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presentation.Rmd

+---
+title: "Prior and Bayse's rule"
+author: "Laurent Modolo"
+date: "`r Sys.Date()`"
+output:
+  beamer_presentation:
+    df_print: tibble
+    fig_caption: no
+    highlight: tango
+    latex_engine: xelatex
+    slide_level: 1
+    theme: metropolis
+  ioslides_presentation:
+    highlight: tango
+  slidy_presentation:
+    highlight: tango
+classoption: aspectratio=169
+---
+
+```{r setup, include=FALSE}
+knitr::opts_chunk$set(echo = FALSE)
+# Load packages
+library(bayesrules)
+library(tidyverse)
+library(janitor)
+
+# Import article data
+data(fake_news)
+```
+
+# Bayes Rules
+
+\begin{center}
+\begin{columns}
+\column{0.5\textwidth}
+
+\includegraphics[width=\textwidth]{img/book_cover.jpeg}
+
+\column{0.5\textwidth}
+
+\href{https://www.bayesrulesbook.com/}{https://www.bayesrulesbook.com/}
+
+\end{columns}
+\end{center}
+
+# Prior probability model
+
+In 1996, Gary Kasparov won three, drew two, and lost one game against the IBM supercomputer Deep Blue.
+Kasparov and Deep Blue were to meet again for in 1997. Let $\pi$ denote Kasparov’s chances of winning any particular game in the re-match.
+
+Given the complexity of chess, machines, and humans, $\pi$ is unknown and can vary or fluctuate over time. Or, in short, $\pi$ is a **random variable**.
+
+To analyse $\pi$ we will start with a prior model which
+
+1. identifies what values $\pi$ can take
+2. assigns a prior weight or probability to each
+3. these probabilities sum to 1
+
+\begin{center}
+\begin{tabular}{ c c c c c }
+ $\pi$ & 0.2 & 0.5 & 0.8 & Total \\ 
+ $f(\pi)$ & 0.10 & 0.25 & 0.65 & 1 
+\end{tabular}
+\end{center}
+
+# Kasparov’s skill level relative to that of Deep Blue
+
+Data $Y$ is the number of the six games in the 1997 re-match that Kasparov wins. Since the chess match outcome isn’t predetermined, $Y$ is a **random variable** that can take any value in $\{0,1,\dots, 6\}$
+
+$Y$ inherently depends upon Kasparov’s win probability $\pi$.
+
+- If $\pi = 0.8$  Kasparov’s victories $Y$ will tend to be high
+- If $\pi = 0.2$  Kasparov’s victories $Y$ will tend to be low
+
+$Y$ *depends upon* or is *conditioned upon* the value of $\pi$
+
+$$f(y|\pi)=P(Y=y|\pi)$$
+
+# Binomial model
+
+1. the outcome of any one game doesn’t influence the outcome of another
+2. Kasparov has an equal probability, $\pi$, of winning any game in the match
+
+This is a common framework in statistical analysis, one which can be represented by the Binomial model.
+
+$$Y|\pi \sim Bin(6,\pi)$$
+In the end, Kasparov only won one of the six games against Deep Blue in 1997
+
+# Binomial model
+
+The **pmf** of a $Bin(6, \pi)$ model is plotted for each possible value of $\pi \in \{0.2,0.5,0.8\}$. The masses marked by the black lines correspond to the eventual observed data, $Y=1$ win.
+
+![Binom chess](img/binom-chess-1.png)
+
+# Binomial likelihood function
+
+The likelihood function $L(\pi|y=1)$ of observing $Y=1$ win in six games for any win probability $\pi \in \{0.2,0.5,0.8\}$. 
+
+\begin{center}
+\includegraphics[width=0.6\textwidth]{img/binom-chess-like-1.png}
+\end{center}
+
+# Probability mass functions vs likelihood functions
+
+When $\pi$ is known, the conditional **pmf** $f(⋅|\pi)$ allows us to compare the probabilities of different possible values of data $Y$ occurring with $\pi$ :
+
+$$f(y_1|\pi) \text{ vs } f(y_2|\pi)$$
+
+When $Y=y$ is known, the **likelihood function** $L(⋅|y)=f(y|⋅)$ allows us to compare the relative likelihood of observing data $y$ under different possible values of $\pi$ :
+
+$$ L(\pi_1|y) \text{ vs } L(\pi_2|y)$$
+
+Thus, $L(⋅|y)$ provides the tool we need to evaluate the relative compatibility of data $Y=y$ with various $\pi$ values.
+
+
+# Computing the posterior probability of $\pi$ given $y$
+
+We want to balance this prior and likelihood information.
