bayes sol

book/_build/html/_sources/chancevar.md

@@ -125,3 +125,5 @@ The range 470 to 530 is therefore the expected sum $\pm$ SE. The sum falls in th
 ```
 
 The normal approximation is justified because of the **central limit theorem**. The central limit theorem states that, when random draws are made replacement from a box, the distribution for the sum of these draws will approximate a normal distribution. This is true even if the contents of the box do not follow a normal curve as long as the number of draws is large.
+
+The central limit theorem is a big deal. 

book/_build/html/_sources/chancevary.ipynb

@@ -2,17 +2,21 @@
  "cells": [
   {
    "cell_type": "markdown",
-   "id": "d0b7dbc4",
+   "id": "70b05281",
    "metadata": {},
    "source": [
     "(chance)=\n",
-    "# Chance Error\n",
+    "# Chance Variability\n",
     "\n",
     "```{admonition} Important Readings\n",
     ":class: seealso\n",
     "- {cite}`freedman2007statistics`, Chapter 16, 17, 18\n",
     "```\n",
     "\n",
+    "In this section, we begin to marry data and probability. This is exciting because it provides the foundation for statistical inference. \n",
+    "\n",
+    "## Chance Errors\n",
+    "\n",
     "A fair coin is expected to land heads 50% of the time. This is expressed by the probability as a long-run frequency,\n",
     "\n",
     "$$\\mathbb{P}(H) = \\frac{ \\text{number of heads}} {\\text{number of tosses}}.$$\n",
@@ -148,7 +152,7 @@
   {
    "cell_type": "code",
    "execution_count": 7,
-   "id": "eb53806e",
+   "id": "39c75ecb",
    "metadata": {
     "tags": [
      "remove-input"
@@ -582,7 +586,7 @@
   },
   {
    "cell_type": "markdown",
-   "id": "74372a1c",
+   "id": "d91040a8",
    "metadata": {},
    "source": [
     "Above, for a sufficiently large number of draws, we see a familiar bell-shaped curve. If the value of the sum is converted to standard units, this will approximate the standard normal curve. \n",
@@ -592,13 +596,14 @@
     "width: 90%\n",
     "name: standardizedprobhistogram\n",
     "---\n",
+    "Probability histogram for the sum of 100 draws from a box with 9 zeros and a single one.\n",
     "```"
    ]
   },
   {
    "cell_type": "code",
    "execution_count": null,
-   "id": "e26dbcfe",
+   "id": "b0a97a52",
    "metadata": {},
    "outputs": [],
    "source": []

book/_build/html/_sources/solutions.md

@@ -185,7 +185,7 @@ A single row will contain the incomes for two partnered individuals, but an $x,y
 ```{solution-end}
 ```
 
-## [Probability](probability)
+## [Probability I](probability)
 
 ```{solution-start} conditional
 :class: dropdown
@@ -203,5 +203,89 @@ $\mathbb{P}(A \mid B) + \mathbb{P}(\text{not }A \mid B)=1$ for any $A,B$.
 Binomial coefficients tell you how many ways you can select $k$ items from a list of $n$, but the $k$ items are not ordered. 
 
