Conditional Probability
Table of contents
Definition
The probability of event \(x\) conditioned on knowing event \(y\) (or of \(x\) given \(y\)) is defined as:
\[p(x\mid y) = \frac{p(x, y)}{p(y)}\]Bayes’ Rule
\[p(x\mid y) = \frac{p(y\mid x)p(x)}{p(y)}\]Example: throwing darts (20 equal regions)
\[p(\text{reg } 5 \mid \text{not reg } 20) = \frac{p(\text{reg } 5, \text{not reg } 20)}{p(\text{not reg } 20)} = \frac{p(\text{reg } 5)}{p(\text{not reg } 20)} = \frac{1/20}{19/20} = \frac{1}{19}\]Example: \(p(Cnt, MT)\) Based on population:
\[p(Cnt) = \begin{bmatrix} p(Cnt = E) \\ p(Cnt = S) \\ p(Cnt = W) \end{bmatrix} = \begin{bmatrix} 0.88 \\ 0.08 \\ 0.04 \end{bmatrix}\]Based on research: conditional probability table:
| \(p(MT\mid Cnt)\) | E | S | W |
| Eng | 0.95 | 0.7 | 0.6 |
| Scot | 0.04 | 0.3 | 0.0 |
| Well | 0.01 | 0.0 | 0.4 |
Calculate joint probability distribution as: \(p(Cnt, MT) = p(MT\mid Cnt) p(Cnt)\)
| \(p(MT, Cnt)\) | E | S | W |
| Eng | \(0.95 \times 0.88\) | \(0.7 \times 0.08\) | \(0.6 \times 0.04\) |
| Scot | \(0.04 \times 0.88\) | \(0.3 \times 0.08\) | \(0.0 \times 0.04\) |
| Well | \(0.01 \times 0.88\) | \(0.0 \times 0.08\) | \(0.4 \times 0.04\) |
Normalization
Discrete probability
\[\sum_x p(x\mid y) = 1\]This means that the sum of all the probabilities of \(x\) given \(y\) is 1. So if \(dom(y) = \{1,2\}\) then:
\[\sum_x p(x\mid y=1) = 1\]And
\[\sum_x p(x\mid y=2) = 1\]