(expert opinion, educated guess,not based on numbers/ not really any data)
Assigning probabilities based on judgement.
-Is most appropriate when one can not realistically assume that the experimental outcomes are equally likely and when little relevant data are available.
~We may use any information available, such as experience or intuition. After considering all available information, a probability value that expresses our "degree of belief" (on a scale from
0 to 1) that the experimental outcome will occur is specified.
B/c subjective probability expresses a person's degree of belief, it is personal. Using the subjective method, different people can be expected to assign different probabilities to the same experimental outcome.
Sample Space and Probabilities
Sample Space and Experimental Outcomes
We can use the letter \(S\) or \(\Omega\) to denote sample space. Supose we have a set of \(n\) experimental outcomes:
\[S = \left\{ E_1, E_2, ..., E_{n-1}, E_n\right\}\]
We can call \(S\) a sample space if:
- \(E_i\) are
mutually exclusive:
- \(E_i\) for all \(i\) are separate outcomes that do not overlap.
- Suppose there are multiple people running for the Presidency, some of these candidates are women, and some of these women are from Texas. If one of the \(E_i\) is that a woman becomes the President, and another \(E_i\) is that someone from Texas becomes the President, \(S\) would not be a sample space, because we could have a woman who is also from Texas win the Presidency.
- \(E_i\) are jointly exhaustive:
- For the experiment with well-defined outcomes, the \(E_i\) in \(S\) need to cover all possible outcomes.
- If you are throwing a six sided dice, there are six possible experimental outcomes, if \(S\) only has five of them, it would not be a sample space.
Thinking about the world in terms of sample space is pretty amazing.
Assigning Probability to Experimental Outcomes
We can assign probabilities to events of a sample space. Since experimental outcomes are themselves events as well, we can assign probabilities to each experimental outcome.
For the mutually exclusive and jointly exhaustive experimental outcomes of the sample space, there are two equirements for assigning probabilities: - Each element of the sample space can not have negative probability of happening, and also can not have more than \(1\) probability of happening, with \(P\) denotes probability, we have: \[0 \le P(E_i) \le 1\] - The probabilities of all the mutually exclusive and jointly exhaustive experimental outcomes in the sample space sum up to \(1\). For an experimental with \(n\) experimental outcomes: \[\Sigma_{i=1}^{n} P(E_i) = 1\]
Heavy Rain | 0.1 |
Light Rain | 0.2 |
No Rain | 0.7 |
tomorrow:1 | Heavy Rain |
tomorrow:2 | Light Rain |
tomorrow:3 | No Rain |
tomorrow:4 | No Rain |
tomorrow:5 | No Rain |
tomorrow:6 | No Rain |
tomorrow:7 | No Rain |
tomorrow:8 | No Rain |
tomorrow:9 | No Rain |
tomorrow:10 | No Rain |
tomorrow:11 | No Rain |
tomorrow:12 | No Rain |
tomorrow:13 | Light Rain |
tomorrow:14 | Heavy Rain |
tomorrow:15 | No Rain |
tomorrow:16 | No Rain |
tomorrow:17 | Light Rain |
tomorrow:18 | No Rain |
tomorrow:19 | No Rain |
tomorrow:20 | No Rain |
Union and Intersection and Complements
Definitions:
- Complement of Event \(A\):
- “Given an event A, the complement of A is defined to be the event consisting of all sample points that are not in A. The complement of A is denoted by \(A^c\).” (AWSCC P185)
- The Union of Events \(A\) and \(B\):
- “The union of A and B is the event containing all sample points belonging to \(A\) or \(B\) or both. The union is denoted by \(A \cup B\).” (AWSCC P186)
- The Intersection of Events \(A\) and \(B\):
- “Given two events \(A\) and \(B\), the intersection of \(A\) and \(B\) is the event containing the sample points belonging to both \(A\) and \(B\). The intersection is denoted by \(A \cap B\).” (AWSCC P187)
Probabilities for Complements and Union
The Probabilities of Complements add up to 1: \[P(A) + P(A^c) = 1\]
The Addition Law: \[P (A \cup B) = P(A) + P(B) - P (A \cap B)\]
If two events \(A\) and \(B\) are mutually exclusive, which means they do not share any experimental outcomes (sample points), then: \(P (A \cap B) = 0\), and \(P (A \cup B) = P(A) + P(B)\).
The Multiplication Law for Indepedent Events: \[P (A \cap B) = P(A) \cdot P(B)\]
If the probability of event \(A\) happening does not change the probability of event \(B\) happening, and vice-versa. The two events are independent. Below we arrive this formulation from conditional probability.
Conditional Probability
We use a straight line \(\mid\) to denote conditional probability. Given \(A\) happens, what is the probability of \(B\) happening?
\[P (A \mid B) = \frac{P(A \cap B)}{P(B)}\]
This says the probability of \(A\) happening given that \(B\) happens is equal to the ratio of the probability that both \(A\) and \(B\) happen divided by the probability of \(B\) happening.
The formula also means that the probability that both \(A\) and \(B\) happens is equal to the probability that \(B\) happens times the probability that \(A\) happens conditional on \(B\) happening: \[ P(A \cap B) = P (A \mid B)\cdot P(B)\]
If \(A\) and \(B\) are independent, that means the probability of \(A\) happening does not change whether \(B\) happens or not, then, \(P (A \mid B) = P(A)\), and: \[ \text{If A and B are independent: } P(A \cap B) = P(A) \cdot P(B)\] This is what we wrote down earlier as well.