrm{Bias}[y]=\mathbb{E}\left[\left(\boldsymbol{y}-\mathbb{E}\left[\boldsymbol{\tilde{y}}\right]\right)^2\right],
\]</div>
<p>and</p>
<div class="math notranslate nohighlight">
\[
\mathrm{var}[\tilde{f}]=\frac{1}{n}\sum_i(\tilde{y}_i-\mathbb{E}\left[\boldsymbol{\tilde{y}}\right])^2.
\mathrm{var}[\tilde{y}]=\frac{1}{n}\sum_i(\tilde{y}_i-\mathbb{E}\left[\boldsymbol{\tilde{y}}\right])^2.
\]</div>
<p>The answer to this exercise should be included in the theory part of the report. This exercise is also part of the weekly exercises of week 37.
Explain what the terms mean and discuss their interpretations.</p>
@@ -1008,11 +1018,11 @@ of code developers and contributors keeps increasing.</p>
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