Thursday, Jan 15

This is an outline of the topics we covered in class. These notes are not a substitute for your own note-taking. I highly recommend that you take your own notes during class. If you ever miss a class for any reason, reach out to another student in class to get a copy of their notes.

Sec 1.3: Conditional Probability

Question: How does evidence (e.g., knowledge of one event occurring) change our knowledge of probabilities for other events?

Roll a fair D6 two times. Let A = \{\text{sum } \geq 10\} and B = \{\text{first roll is } 6\}. A feels more likely if we already know B has occurred.

The conditional probability of A given B is: , \Pr(A \mid B) = \frac{\Pr(A\cap B)}{\Pr(B)}

\Pr(A \mid B) tells the proportion of B which is overlapped by A.

Two overlapping circles representing events A and B sit inside a rectangle representing the sample space \Omega. The circle labeled B is shaded. The portion of that circle which is overlapped by the A circle is also filled in with slanted lines.

\begin{tikzpicture} \def\firstcircle{(180:1.75cm) circle (2.5cm)} \def\secondcircle{(0:1.75cm) circle (2.5cm)} \fill [gray!30] \secondcircle; \begin{scope} \clip \firstcircle; \clip \secondcircle; \fill [pattern=north east lines] \firstcircle; \end{scope} \draw \firstcircle node[text=black] {$A$}; \draw \secondcircle node[text=black] {$B$}; \draw (-5, -3) rectangle (5, 3) node [text=black,right] {$\Omega$}; \end{tikzpicture}

Continuing from the previous example, |\Omega| = 36. A \amp = \{(4, 6), (5, 5), (5, 6), (6, 4), (6, 5), (6, 6)\} B \amp = \{(6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)\} A \cap B \amp = \{(6, 4), (6, 5), (6, 6)\} So \Pr(A) = \frac{6}{36}, \Pr(B) = \frac{6}{36}, \text{and } \Pr(A\cap B) = \frac{3}{36}. Then: \Pr(A \mid B) = \frac{\Pr(A\cap B)}{\Pr(B)} = \frac{3/36}{6/36} = \frac{3}{6} = \frac{1}{2}. Notice that \Pr(A\mid B) is significantly larger than \Pr(A).

Diagnostic Testing

Setup: A patient takes a diagnostic test. Let P be the event that they test positive. Let D be the event that they have the disease.

The sensitivity of a diagnostic test is \Pr(P \mid D). The specificity of a diagnostic test is \Pr(P^c \mid D^c).

But, what the patient really wants to know is \Pr(D \mid P).

A disease has a prevalence of 1%. A test has sensitivity of 90% and specificity of 91%. For a patient who gets a positive test result, what is the probability that they have the disease?

\text{A) } 9/10 \amp \amp \text{B) } 8/10 \amp \amp \text{C) } 1/10 \amp \amp \text{D) } 1/100

Bayes' Theorem (v1)

For events A, B with nonzero probability: \Pr(B \mid A) = \frac{\Pr(A \mid B)\Pr(B)}{\Pr(A)}

For example, if a patient sees a positive diagnostic test result, they might try to calculate: \Pr(D \mid P) = \frac{\Pr(P \mid D)\Pr(D)}{\Pr(P)} \Pr(P \mid D) is the sensitivity. \Pr(D) could be the prevalence. We don't have direct access to \Pr(P).

Observation: \Omega = D \cup D^c, so P = (P\cap D) \cup (P\cap D^c).

\begin{tikzpicture} \def\firstcircle{(0, 0) circle (2)} \def\leftside{(-3, -3) rectangle (-0.5, 3)} \def\rightside{(-0.5, -3) rectangle (4, 3)} \begin{scope} \clip\leftside; \fill [gray!50] \firstcircle; \end{scope} \begin{scope} \clip\rightside; \fill [pattern=north east lines] \firstcircle; \end{scope} \draw (0, 0) circle (2); \node at (2.5, 0) {$P$}; \draw (-0.5, 3) to (-0.5, -3); \node at (-1.5, -3.5) {$D$}; \node at (1.5, -3.5) {$D^c$}; \draw (-3, -3) rectangle (4, 3) node [text=black,right] {$\Omega$}; \end{tikzpicture}

Observation 2: \Pr(P \mid D) \amp \frac{\Pr(P\cap D)}{\Pr(D)} \amp \amp \Rightarrow \amp \Pr(P \cap D) \amp = \Pr(P\mid D)\Pr(D) \Pr(P \mid D^c) \amp \frac{\Pr(P\cap D^c)}{\Pr(D^c)} \amp \amp \Rightarrow \amp \Pr(P \cap D^c) \amp = \Pr(P\mid D^c)\Pr(D^c) So: \Pr(P) = \Pr(P\mid D)\Pr(D) + \Pr(P\mid D^c)\Pr(D^c)

Bayes' Theorem (v2)

\Pr(B \mid A) = \frac{\Pr(A \mid B)\Pr(B)}{\Pr(A \mid B)\Pr(B) + \Pr(A \mid B^c)\Pr(B^c)}

Continuing from the previous example: \Pr(D\mid P) \amp = \frac{\Pr(P\mid D)\Pr(D)}{\Pr(P\mid D)\Pr(D) + \Pr(P\mid D^c)\Pr(D^c)} \amp = \frac{(0.9)(0.01)}{(0.9)(0.01) + (1 - 0.91)(1 - 0.01)} \amp \approx 0.092 What if the patient got a negative test result instead? In that case, what is the probaiblity they do not have the disease? \Pr(D^c\mid P^c) \amp = \frac{\Pr(P^c\mid D^c)\Pr(D^c)}{\Pr(P^c\mid D^c)\Pr(D^c) + \Pr(P^c\mid D)\Pr(D)} \amp = \frac{(0.91)(1 - 0.01)}{(0.91)(1 - 0.01) + (1 - 0.9)(0.01)} \amp \approx 0.999

Sec 1.4: Independent Events

Question: \Pr(A \mid B) is supposed to capture how information about B affects the probability of A. What if it doesn't?

Events A, B are independent if \Pr(A \mid B) = \Pr(A).

Observation: If A, B have nonzero probability and are independent, then: \Pr(A\mid B) \amp = \Pr(A) \frac{\Pr(A\cap B)}{\Pr(B)} \amp = \Pr(A) \Pr(A\cap B) \amp = \Pr(A)\Pr(B) We can take this last equation as a definition of independence.

Continuing , recall A = \{\text{sum} \geq 10\} and B = \{\text{1st roll is } 6\}. We found that \Pr(A \mid B) \neq \Pr(A), so A and B are not independent.

Now consider the event C = \{\text{sum} = 7\}. We have: C \amp = \{(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)\} \Pr(C) \amp = \frac{6}{36} = \frac{1}{6} B\cap C \amp = \{(6, 1)\} \Pr(B\cap C) \amp = \frac{1}{36} \text{therefore: } \Pr(C\mid B) \amp = \frac{\Pr(C\cap B)}{\Pr(B)} = \frac{1/36}{1/6} = \frac{1}{6} = \Pr(C) Therefore events B, C are independent.