Question 4

(ns assignments.hw2.q4
  (:require
    [assignments.hw2.utils :refer :all]
    [fastmath.core :as m]))

4.1) If P(A|B) < P(A), show that P(B|A) < P(B).

To prove this, we’ll use the definition of conditional probability and algebraic manipulation:

\[P(A|B) = P(A ∩ B) / P(B)\]

\[P(B|A) = P(A ∩ B) / P(A)\]

Given: \(P(A|B) < P(A)\)

Step 1: Express the inequality using the definition of conditional probability

\[P(A ∩ B) / P(B) < P(A)\]

Step 2: Multiply both sides by \(P(B)\)

\[P(A ∩ B) < P(A) * P(B)\]

Step 3: Divide both sides by \(P(A)\)

\[P(A ∩ B) / P(A) < P(B)\]

Step 4: Recognize the left side as \(P(B|A)\)

\[P(B|A) < P(B)\]

(answer
  "Thus, we have shown that if $P(A|B) < P(A)$, then $P(B|A) < P(B)$.")

Thus, we have shown that if \(P(A|B) < P(A)\), then \(P(B|A) < P(B)\).

4.2) If A and B are mutually exclusive, find P(A|B). If A and B are independent, find P(A|B).

Case 1: A and B are mutually exclusive

Mutually exclusive events have no overlap, meaning \(A ∩ B = ∅\)

\[P(A|B) = P(A ∩ B) / P(B) = 0 / P(B) = 0\]

Case 2: A and B are independent

For independent events, \(P(A ∩ B) = P(A) * P(B)\)

\[P(A|B) = P(A ∩ B) / P(B) = (P(A) * P(B)) / P(B) = P(A)\]

(answer "If A and B are mutually exclusive: $P(A|B) = 0$")

If A and B are mutually exclusive: \(P(A|B) = 0\)

(answer "If A and B are independent: $P(A|B) = P(A)$")

If A and B are independent: \(P(A|B) = P(A)\)

Implementation:

(defn demonstrate-conditional-probability-inequality [p-a p-b p-a-and-b]
  (let [p-a-given-b (/ p-a-and-b p-b)
        p-b-given-a (/ p-a-and-b p-a)]
    (md (str "Given:"))
    (md (str "P(A) = " p-a))
    (md (str "P(B) = " p-b))
    (md (str "P(A ∩ B) = " p-a-and-b))
    (md (str "P(A|B) = " (m/approx p-a-given-b 4)))
    (md (str "P(B|A) = " (m/approx p-b-given-a 4)))
    (md (str "P(A|B) < P(A): " (< p-a-given-b p-a)))
    (md (str "P(B|A) < P(B): " (< p-b-given-a p-b)))))

Demonstration of the proof:

P(B|A) < P(B): true

In this example:

P(A) = 0.5
P(B) = 0.4
P(A ∩ B) = 0.1
P(A|B) = P(A ∩ B) / P(B) = 0.1 / 0.4 = 0.25

Since P(A|B) = 0.25 < P(A) = 0.5, this example satisfies the condition P(A|B) < P(A).

(defn conditional-probability [p-a p-b intersection type]
  (case type
    :mutually-exclusive 0
    :independent p-a
    (/ intersection p-b)))

Demonstration for mutually exclusive events:

(let [p-a 0.3 p-b 0.4]
  (answer (str "P(A|B) for mutually exclusive events: "
               (conditional-probability p-a p-b 0 :mutually-exclusive))))

P(A|B) for mutually exclusive events: 0

Demonstration for independent events:

(let [p-a 0.3 p-b 0.4]
  (answer (str "P(A|B) for independent events: "
               (conditional-probability p-a p-b (* p-a p-b) :independent))))

P(A|B) for independent events: 0.3

Conclusion:

We have demonstrated both algebraically and numerically that: 1. If P(A|B) < P(A), then P(B|A) < P(B). 2. For mutually exclusive events, P(A|B) = 0. 3. For independent events, P(A|B) = P(A).

These results are fundamental in understanding the relationships between events in probability theory.

source: src/assignments/hw2/q4.clj