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session 2: conditional probability

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lectures_beamer/S1E2_Inference.tex
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\section{Conditional probability}
\interludeframe{\textbf{\large (Conditional) Probability}}
% Plausible reasoning
% Logic game: lights on/off
% Probability as logic
% Contingency table
% Probability
% Conditional probability
% Factorization of the joint probability
% Marginal probability
% Product rule
% Frequentist view: contingency table
% Product rule
% Conditional probability
% TODO: Propagation errors to measurement as marginalization
% TODO: Logic game: lights on/off
\input{include/probability.tex}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
......@@ -67,18 +69,13 @@
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% 100 min
\section{Gaussian Distribution}
\interludeframe{\textbf{\large Gaussian (Normal) distribution}}
% Distribution of Jointly Gaussian variables
% Conditional distribution of Gaussian variables
% TODO: 2D example: visual
\input{include/mvnormal_distribution_conditioning.tex}
\section{Update step}
\interludeframe{\textbf{\large Update step}}
% Noisy iterated map with measurement
% TODO: bayesian fomrulation of the inference problem
% Distribution of Jointly Gaussian variables
% Conditional distribution of Gaussian variables
% Update step
% Kalman filter
\input{include/update_step.tex}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
......@@ -101,10 +98,10 @@
% 60 min
\section{Parameter estimation}
\interludeframe{\textbf{\large Parameter estimation}}
% Online parameter update
% Augmented state
% Maximum likelihood
% Batch regression of historical data: functional analysis, etc.
% TODO: Online parameter update
% TODO: Augmented state
% TODO: Maximum likelihood
% TODO: Batch regression of historical data: functional analysis, etc.
% \input{include/parameter_estimation.tex}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
......
lectures_beamer/include/mvnormal_distribution_conditioning.tex deleted 100644 → 0
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\begin{frame}{Distribution of Jointly Gaussian variables}
\[
\gaussian(\bm{x}| \bm{m}, \tens{P}) = \frac{1}{\sqrt{(2\pi)^n\det{\tens{P}}}}
\exp{-\frac{1}{2}\left(\bm{x} - \bm{m}\right)^\trp \tens{P}^{-1}\left(\bm{x} - \bm{m}\right)}
\]
\begin{columns}[t]
\column{0.5\textwidth}
\[
\begin{bmatrix} \bm{x} \\ \bm{y}\end{bmatrix}
\sim \gaussian\left(\begin{bmatrix}\bm{a} \\ \bm{b}\end{bmatrix},
\begin{bmatrix}\tens{A} & \tens{C} \\ \tens{C}^\trp & \tens{B}\end{bmatrix}\right)
\]
\begin{align*}
\bm{x} &\sim \gaussian(\bm{a},\tens{A})\\
\bm{y} &\sim \gaussian(\bm{b},\tens{B})\\
\bm{x} | \bm{y} &\sim \gaussian(\bm{a} + \tens{C}\tens{B}^{-1}(\bm{y} - \bm{b}),
\tens{A} - \tens{C}\tens{B}^{-1}\tens{C}^\trp)\\
\bm{y} | \bm{x} &\sim \gaussian(\bm{b} + \tens{C}^\trp\tens{A}^{-1}(\bm{x} - \bm{a}),
\tens{B} - \tens{C}^\trp\tens{A}^{-1}\tens{C})
\end{align*}
\column{0.5\textwidth}
\begin{align*}
\bm{x} &\sim \gaussian(\bm{m}, \tens{P})\\
\bm{y} | \bm{x} &\sim \gaussian(\tens{H}\bm{x} + \bm{u}, \tens{R})\\
\end{align*}
\begin{align*}
\begin{bmatrix} \bm{x} \\ \bm{y}\end{bmatrix} &
\sim \gaussian\left(\begin{bmatrix}\bm{m} \\ \tens{H}\bm{x} + \bm{u}\end{bmatrix},
\begin{bmatrix}\tens{P} & \tens{P}\tens{H}^\trp \\ \tens{H}\tens{P} & \tens{H}\tens{P}\tens{H}^\trp + \tens{R}\end{bmatrix}\right)\\
\bm{y} | \bm{x} &\sim \gaussian(\tens{H}\bm{x} + \bm{u}, \tens{H}\tens{P}\tens{H}^\trp + \tens{R})
\end{align*}
\end{columns}
\note<1>{
- Schur complement
- An interpretation of the conditional mean?