 The president/VP pair is ordered and the co-preseident pair is not. Only the co-president pairings are counted by $\binom{3}{2}$. There will be twice as many president/VP pairings because (Alice, Bob) and (Bob, Alice) are considered distinct. 
+```{solution-end}
+```
+
+## [Probability II](bayes)
+
+
+```{solution-start} boxes
+:class: dropdown
+```
+
+1. $\mathbb{P}(H) = 0.5$
+2. $\mathbb{P}(\text{box with two Ts} \mid H) = 0$ because the box must have an $H$ to draw an $H$. 
+3. $$\mathbb{P}(\text{box with two Hs} \mid H) = \frac{ \mathbb{P}(H \mid \text{box with two Hs}) \mathbb{P}(\text{box with two Hs})}{\mathbb{P}(H)} $$
+
+$$= \frac{1\times 0.25}{0.5} = 0.25$$
+
+4. To find $\mathbb{P}(\text{H on second draw} \mid \text{H on first draw})$, the subtlety is that these are not independent. The unconditional probability is $\mathbb{P}(\text{H on second draw}) = 0.5$. This allows for the possibility of a box with two $T$s. An $H$ on the first draw reveals that the box does not have two $T$s, which will push our probability up. The intuititive answer of 0.5 is therefore *wrong*. For a similar problem, see also the famously difficult [boy girl paradox](https://en.wikipedia.org/wiki/Boy_or_girl_paradox).
+
+
+**4 - Solution 1**
+The first draw reveals there is an $H$ in the box. By part 3, we know there is a 0.5 probability the box contains two $H$s. The probability of an $H$ on the second draw is  
+
+
+
+$$ \underbrace{\frac{1}{2}}_{a} \cdot \overbrace{1}^{b} + \underbrace{\frac{1}{2}}_{c} \cdot \overbrace{\frac{1}{2}}^{b} = \frac{3}{4}.$$
+
+
+| Term | Probability of ... given $H$ on first draw               |
+|------|----------------------------------------------------------|
+| $a$  | box with two $H$s                                        |
+| $b$  | $H$ given two two-$H$ box                                |
+| $c$  | box with one $H$ and one $T$                             |
+| $d$  | $H$ given two one-$H$-one-$T$ box                        | 
+
+Each ticket is marked $H$ or $T$ independently and with equal chance, so $d$ is $\frac{1}{2}$. 
+
+
+**4 - Solution 2 (my favorite)**
+
+This solution is remarkably similar to the previous, but with a different interpretation. The first draw reveals there is an $H$ in the box. Half of the time you will draw the same ticket on your second draw. The other half of the time you draw the other ticket. 
+
+The conditional probability of an $H$ on the second draw given an $H$ on the first draw can be expanded as the sum of the probability of *the same ticket and $H$* and the probability of *a different ticket and $H$*, all conditional on $H$ on the first draw. This is
+
+$$ \underbrace{\frac{1}{2}}_{\alpha} \cdot \overbrace{1}^{\beta} + \underbrace{\frac{1}{2}}_{\gamma} \cdot \overbrace{\frac{1}{2}}^{\delta} = \frac{3}{4}.$$
+
+| Term     | Probability of ... given $H$ on first draw             |
+|----------|--------------------------------------------------------|
+| $\alpha$ | the same ticket (an $H$).                              |
+| $\beta$  | $H$ given the same ticket.                             |
+| $\gamma$ | drawing the other ticket (which can be $H$ or $T$)     |
+| $\delta$ | $H$ given the other ticket                             |
+
+
+Each ticket is marked $H$ or $T$ independently and with equal chance, so $\delta$ is $\frac{1}{2}$. 
+
+**4 - Solution 3 (the worst)**
+
+Now we attempt to solve this by applying Bayes Theorem directly. It's important to get the notation clear. Let $H_2$ denote getting an $H$ on the second draw and $H_1$ is an $H$ on the first. We want to find $\mathbb{P}(H_2 \mid H_1)$. Bayes Theorem tells us
+
+$$\mathbb{P}(H_2 \mid H_1) = \frac{\mathbb{P}(H_1 \mid H_2) \mathbb{P}(H_2)}{\mathbb{P}(H_1)}.$$
+
+This doesn't do much to simplify the problem because $\mathbb{P}(H_1 \mid H_2)$ is no easier to solve for and we'll end up repeating calculations from Solution 1 or Solution 2 to find $\mathbb{P}(H_1 \mid H_2) \mathbb{P}(H_2) = \mathbb{P}(H_2 \text{ and } H_1).$ It's tempting to write this as $\mathbb{P}(H_1) \times \mathbb{P}(H_2)$, but these are dependent events so we can't. A tree is helpful. 
+
+For a single draw. 
+```{figure} images/tikz/HTrandomboxTree.svg
+:width: 70%
+:name: HTrandomboxTree2
+```
+
+For two draws. 
+```{figure} images/tikz/HTrandomboxTree2Draws.svg
+:width: 70%
+:name: HTrandomboxTree2Draws
+```
+
+Therefore the probability of $H_1$ and $H_2$ is $0.5\times 0.25 + 0.25 \times 1 = \frac{3}{8}$. Plugging this into the formula for $\mathbb{P}(H_2 \mid H_1)$, we get 
+
+$$\dfrac{\frac{3}{8}}{\frac{1}{2}} = \frac{3}{4}.$$ 
+
+It follows that $\mathbb{P}(H_1 \mid H_2)$ is also $\frac{3}{4}$ because the unconditional probabilities of $H_1$ and $H_2$ are both $\frac{1}{2}$.
+
+[Here is a Google Sheets simulation](https://docs.google.com/spreadsheets/d/1xlryzoPWZ05K4SeHVzZHiCC4yQ-XhLVaZChy-gYoWRc/edit?usp=sharing).
+
+
 ```{solution-end}
 ```

