Module 6 notes in progress

OMSCS/Courses/AI/Module_06/Module 6 - Bayes Nets.md

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+---
+tags:
+  - OMSCS
+  - AI
+  - Probability
+  - Bayes
+---
+# Module 6 - Bayes Nets
+
+## Bayes Rule Graphically
+![[Pasted image 20240217144613.png]]
+
+Diagnostic Reasoning
+- $P(A | B)$
+- $P(A|\neg B)$
+
+You need 3 parameters to specify a bayes network
+- $P(A)$
+- $P(B|A)$
+- $P(B|\neg A)$
+
+## Bayes Rule Refresher
+$$
+P(A|B)=\frac{P(B|A)P(A)}{P(B)}
+$$
+
+$$
+P(\neg A | B) = \frac{P(B|\neg A)P(\neg A)}{P(B)}
+$$
+
+$$
+P(A|B)+P(\neg A|B)=1
+$$
+
+- $P'(A|B)=P(B|A)P(A)$
+- $P'(\neg A|B)=P(B|\neg A)P(\neg A)$
+- For some normalizer value $\eta$
+	- $P(A|B)=\eta P'(A|B)$
+	- $P(\neg A|B)=\eta P'(\neg A|B)$
+	- $\eta=(P'(A|B)+P'(\neg A|B))^{-1}$
+- The advantage of using these pseudo probabilities is that you don't have to calculate $P(B)$ directly, which can be pretty complicated in some problems.
+
+### 2-Test Cancer Example
+![[Pasted image 20240217150019.png]]
+- $P(C)=0.01$
+- $P(\neg C)=0.99$
+- $P(T_x=+|C)=0.9$
+- $P(T_x=-|C)=0.1$
+- $P(T_x=-|\neg C)=0.8$
+- $P(T_x=+|\neg C)=0.2$
+- $P(C|T_1=+, T_2=+)=?$
+
+Working out
+- Calculating $P(++|C)$
+	- $P(T_1=+,T_2=+|C)=P(++|C)$
+	- $P(++|C)=P(T_1=+|C)P(T_2=+|C)=0.9^2$
+	- $P(++|C)=0.81$
+- Calculating $P(++|\neg C)$
+	- $P(T_1=+,T_2=+|\neg C)=P(++|\neg C)$
+	- $P(++|\neg C)=P(T_1=+|\neg C)P(T_2=+|\neg C)=0.2^2$
+	- $P(++|\neg C)=0.04$
+- $P(C)=0.01$
+- $P(++)$ $\leftarrow$ we can avoid calculating this
+	- Calculating $P'(C|++)$
+		- $P'(C|++)=P(++|C)P(C)$
+		- $P'(C|++)=(0.81)(0.01)$
+		- $P'(C|++)=0.0081$
+	- Calculating $P'(\neg C|++)$
+		- $P'(\neg C|++)=P(++|\neg C)P(\neg C)$
+		- $P'(\neg C|++)=(0.04)(0.99)$
+		- $P'(\neg C|++)=0.0396$
+	- $\eta = (0.0081+0.0396)^{-1} = 20.9643...$
+- Calculating $P(C|++)$
+	- $P(C|++)=\eta P'(C|++)$
+	- $P(C|++)=(20.9643...)(0.0081)$
+	- $P(C|++)=0.1698$
+
+I initially got the wrong answer for this due to using the wrong value of $P(T_1=+|\neg C)$ in the "Calculating $P(++|\neg C)$" step. I used 0.1 instead of 0.2, which multiplied the errors downstream.