- Do exercise 3.5 from the book
}
\end{frame}
\ No newline at end of file
lectures_beamer/include/probability.tex
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\begin{frame}{Probability}
\includegraphics{Jaynes_Probability_Logic_of_Science.png}
\begin{columns}[c]
\column{0.5\textwidth}
\raggedright
The actual science of logic is conversant at present only with things either certain, impossible,
or entirely doubtful, none of which (fortunately) we have to reason on.
Therefore the true logic for this world is the calculus of Probabilities, which takes account of the
magnitude of the probability which is, or ought to be, in a reasonable man’s mind.
\\ \vspace{-1em}\hrulefill \\
James Clerk Maxwell (1850)
\column{0.5\textwidth}
\includegraphics{Jaynes_Probability_Logic_of_Science.png}
\end{columns}
Probability is a numerical between value assigned to a plausibility.
Impossibility is assigned the probability 0 and certainty is assigned the probability 1.
\note<1>{
We will see how Bayesian probability includes classical logic as a particular case and extends it to
include reasoning on plausible events.
}
\end{frame}
\begin{frame}{Propositional logic}
Works with propositions ($A$, $B$, etc.) that can be true or false.
\begin{align*}
A &\equiv \text{It rains at 10:00}\\
B &\equiv \text{We see a blue sky}
\end{align*}
\begin{columns}[t]
\begin{column}{0.75\textwidth}
Logic
\begin{itemize}
\item negation $\lnot{A}$: true if $A$ is false
\item conjunction $A \land B$: true if $A$ and $B$ are both true
\item disjunction $A \lor B$: true if any of (or both) $A$ or $B$ are true
\item implication $A \implies B$: says that $A \land \lnot{B}$ is false ($\lnot{A} \lor B$ is true)
\end{itemize}
\end{column}
\hspace{-2em}
\vrule
\hspace{2em}
\begin{column}{0.25\textwidth}
Boolean
\begin{itemize}
\item $\bar{A}$
\item $AB$
\item $A + B$
\item $A \implies B$
\end{itemize}
\end{column}
\end{columns}
\note<1>{
State several propositions and determine their truth value.
Does the value depend on the context?
}
\end{frame}
\begin{frame}{Implication (logical)}
$A \implies B$: says that $A\bar{B}$ is false ($\bar{A}+B$ is true).
Means \textbf{only that} $A = AB$ (the truth value of $A$ is the same as the truth value of $AB$).
\vspace{-1em}
\begin{table}
\begin{tabular}{cccc}
\toprule
A & $\implies$ & B \\
\midrule
T & T & T\\
T & F & F\\
F & T & T\\
F & T & F\\
\bottomrule
\end{tabular}
\end{table}
\begin{itemize}
\item All true propositions logically imply all other true propositions.
\item It doesn't mean: $B$ deducible from $A$.
This depends on the background information.
\item It doesn't mean: $B$ is caused by $A$.
Causal physical consequence can be effective only at a later time.
The rain at 10:00 is not the physical cause of the clouds before 10:00.
Implication doesn't follow the uncertain causal direction clouds\textrightarrow rain, but rather the
certain non-causal rain\textrightarrow clouds.
\end{itemize}
\end{frame}
\begin{frame}{Conditional probability}
\framesubtitle{Definition}
\[
\mathcal{p}(A | B)
\]
The plausibility (probability) of a proposition $A$ depends, in general, on whether
we known/assume/believe some other proposition $B$ is true.
Example:
\begin{align*}
A &\equiv \text{My english pronunciation is perfect}\\
B &\equiv \text{I grew up in Argentina}\\
C &\equiv \text{I grew up in England}\\
&\text{which is bigger? } \mathcal{p}(A|B),\, \mathcal{p}(A|C)
\end{align*}
\note<1>{
All probabilities are conditional probabilities, because there is always some assumptions we made or knowledge
we have that is relevant for the assignation.
Think of "The probability of a 6 in a fair die is $\frac{1}{6}$".
What's the background information?
}
\end{frame}
\begin{frame}{Product rule}
\framesubtitle{Plausibility (probability)}
To evaluate $\mathcal{p}(AB|C)$ we can proceed as follows:
\begin{columns}
\column{0.5\textwidth}
\begin{enumerate}
\item determine the probability of $A$ true given information $C$. $\quad \to \mathcal{p}(A|C)$
\item based on that, determine the probability of $B$ true. $\quad \to \mathcal{p}(B|AC)$
\end{enumerate}
\column{0.5\textwidth}
\begin{enumerate}
\item determine the probability of $B$ true given information $C$. $\quad \to \mathcal{p}(B|C)$
\item based on that, determine the probability of $A$ true. $\quad \to \mathcal{p}(A|BC)$
\end{enumerate}
\end{columns}
Both paths should give the same
\[
\mathcal{p}(AB|C) = \mathcal{p}(A|BC) \mathcal{p}(B|C) = \mathcal{p}(B|AC) \mathcal{p}(A|C)
\]
\note<1>{
$\mathcal{p}(AB|C)$ is called the joint probability of $A$ and $B$.
