L7 Conditional Probability (WMS 2.7-10)

If possible, be prepared next lecture with idea for research project
Typically, don't count to determine P(E); estimate from sample

Conditional probability

Definition: P(B)=\frac{P(A\cap B)}{P(B)}
This is how online stores (e.g. Ebay, Amazon, Google) figure out what to advertise: given that you purchased a textbook, how likely are you to want a Lego set or motorcycle helmet?
Story problem keywords: "given", "conditional on", "among", or "out of"
Example 1: Among set S of workers in particular industry, unemployment rate P(U) = .10, women P(W) = .25, intersection P(U\cap W) = .05
- Rectangular Venn diagram
- Unemployment rate among women P(W) =\frac{.05}{.25} = .20
- Fraction of unemployed who are women P(U) =\frac{.05}{.10} = .50
- Practice:
  1. Unemployment rate among men P(\bar W) =\frac{.05}{.75} =\frac{1}{15}\approx .07
  2. Fraction of unemployed who are men P(U) =\frac{.05}{.10} = .50 = 1 - P(W|U)

Independence

Definition: P(A|B) = P(A), P(B|A) = P(B) (equivalent to P(A\cap B) = P(A)P(B))
What is the probability of a person being unemployed? P(A) = .10; what if it's raining outside? Then the probability of being unemployed is P(A|B) = .10.
Surgeon joke (failing to account for independence): the bad news is that this type of surgery is successful only 25% of the time. The good news is that the last three patients all died.

Event decomposition:

If E_{1},\ldots,E_{k} are mutually exclusive and collectively exhaustive then P(A) = \sum_{k=1}^n P(A\cap E_{k})
Example 1: 30% of web traffic comes from a Google add (G), 30% from online newspaper (N), and 40% from a product reviewer's blog (R). 40% of Google traffic, 20% of newspaper traffic, and 30% of reviewer traffic end in a sale (S). What fraction of overall traffic ends in a purchase?
- Step 1: draw event tree (first web source, then purchase decision)
- Step 2: translate question into notation. P(G) = .3, P(N) = .3, P(T) = .4, P(G) = .4, P(N) = .2, P(R) = .3, wish to find P(S)
- $P(S) = P(S\cap G) + P(S\cap N) + P(S\cap R)$ $= P(G)P(S|G) + P(N)P(S|N) + P(R)P(S|R)$ = .3\times .4 + .3\times .2 + .4\times .3 = .12 + .06 + .12 = .3

S and R are independent, since P(S|R) = P(S) = .3. Is S independent of G? Of N?

Bayes' Rule
- ```
P(A\cap B) = \begin{cases}P(A|B)P(B)\\ P(B|A)P(A)\end{cases}
```
- Therefore, can derive P(A|B) from P(B|A), or vice versa.
- P(B) =\frac{P(B|A)P(A)}{P(B)} =\frac{P(B|A)P(A)}{P(B|A)P(A) + P(B|\bar A)P(\bar A)}
- Practice: find and interpret P(G|S), P(R|S), P(N|S) =\frac{P(N\cap S)}{P(S)} =\frac{.06}{.3} = .2 (mere coincidence that P(N|S) = P(S|N))
Warning: think carefully about difference between P(A|B), P(A), and P(B|A). Be sure you know which you really want.
Note: It's possible for composite probabilities and conditional probabilities to tell rather opposite stories
- Charig et al. (1986): Kidney stone treatment B looked more effective, but A was more actually effective more effective both with small stones and large stones (but stone size matters, and treatments A and B had been used disproportionately on large and small stones, respectively)

Kidney stone size	Treatment A	Treatment B
Small	81/87=93%	234/270=87%
Large	192/263=73%	55/80=69%
Both	273/350=78%	289/350=83%

MLB batting averages: David Justice was better in 1995 and 1996 but Derek Jeter was better in 1995-96. Who is better batter?

Batter	1995	1996	1995-96
Derek Jeter	12/48=.250	183/582=.314	195/630=.310
David Justice	104/411=.253	45/140=.321	149/551=.270
Either could be. Likely depends on which is more predictive of 1997 (depends on other assumptions)

Israel covid data: August 2021 (https://www.covid-datascience.com/post/israeli-data-how-can-efficacy-vs-severe-disease-be-strong-when-60-of-hospitalized-are-vaccinated)
1. When covid Delta variant hit, Israeli hospitals filled up with covid cases: 214 that were unvaccinated and 301 that were vaccinated. Since 60% were vaccinated, superficial conclusion is that vaccines make covid worse, not better!
2. But 60%=P(vax|cv). We really want to know P(cv|vax) (actually, want to compare P(cv|vax) and P(cv|no vax))
3. $P(cv|vax) = 301/5,634,634 = 5.3*10^{- 5}$ P(\text{novax}) =\frac{214}{1,302,912} = 16.4*10^{- 5} Vaccinated only catch covid \frac{5.3}{16.4} = 32\% as often (i.e. vaccine 68% effective)

iv. Nearly 80% of Israelis over age 12 were vaccinated against covid, so if it were unrelated random draw, 80% of covid patients should have been vaccinated; lower rate than 80% supports hypothesis that treatment helped.

v. Put differently, so many Israelis were vaccinated that even though those vaccinated only got covid 68% as often, there were more vaccinated covid cases than unvaccinated covid cases.

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L7 Conditional Probability (WMS 2.7-10)

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