The Bayesian view of probability is quite objective and also more general than the frequentist’s view. It doesn’t rely on repeatition of events.

one of the most important concepts in all of probability theory — that of conditional probability.

– Sheldon M. Ross

## Topics

• Association Rules
• Conditional Probability
• Naive Bayes

## Association Rules

Association Rules

$$\text{Milk} \Rightarrow \text{Croissant} [ \text{support} = 2/5, \text{confidence} = 2/3 ]$$

• A measure of co-occurrence

## Prize in Boxes

Three boxes M, N, O:

• Only one of them refers to a prize.
• The participant claims one of them, e.g., M.
• The host removes one of the empty boxes (N) from the other two boxes (N, O).
• Now we have only two candidate boxes for the participants, M, O.
• The participant is asked to reclaim a box.

Question:

• Should the participant switch?

## Frequentist vs Bayesian

Frequentists:

• Probability is based on the repetition of events.
• Without repetition, the probability is unknown.
• Events are random.
• Make predictions based on probabilities.
• Relies on NHST to validate our models.

Bayesian:

• Probability is objective (educated guess). It is a conceptual tool to describe our degree of certainty.
• Probability is not necessarily a one-to-one map of occurrences of events.
• Data (such as a previous reoccurring event) is used to update our beliefs.
• Parameters of models are random.

## Joint Probability

Joint Probability $$P(A\cap B)$$ also denoted as $P(A, B)$

Examples of joint probabilities:

• $P(A, B)=0$: Event $A$ and event $B$ are so different that they will never happen together. In this case, $P(A\cup B) = P(A) + P(B)$.
• $P(A, B) = P(A)P(B)$: $A$ and $B$ are independent of each other.
• $P(A) + P(B) = 1$.

## Conditional Probability and Bayes’s Theorem

$P(A\vert B)$: the probability of event $A$ if event $B$ occurred.

• It’s a rescaling/(re)normailization of $P(A)$: $P(A\cap B) = P(A\vert B)P(B)$;
• We didn’t specify $A$ and $B$: $P(B\cap A) = P(B\vert A)P(A)$ also holds;
• Apply $P(A\cap B) = P(B \cap A)$: $P(A\vert B)P(B) = P(B\vert A)P(A)$

In

$$\begin{equation}P(A\vert B) = \frac{P(B\vert A)P(A)}{P(B)}\end{equation}$$

• $P(A)$: prior
• $P(A\vert B)$: posterior
• $P(B\vert A)$: likelihood

$A\to H$ (A theory) & $B\to D$ (some data points)

• $P(H\vert D) = \frac{P(D\vert H)P(H)}{P(D)}$

## Solutions to the Prize in Boxes Problem

• $P(\text{Win}\vert \text{Switch}, \text{Wrong Box}) = 1$: The participant made a wrong choice at the first attempt, then switched.
• $P(\text{Win}\vert \text{Switch}, \text{Right Box}) = 0$
• $P(\text{Win}\vert \text{NoSwitch}, \text{Wrong Box})=1$
• $P(\text{Wrong Box}) = 2/3$
• $P(\text{Right Box}) = 1/3$
• $P(\text{Win}\vert \text{Switch})$: We will solve this.

Applying Bayes’ theorem,

\begin{align} P(\text{Win}\vert \text{Switch}) =& P(\text{Win}\vert \text{Switch}, \text{Wrong Box}) P(\text{Wrong Box}) + P(\text{Win}\vert \text{Switch}, \text{Right Box}) P(\text{Right Box}) \\ =& 2/3 \nonumber \end{align}

\begin{align} P(\text{Win}\vert \text{NoSwitch}) =& P(\text{Win}\vert \text{NoSwitch}, \text{Wrong Box}) P(\text{Wrong Box}) + P(\text{Win}\vert \text{NoSwitch}, \text{Right Box}) P(\text{Right Box}) \\ =& 1/3 \nonumber \end{align}

We update our probability perception when we have new data.

## Rare Disease

We are about to test for a rare disease:

• Prevalence: $P(D) = 0.001$
• Test method: $P(+\vert D) = 0.99$
• Test method: $P(+\vert H) = 0.005$

The positive rate for a random person is

$$P(+) = P(+\vert D)P(D) + P(+\vert H) P(H) = 0.006.$$

If a test is positive, the probability of the person being test has the disease is

$$P(D\vert +) = \frac{P(+\vert D) P(D)}{P(+)} = 0.99\times 0.001/0.006 = 0.165.$$

$$P(H\vert +) = \frac{P(+\vert H) P(H)}{P(+)} = 0.005\times (1-0.001)/0.006 = 0.8325.$$

## Naive Bayes

• Chapter 16 of Introduction to Probability and Statistics for Engineers and Scientists (6th ed.) by Sheldon Ross.
• Naive Bayes

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