The probability that both propositions are true.
}
\end{frame}
\begin{frame}{Sum rule}
\framesubtitle{Plausibility (probability)}
\[
\mathcal{p}(A|C) + \mathcal{p}(\bar{A}|C) = 1 \;\rightarrow\; \sum_{i}{\mathcal{p}(A_i|C)} = 1
\;\rightarrow\; \int_{-\infty}^{\infty}{\mathcal{p}(A=a|C) \ud a} = 1
\]
States that the probability distributes totally over the plausible propositions.
Particular case of (using $B = \bar{A}$):
\[
\mathcal{p}(A + B | C) = \mathcal{p}(A|C) + \mathcal{p}(B|C) - \mathcal{p}(AB|C)
\]
\end{frame}
\begin{frame}{Sillogisms}
\only<1>{
\begin{figure}
\includegraphics{burglar.png}\\
\source{Luis Prade and Gan Khoon Lay at \url{https://thenounproject.com}}
\end{figure}
A dark night, a policeman walks down an apparently deserted street.
Suddenly burglar alarm, across the street a jewelry store with a broken window.
A masked man crawls out of the broken window, carrying a bag which turns out to be full of expensive jewelry.
The policeman doesn't hesitate at all in deciding that this man is dishonest.
Is guilt implied by evidence?
}
\only<2->{
Assume that $A \implies B$ is true.
\only<3>{$C \equiv A \implies B$}
\begin{columns}[t]
\begin{column}{0.5\textwidth}
Strong
\vspace{0.5em}
\begin{itemize}
\item if $A$ is true, then $B$ is true
\only<2>{\begin{align*}
A &\equiv \text{rain at 10:00}, \, B \equiv \text{clouds before 10:00}\\
A &\equiv \text{man is burglar}, \, B \equiv \text{robbed jewelry store}
\end{align*}}
\only<3>{\[
\mathcal{p}(B|AC) = \frac{\mathcal{p}(AB|C)}{\mathcal{p}(A|C)} = \frac{\mathcal{p}(A|C)}{\mathcal{p}(A|C)} = 1
\]}
\item if $B$ is false, then $A$ is false
\only<2>{\begin{align*}
\bar{B} &\equiv \text{no clouds before 10:00}, \, \bar{A} \equiv \text{no rain at 10:00}\\
\bar{B} &\equiv \text{didn't rob jewelry store}, \, \bar{A} \equiv \text{man isn't burglar}
\end{align*}}
\only<3>{\[
\mathcal{p}(A|\bar{B}C) = \frac{\overbrace{\mathcal{p}(A\bar{B}|C)}^{=0}}{\mathcal{p}(\bar{B}|C)} = 0
\]}
\end{itemize}
\end{column}
\hspace{-0.5em}
\vrule
\hspace{0.5em}
\begin{column}{0.5\textwidth}
Weak
\vspace{0.5em}
\begin{itemize}
\item if $B$ is true, then $A$ is more plausible
\only<3>{\begin{align*}
\mathcal{p}(A|BC) &\geq \mathcal{p}(A|C)\quad\mathcal{p}(\bar{A}|BC) \leq \mathcal{p}(\bar{A}|C)\\
\frac{\overbrace{\mathcal{p}(B|AC)}^{=1}}{\mathcal{p}(B|C)} &\geq 1
\end{align*}}
\only<2>{
Evidence doesn't prove $A$, verification of its consequence gives more confidence in $A$
\begin{align*}
B &\equiv \text{clouds before 10:00}, \, A \equiv \text{rain at 10:00}\\
B &\equiv \text{robbed jewelry store}, \, A \equiv \text{man is burglar}
\end{align*}}
\item if $A$ is false, then $B$ less plausible
\only<3>{\begin{align*}
\mathcal{p}(B|\bar{A}C) &\leq \mathcal{p}(B|C)\\
\frac{\mathcal{p}(\bar{A}|BC)}{\mathcal{p}(\bar{A}|C)} &\leq 1
\end{align*}}
\only<2>{
Evidence doesn't disprove $B$, one of the reasons eliminated, less confidence in $B$
\begin{align*}
\bar{A} &\equiv \text{no rain at 10:00}, \, B \equiv \text{clouds before 10:00}\\
\bar{A} &\equiv \text{man isn't burglar}, \, B \equiv \text{robbed jewelry store}
\end{align*}}
\end{itemize}
\end{column}
\end{columns}
}
\note<1>{
Is it a logical deduction?
