Lecture 1. E) More on Probability Functions
More on Probability Functions
Finally, let us define some properties of [math]P\left(\cdot\right)[/math]:
- [math]P\left(B\right)=1-P\left(B^{C}\right)[/math],
which implies that
- [math]P\left(\emptyset\right)=0[/math], since [math]P\left(\emptyset\right)=1-\underset{=1}{\underbrace{P\left(S\right)}}.[/math]
Also,
[math]P\left(A\cup B\right)=P\left(A\right)+P\left(B\right)-P\left(A\cap B\right)[/math], which implies that
- [math]P\left(A\cup B\right)\leq P\left(A\right)+P\left(B\right)[/math] and
- [math]P\left(A\cap B\right)\geq P\left(A\right)+P\left(B\right)-1.[/math]
Conditional probability
If A and B are events (remember, [math]A[/math] and/or [math]B[/math] could be [math]\emptyset[/math] or [math]\left\{ Heads,Tails\right\}[/math] for example, in the coin tossing example), and [math]P\left(B\right)\gt 0[/math], then the conditional probability of [math]A[/math] given [math]B[/math], denoted as [math]P\left(A|B\right)[/math], is
[math]P\left(A|B\right)=\frac{P\left(A\cap B\right)}{P\left(B\right)}[/math].
This definition is intuitive. Suppose that, for a certain region, we obtain the probability of a given farmland having vineyards and/or cork trees, as in the table below:
| Cork Trees | |||
|---|---|---|---|
| Yes | No | ||
| Vineyard | Yes | 20% | 5% |
| No | 15% | 60% | |
For example, the table above states that most fields have no vineyards and no cork trees, while 20% of them have both. Suppose we would like to ask “what is the probability of observing cork trees in a field that has vineyards?” Looking at the first row (corresponding to fields that have vineyards), it appears that only [math]20\%[/math] of the fields that have vineyards also have cork trees. However, the row sums up to 25%. The conditional probability of observing cork trees in a field that has vineyards is [math]20\%/25\%=0.8[/math].
Independence
To wrap up, one more definition: Two events, [math]A[/math] and [math]B[/math], are independent if [math]P\left(A\cap B\right)=P\left(A\right).P\left(B\right)[/math].
Clearly, the probabilities in Table i. are not independent. Consider a new set of probabilities:
| Vineyard | |||
|---|---|---|---|
| Yes | No | ||
| Cork Trees | Yes | 25% | 25% |
| No | 25% | 25% | |
First, notice that the probability of a vineyard is constant, independently of whether cork trees exist or not; and the converse is also true. For example,
[math]\begin{aligned} P\left(V\cap C\right)= & 25\%=P\left(V\right).P\left(C\right)=50\%.50\%\end{aligned}[/math].