book/_build/html/bayes.html

@@ -77,7 +77,7 @@
     <link rel="shortcut icon" href="_static/norm_favicon.ico"/>
     <link rel="index" title="Index" href="genindex.html" />
     <link rel="search" title="Search" href="search.html" />
-    <link rel="next" title="Chance Error" href="chancevar.html" />
+    <link rel="next" title="Chance Variability" href="chancevary.html" />
     <link rel="prev" title="Probability" href="probability.html" />
   <meta name="viewport" content="width=device-width, initial-scale=1"/>
   <meta name="docsearch:language" content="en"/>
@@ -195,7 +195,7 @@
 <li class="toctree-l1"><a class="reference internal" href="regression.html">Regression</a></li>
 <li class="toctree-l1"><a class="reference internal" href="probability.html">Probability</a></li>
 <li class="toctree-l1 current active"><a class="current reference internal" href="#">Probability II</a></li>
-<li class="toctree-l1"><a class="reference internal" href="chancevar.html">Chance Error</a></li>
+<li class="toctree-l1"><a class="reference internal" href="chancevary.html">Chance Variability</a></li>
 <li class="toctree-l1"><a class="reference internal" href="bibliography.html">Bibliography</a></li>
 </ul>
 <p aria-level="2" class="caption" role="heading"><span class="caption-text">Google Sheets (optional)</span></p>
@@ -692,11 +692,11 @@ <h2>Exercises<a class="headerlink" href="#exercises" title="Permalink to this he
       </div>
     </a>
     <a class="right-next"
-       href="chancevar.html"
+       href="chancevary.html"
        title="next page">
       <div class="prev-next-info">
         <p class="prev-next-subtitle">next</p>
-        <p class="prev-next-title">Chance Error</p>
+        <p class="prev-next-title">Chance Variability</p>
       </div>
       <i class="fa-solid fa-angle-right"></i>
     </a>

book/_build/html/centerspread.html

@@ -195,7 +195,7 @@
 <li class="toctree-l1"><a class="reference internal" href="regression.html">Regression</a></li>
 <li class="toctree-l1"><a class="reference internal" href="probability.html">Probability</a></li>
 <li class="toctree-l1"><a class="reference internal" href="bayes.html">Probability II</a></li>
-<li class="toctree-l1"><a class="reference internal" href="chancevar.html">Chance Error</a></li>
+<li class="toctree-l1"><a class="reference internal" href="chancevary.html">Chance Variability</a></li>
 <li class="toctree-l1"><a class="reference internal" href="bibliography.html">Bibliography</a></li>
 </ul>
 <p aria-level="2" class="caption" role="heading"><span class="caption-text">Google Sheets (optional)</span></p>

book/_build/html/chancevar.html

@@ -77,8 +77,6 @@
     <link rel="shortcut icon" href="_static/norm_favicon.ico"/>
     <link rel="index" title="Index" href="genindex.html" />
     <link rel="search" title="Search" href="search.html" />
-    <link rel="next" title="Bibliography" href="bibliography.html" />
-    <link rel="prev" title="Probability II" href="bayes.html" />
   <meta name="viewport" content="width=device-width, initial-scale=1"/>
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   </head>
@@ -181,7 +179,7 @@
             </li>
         </ul>
         <p aria-level="2" class="caption" role="heading"><span class="caption-text">Main Text</span></p>
-<ul class="current nav bd-sidenav">
+<ul class="nav bd-sidenav">
 <li class="toctree-l1"><a class="reference internal" href="gettingstarted.html">Getting Started</a></li>
 <li class="toctree-l1"><a class="reference internal" href="experiments.html">Experiments</a></li>
 