+
+|  | prior | + | + | P' | $P(C\|++)$ |
+| ---- | ---- | ---- | ---- | ---- | ---- |
+| $C$ | 0.01 | 0.9 | 0.9 | 0.0081 | 0.1698 |
+| $\neg C$ | 0.99 | 0.2 | 0.2 | 0.0396 | 0.8301 |
+### 2-Test Cancer Example 2
+
+- $P(C)=0.01$
+- $P(\neg C)=0.99$
+- $P(T_x=+|C)=0.9$
+- $P(T_x=-|C)=0.1$
+- $P(T_x=-|\neg C)=0.8$
+- $P(T_x=+|\neg C)=0.2$
+- $P(C|T_1=+, T_2=-)=?$
+
+Working out
+- Calculating $P(+-|C)$
+	- $P(T_1=+,T_2=-|C)=P(+-|C)$
+	- $P(+-|C)=P(T_1=+|C)P(T_2=-|C)=(0.9)(0.1)$
+	- $P(+-|C)=0.09$
+- Calculating $P(+-|\neg C)$
+	- $P(T_1=+,T_2=-|\neg C)=P(+-|\neg C)$
+	- $P(+-|\neg C)=P(T_1=+|\neg C)P(T_2=-|\neg C)=(0.8)(0.2)$
+	- $P(+-|\neg C)=0.16$
+- $P(C)=0.01$
+- $P(+-)$ $\leftarrow$ we can avoid calculating this
+	- Calculating $P'(C|+-)$
+		- $P'(C|+-)=P(+-|C)P(C)$
+		- $P'(C|+-)=(0.09)(0.01)$
+		- $P'(C|++)=0.0009$
+	- Calculating $P'(\neg C|+-)$
+		- $P'(\neg C|+-)=P(+-|\neg C)P(\neg C)$
+		- $P'(\neg C|+-)=(0.16)(0.99)$
+		- $P'(\neg C|+-)=0.1584$
+	- $\eta = (0.1584+0.0009)^{-1} = 6.2774...$
+- Calculating $P(C|+-)$
+	- $P(C|+-)=\eta P'(C|+-)$
+	- $P(C|+-)=(6.2774...)(0.0009)$
+	- $P(C|++)=0.005649...$
+
+|  | prior | + | - | P' | $P(C\|++)$ |
+| ---- | ---- | ---- | ---- | ---- | ---- |
+| $C$ | 0.01 | 0.9 | 0.1 | 0.0009 | 0.0056 |
+| $\neg C$ | 0.99 | 0.2 | 0.8 | 0.1584 | 0.9943 |
+
+## Conditional Independence
+Useful assumptions in the examples above.
+- Both $T_x$ variables are conditionally independent
+- $P(T_2|C,T_1)=P(T_2|C)$ is one way of specifying conditional independence. 
+
+![[Pasted image 20240217153405.png]]
+
+$B \perp C \space | \space A \ne B \perp C$
+
+> Getting a positive test result about cancer (B) increases the probability of having cancer (A) above the prior probability for A, which changes the probability that another test will result in a positive (C).
+
+### Cancer Example 3
+
+- $P(C)=0.01$
+- $P(\neg C)=0.99$
+- $P(T_x=+|C)=0.9$
+- $P(T_x=-|C)=0.1$
+- $P(T_x=-|\neg C)=0.8$
+- $P(T_x=+|\neg C)=0.2$
+- $P(T_2=+ \space | \space T_1=+)=?$
+
+Ok so the theory here is that by having a positive result for $T_1$, that raises the probability of $C$ above it's prior value, which affects the calculations for $T_2$.
+
+> For this problem, we want to apply the principle of total probability.
+
+- $P(T_2=+ \space | \space T_1=+) = P(+_2 | +_1)$
+- $P(+_2 | +_1)=P(+_2|+_1, C)P(C|+_1)+P(+_2|+_1, \neg C)P(\neg C | +_1)$
+- $P(C|+_1)=0.043$
+- $P(\neg C|+_1)=1-0.043=0.957$
+- Thanks to conditional independence
+	- $P(+_2|+_1, C)=P(+_2|C)=0.9$
+	- $P(+_2|+_1, \neg C)=P(+_2|\neg C)=0.2$
+- $P(+_2 | +_1)=(0.9)(0.043)+(0.2)(0.957)$
+- $P(+_2 | +_1)=0.2301$
+- Baseline: $P(+_x)=0.207$
+
+## Absolute and Conditional Independence
+- $A \perp B$ - means that A and B have "absolute independence"
+- $A \perp B \space | \space C$ - means that A and B are "conditionally independent" on C
+- Absolute independence does not imply conditional independence.