Alternative hypothesis?
Is there validity in the reasoning?
}
\note<2>{
The implication does not imply causation or deduction.
The examples are just for illustration of the sillogisms.
}
\end{frame}
\begin{frame}{Waeker sillogism}
Assume that "if $A$ is true, then $B$ is more plausible" is true.
if $B$ is true, then $A$ more plausible.
Assume that if "a man is robbing a jewelry store" it is more plausible that he will be "found in the suspicious situation".
If he is actually found, then it is more plausible that he is robbing the jewelry.
We assume that
\[
\mathcal{p}(B | AZ) \geq \mathcal{p}(B | Z)
\]
then the product rule gives the result
\[
\mathcal{p}(A|BZ) = \frac{\mathcal{p}(B|AZ)}{\mathcal{p}(B|Z)}\mathcal{p}(A|Z) \geq \mathcal{p}(A|Z)
\]
\note<1>{
I used $Z$ for the context information, to avoid confusion with the $C$ used before.
}
\end{frame}
lectures_beamer/include/update_step.tex
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\begin{frame}{Bayesina inference}
Predicton step in bayesian form
Updat step in bayesian form
\end{frame}
\begin{frame}{Distribution of Jointly Gaussian variables}
\[
\gaussian(\bm{x}| \bm{m}, \tens{P}) = \frac{1}{\sqrt{(2\pi)^n\det{\tens{P}}}}
\exp{-\frac{1}{2}\left(\bm{x} - \bm{m}\right)^\trp \tens{P}^{-1}\left(\bm{x} - \bm{m}\right)}
\]
\begin{columns}[t]
\column{0.5\textwidth}
\[
\begin{bmatrix} \bm{x} \\ \bm{y}\end{bmatrix}
\sim \gaussian\left(\begin{bmatrix}\bm{a} \\ \bm{b}\end{bmatrix},
\begin{bmatrix}\tens{A} & \tens{C} \\ \tens{C}^\trp & \tens{B}\end{bmatrix}\right)
\]
\begin{align*}
\bm{x} &\sim \gaussian(\bm{a},\tens{A})\\
\bm{y} &\sim \gaussian(\bm{b},\tens{B})\\
\bm{x} | \bm{y} &\sim \gaussian(\bm{a} + \tens{C}\tens{B}^{-1}(\bm{y} - \bm{b}),
\tens{A} - \tens{C}\tens{B}^{-1}\tens{C}^\trp)\\
\bm{y} | \bm{x} &\sim \gaussian(\bm{b} + \tens{C}^\trp\tens{A}^{-1}(\bm{x} - \bm{a}),
\tens{B} - \tens{C}^\trp\tens{A}^{-1}\tens{C})
\end{align*}
\column{0.5\textwidth}
\begin{align*}
\bm{x} &\sim \gaussian(\bm{m}, \tens{P})\\
\bm{y} | \bm{x} &\sim \gaussian(\tens{H}\bm{x} + \bm{u}, \tens{R})\\
\end{align*}
\begin{align*}
\begin{bmatrix} \bm{x} \\ \bm{y}\end{bmatrix} &
\sim \gaussian\left(\begin{bmatrix}\bm{m} \\ \tens{H}\bm{x} + \bm{u}\end{bmatrix},
\begin{bmatrix}\tens{P} & \tens{P}\tens{H}^\trp \\ \tens{H}\tens{P} & \tens{H}\tens{P}\tens{H}^\trp + \tens{R}\end{bmatrix}\right)\\
\bm{y} | \bm{x} &\sim \gaussian(\tens{H}\bm{x} + \bm{u}, \tens{H}\tens{P}\tens{H}^\trp + \tens{R})
\end{align*}
\end{columns}
\note<1>{
- Schur complement
- An interpretation of the conditional mean?
- Do exercise 3.5 from the book
}
\end{frame}
\begin{frame}{Update step}
\begin{align*}
\mathcal{p}(\bm{x}_k|\bm{y}_{1:k}) &\propto \mathcal{p}(\bm{y}_k|\bm{x}_{k}) \mathcal{p}(\bm{x}_k|\bm{y}_{1:{k-1}})\\
......
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