@@ -195,7 +193,7 @@
 <li class="toctree-l1"><a class="reference internal" href="regression.html">Regression</a></li>
 <li class="toctree-l1"><a class="reference internal" href="probability.html">Probability</a></li>
 <li class="toctree-l1"><a class="reference internal" href="bayes.html">Probability II</a></li>
-<li class="toctree-l1 current active"><a class="current reference internal" href="#">Chance Error</a></li>
+<li class="toctree-l1"><a class="reference internal" href="chancevary.html">Chance Variability</a></li>
 <li class="toctree-l1"><a class="reference internal" href="bibliography.html">Bibliography</a></li>
 </ul>
 <p aria-level="2" class="caption" role="heading"><span class="caption-text">Google Sheets (optional)</span></p>
@@ -465,7 +463,7 @@ <h2> Contents </h2>
 <figure class="align-default" id="chanceerrorhist">
 <a class="reference internal image-reference" href="_images/chanceErrorHist.svg"><img alt="_images/chanceErrorHist.svg" src="_images/chanceErrorHist.svg" width="90%" /></a>
 <figcaption>
-<p><span class="caption-number">Fig. 44 </span><span class="caption-text">This shows the results from 10 and 100 tosses of a coin, repeated 1000 times each. The histogram for 100 tosses is more spread out in absolute terms. But, in percentage terms, more of the 1000 trials yield close to 50% heads.</span><a class="headerlink" href="#chanceerrorhist" title="Permalink to this image">#</a></p>
+<p><span class="caption-text">This shows the results from 10 and 100 tosses of a coin, repeated 1000 times each. The histogram for 100 tosses is more spread out in absolute terms. But, in percentage terms, more of the 1000 trials yield close to 50% heads.</span><a class="headerlink" href="#chanceerrorhist" title="Permalink to this image">#</a></p>
 </figcaption>
 </figure>
 <p>In absolute terms, the chance error will increase, but more slowly than the number of tosses. In fact, if the number of tosses increases 2x, the chance error increases only by a factor of <span class="math notranslate nohighlight">\(\sqrt{2}\)</span>. If the number of tosses increases 100x, the chance error increases only by a factor of 10.</p>
@@ -475,7 +473,7 @@ <h2>Sums from Boxes<a class="headerlink" href="#sums-from-boxes" title="Permalin
 <figure class="align-default" id="box01">
 <a class="reference internal image-reference" href="_images/box01.svg"><img alt="_images/box01.svg" src="_images/box01.svg" width="22%" /></a>
 <figcaption>
-<p><span class="caption-number">Fig. 45 </span><span class="caption-text">This is a 0-1 box representing a fair coin flip.</span><a class="headerlink" href="#box01" title="Permalink to this image">#</a></p>
+<p><span class="caption-text">This is a 0-1 box representing a fair coin flip.</span><a class="headerlink" href="#box01" title="Permalink to this image">#</a></p>
 </figcaption>
 </figure>
 <p>There is chance variability based on the <em>number of draws</em> and the actual chance inherent in a process. Suppose 100 draws are made with replacement from one of the boxes below. You will win $1 if you can guess the sum within 10.</p>
@@ -544,6 +542,7 @@ <h2>Normal Approximation<a class="headerlink" href="#normal-approximation" title
 <p class="sd-card-text">The range 470 to 530 is therefore the expected sum <span class="math notranslate nohighlight">\(\pm\)</span> SE. The sum falls in this range 68% of the time.</p>
 </div>
 </details><p>The normal approximation is justified because of the <strong>central limit theorem</strong>. The central limit theorem states that, when random draws are made replacement from a box, the distribution for the sum of these draws will approximate a normal distribution. This is true even if the contents of the box do not follow a normal curve as long as the number of draws is large.</p>
+<p>The central limit theorem is a big deal.</p>
 <hr class="footnotes docutils" />
 <aside class="footnote brackets" id="id3" role="note">
 <span class="label"><span class="fn-bracket">[</span><a role="doc-backlink" href="#id2">1</a><span class="fn-bracket">]</span></span>
@@ -582,24 +581,6 @@ <h2>Normal Approximation<a class="headerlink" href="#normal-approximation" title
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