+- Conditional independence does not imply absolute independence.
+
+![[Pasted image 20240217160022.png]]
+
+- $P(R | S) = P(R)$
+	- We don't know anything about $H$, so $S$ has no effect on $R$.
+
+### Quiz 1
+
+- $P(R|H,S)=?$
+
+> Using Bayes Rule we can transform this question.
+
+$$
+P(R|H,S)=\frac{P(H|R,S)P(R|S)}{P(H|S)}
+$$
+
+Where the hell did $P(H|S)$ come from?
+
+> thanks to the rules of conditional independence: $P(R|S)=P(R)$
+> 
+> thanks to the rules of total probability:
+> $P(H|S)=P(H|R,S)P(R)+P(H|\neg R,S)P(\neg R)$
+>
+> Substituting these values in to the equation above.
+$$
+\frac{P(H|R,S)P(R|S)}{P(H|S)}=\frac{P(H|R,S)P(R)}{P(H|R,S)P(R)+P(H|\neg R,S)P(\neg R)}
+$$
+
+### Quiz 2
+- $P(R|H)=?$
+
+Using bayes theorem
+
+$$
+P(R|H)=\frac{P(H|R)P(R)}{P(H)}
+$$
+
+- $P(H|R)$
+	- $= P(H|\neg S,R)P(\neg S) + P(H|S,R)P(S)$
+	- $=(0.9)(0.3)+(1)(0.7)$
+	- $=0.97$
+- $P(R) = 0.01$
+- $P(H)$
+	- $= P(H|S,R)P(S, R)$
+		- $=P(H|S,R)P(S)P(R)$
+		- $=(1)(0.7)(0.01)$
+		- $=0.007$
+	- $+ \space P(H|\neg S, R)P(\neg S, R)$
+		- $=P(H|\neg S, R)P(\neg S)P(R)$
+		- $= (0.9)(0.3)(0.01)$
+		- $=0.0027$
+	- $+ \space P(H|S,\neg R)P(S, \neg R)$
+		- $=P(H|S,\neg R)P(S)P(\neg R)$
+		- $=(0.7)(0.7)(0.99)$
+		- $=0.4851$
+	- $+ \space P(H|\neg S, \neg R)P(\neg S, \neg R)$
+		- $=P(H|\neg S, \neg R)P(\neg S)P(\neg R)$
+		- $=(0.1)(0.3)(0.99)$
+		- $=0.0297$
+	- $=0.007+0.0027+0.4851+0.0297$
+	- $=0.5245$
+- $\frac{P(H|R)P(R)}{P(H)}$
+	- $=(0.97)(0.01)(0.5245)^{-1}$
+	- $=(0.97)(0.01)(1.9065...)$
+	- $=0.01849...$
+
+Using the "pseudo-probabilities" method
+
+$$
+P(R|H)=\frac{P(H|R)P(R)}{P(H|R)P(R)+P(H|\neg R)P(\neg R)}
+$$
+
+- $P(H|R)$
+	- $= P(H|\neg S, R)P(\neg S) + P(H|S,R)P(S)$
+	- $=(0.9)(0.3)+(1)(0.7)$
+	- $=0.97$
+- $P(H|\neg R)$
+	- $= P(H|\neg S, \neg R)P(\neg S) + P(H|S,\neg R)P(S)$
+	- $= (0.1)(0.3) + (0.7)(0.7)$
+	- $= 0.03+0.49$
+	- $=0.52$
+- $P'(R|H)=P(H|R)P(R)$
+	- $=(0.97)(0.01)$
+	- $=0.0097$
+- $P'(\neg R|H)=P(H|\neg R)P(\neg R)$
+	- $=(0.52)(0.99)$
+	- $=0.5148$
+- $\eta = (P'(R|H)+P'(\neg R|H))^{-1}$
+	- $=(0.0097+0.5148)^{-1}$
+	- $=0.5245^{-1}$
+	- $=1.9065...$
+- $P(R|H)=\eta P'(R|H)$
+	- $=(1.9065...)(0.0097)$
+	- $=0.01849...$