+
+$$f(\pi | y) \propto f(\pi)L(\pi|y)$$
+$L(\pi|y)$ is not a probability so we need to add a **normalizing constant**
+
+# Baye's rule
+
+Bayes’ Rule, requires three pieces of information: the **prior**, **likelihood**, and a **normalizing constant**.
+
+
+$$f(\pi | y) = \frac{f(\pi)L(\pi|y)}{\text{normalizing constant}} = \frac{P(\pi \cap y)}{\text{normalizing constant}}$$
+
+$P(\pi \cap y)$ is, the joint probability of observing both $\pi$ and $y$
+
+$$\text{posterior} = \frac{\text{prior} ⋅ \text{likelihood}}{\text{normalizing constant}}$$
+
+# Normalizing constant
+
+We must determine the **total probability** that Kasparov would win $Y=1$ game across all possible win probabilities $\pi$, $f(y=1)$.
+
+$$f(y = 1) = \sum_{\pi \in \{0.2,0.5,0.8\}} L(\pi | y=1) f(\pi)$$
+
+# Porterior probability model
+
+$$f(\pi | y=1) = \frac{f(\pi)L(\pi|y=1)}{f(y = 1)} \;\; \text{ for } \pi \in \{0.2,0.5,0.8\}$$
+
+![posterio_likelhood_posterio_chess](img/chesssummary-1.png)
+
+# Proportionality
+
+$$f(\pi | y) = \frac{f(\pi)L(\pi|y)}{f(y)} \propto f(\pi)L(\pi|y)$$
+
+![ch2scaled-1.png](img/ch2scaled-1.png)
+
+# Beta-Binomial model
+
+In the previous example our prior was an over-simplification:
+
+- the possible value of $\pi$ are continuous between $[0,1]$
+- We can use the $Beta$ distribution with parameter $\alpha$  and $\beta$ to model our prior
+
+**Alison Bechdel’s 1985 comic rule:**
+only see a movie if it satisfies the following three rules (Bechdel 1986):
+
+- the movie has to have at least two women in it
+- these two women talk to each other; and
+- they talk about something besides a man.
+
+What percentage of all recent movies do you think pass the Bechdel test? Is it closer to 10%, 50%, 80%, or 100%?
+
+# Three prior models for the proportion of films that pass the Bechdel test.
+
+![ch4-bechdel-priors-1.png](img/ch4-bechdel-priors-1.png)
+
+$$Y | \pi \sim \text{Bin}(n, \pi)$$
+$$\pi \sim \text{Beta}(\alpha, \beta)$$
+
+# Different priors, different posteriors
+
+We sample $n=20$ and $Y=9$ (45%) pass the Bechdel test.
+
+![unnamed-chunk-134-1.png](img/unnamed-chunk-134-1.png)
+
+likelihood functions are not pdfs, scaling the likelihood function here is for simplifying the visual comparisons between the prior vs data evidence about $\pi$.
+
+# Different priors, different posteriors
+
+We sample $n=20$ and $Y=9$ (45%) pass the Bechdel test.
+
+![bechdel-post-ch4-1.png](img/bechdel-post-ch4-1.png)
+
+likelihood functions are not pdfs, scaling the likelihood function here is for simplifying the visual comparisons between the prior vs data evidence about $\pi$.
+
+# Different data, different posteriors
+
+- $n=13$ movies from the year 1991, among which $Y=6$ (about 46%) pass the Bechdel test
+- $n=63$ movies from 2000, among which $Y=29$ (about 46%) pass the Bechdel test
+- $n=99$ movies from 2013, among which $Y=46$ (about 46%) pass the Bechdel test
+
+![unnamed-chunk-138-1.png](img/unnamed-chunk-138-1.png)
+
+# Different data, different posteriors
+
+- $n=13$ movies from the year 1991, among which $Y=6$ (about 46%) pass the Bechdel test
+- $n=63$ movies from 2000, among which $Y=29$ (about 46%) pass the Bechdel test
+- $n=99$ movies from 2013, among which $Y=46$ (about 46%) pass the Bechdel test
+
+![bechdel-data-ch4-1.png](img/bechdel-data-ch4-1.png)
+
+# Different priors, different data, different posteriors
+
+\begin{center}
+\includegraphics[width=\textwidth]{img/bechdel-combined-ch4-1.png}
+\end{center}
+
+
+# Bayesian knowledge building
+
+\begin{center}
+\includegraphics[width=0.6\textwidth]{img/bayes_diagram.png}
+
+ A Bayesian knowledge-building diagram
+\end{center}

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