Probability Axioms, Conditional Probability and Bayes
Statistics I · Lecture 5
Today
- Axioms and corollaries
- Conditional probability and independence
- Law of Total Probability and Bayes theorem
- Practice, from 🟢 to 🟠
Probability Axioms
Probability
Andrey Kolmogorov defined a set of characteristics that any probability \(P\) measure should have, these are called the Kolmogorov’s axioms (1933):
- \(P(A)\in\mathbb{R}\) and \(P(A)\geq 0\), for any event \(A\subseteq\Omega\).
- \(P(\Omega)=1\)
- For any \(A\) and \(B\) disjoint, \(P(A)+P(B)=P(A\cup B)\)
Corollary
Let \(A\) and \(B\) be some events in \(\Omega\)
- \(P(\emptyset)=0\)
- \(P(A^c)=1-P(A)\)
- \(A\subset B\Rightarrow P(A)\leq P(B)\)
- \(0\leq P(A)\leq 1,\ \forall A\)
- \(P(A\setminus B)=P(A)-P(A\cap B)\)
Corollary
- \(P(A\cup B)= P(A)+P(B)-P(A\cap B)\)
- \[P\left(\bigcup^n_{i=1}A_i\right)=\sum_{i=1}^n P(A_i)-\sum_{i\neq j} P(A_i\cap A_j)+\\ \sum_{i\neq j\neq k} P(A_i\cap A_j\cap A_k)+...+(-1)^{n-1}P\left(\bigcap_{i=1}^n A_i\right)\]
Conditional Probability
Conditional probability, as the wording implies, means the probability of something happening given something else has happened. Now, note “something” here makes reference to an event.
\[P(A|B)\]
It reads the probability of \(A\), given \(B\).
Conditional Probability
Note that if we think on sets, saying given \(B\) we are immediately excluding everything that could have happened if \(B\) did not happen, and therefore our Universal set is no longer \(\Omega\), but \(B\).
What we are looking for are, among the events that live in \(B\), how many of those live in \(A\) (because those would trigger event \(A\)). Actually, we are interested on the relative measure of those outcomes, compared to the whole size of \(B\): \[P(A|B)=\frac{P(A\cap B)}{P(B)}\]
Example
Consider a factory that makes 10 wrenches :wrench:. Among those, we know that 2 have imperfections. Suppose you intend to remove, randomly, 2 :wrench: from the lot (of 10). Consider the following events:
\(A = \{\text{The first :wrench: is faulty}\}\) \(B = \{\text{The second :wrench: is faulty}\}\)
What if we want to compute \(P(B)\)? For a correct assessment for \(B\), we would better have some information on the realization of \(A\)!
Example
If the fist :wrench: was faulty, then \(A\) happened. If the first :wrench: was ok, then \(A^c\) happened, and therefore we can compute \(P(B|A)\) and \(P(B|A^c)\). We are assuming that we are removing these :wrench: without replacing them.
Let’s see why this last detail (replacing the :wrench:) is so relevant before going on.
Example: With replacement
Initial set:
| 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |
|---|---|---|---|---|---|---|---|---|---|
| :wrench: | :wrench: | :collision: | :wrench: | :collision: | :wrench: | :wrench: | :wrench: | :wrench: | :wrench: |
Remove one (if randomly you do not know which), but after you remove you can see what happened, let’s take out 7.
| 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |
|---|---|---|---|---|---|---|---|---|---|
| :wrench: | :wrench: | :collision: | :wrench: | :collision: | :wrench: | :wrench: | :wrench: | :wrench: |
We observe, and voilá it was a fine wrench :wrench:. If we replace it though, we would be picking from
| 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |
|---|---|---|---|---|---|---|---|---|---|
| :wrench: | :wrench: | :collision: | :wrench: | :collision: | :wrench: | :wrench: | :wrench: | :wrench: | :wrench: |
That is in the exact same conditions we made our first choice, and therefore what happens with the first pick is irrelevant: These events are now independent!
Example: with replacement
With \(A\), we know that when we pick the second wrench (\(B\)), in the box there are 2 :collision: and 8 :wrench:.
With \(A^c\), we know that when we pick the second wrench (\(B\)), in the box there are 2 :collision: and 8 :wrench:.
The probability of getting a :collision: is the same in each scenario! \(P(B|A)= P(B|A^c)\)!
\[P(B|A)=\frac{2}{10}=0.2\ \text{and}\ P(B|A^c)=\frac{2}{10}=0.2\]
Example: no replacement
Initial set:
| 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |
|---|---|---|---|---|---|---|---|---|---|
| :wrench: | :wrench: | :collision: | :wrench: | :collision: | :wrench: | :wrench: | :wrench: | :wrench: | :wrench: |
Remove one (if randomly you do not know which is broken)
| 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |
|---|---|---|---|---|---|---|---|---|---|
| :wrench: | :wrench: | :collision: | :wrench: | :wrench: | :wrench: | :wrench: | :wrench: | :wrench: |
We observe, and voilá it was a broken wrench :collision: \(A\) happened!
| 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |
|---|---|---|---|---|---|---|---|---|---|
| :wrench: | :wrench: | :collision: | :wrench: | :collision: | :wrench: | :wrench: | :wrench: | :wrench: |
We observe, and voilá it was a fine wrench :wrench: \(A^c\) happened!
Example: no replacement
With \(A\), we know that when we pick the second wrench (\(B\)), in the box there are 1 :collision: and 9 :wrench:.
With \(A^c\), we know that when we pick the second wrench (\(B\)), in the box there are 2 :collision: and 9 :wrench:.
The probability of getting a :collision: is different in each scenario! \(P(B|A)\neq P(B|A^c)\)!
\[P(B|A)=\frac{1}{9}=0.111\ \text{and}\ P(B|A^c)=\frac{2}{9}=0.222\]
Conclusion Conditional Probability
So formally
Corollary
It follows that the identity \[P(B|A)=\frac{P(A\cap B)}{P(A)}\] or \[P(A\cap B)=P(B|A)P(A)\] with \(P(A)\neq 0\) also holds true.
Corollary
Now let’s think on \(P(A\cap B \cap C)\):
- \(P(A\cap B \cap C)=P(A\cap (B \cap C))\)
- \(P(A\cap B \cap C)=P(A|B\cap C)P(B\cap C)\)
- \(P(A\cap B \cap C)=P(A|B\cap C)P(B|C) P(C)\)
Corollary
Note that given the commutativity of the intersection, we could have obtained also:
- \(P(A\cap B \cap C)=P(A|B\cap C)P(C|B) P(B)\)
- \(P(A\cap B \cap C)=P(B|A\cap C)P(A|C) P(C)\)
- \(P(A\cap B \cap C)=P(B|A\cap C)P(C|A) P(A)\)
- \(P(A\cap B \cap C)=P(C|A\cap B)P(A|B) P(B)\)
- \(P(A\cap B \cap C)=P(C|A\cap B)P(B|A) P(A)\)
And to make sense of all of this we need \(P(X)>0\), \(P(X\cap Y)>0\) with \(X,Y\in\{A,B,C\}\).
Example
Consider a region with 1,000 adults. Their job data is captured by the following table:
| Employed | Unemployed | Total | |
|---|---|---|---|
| Women | 470 | 55 | 525 |
| Men | 430 | 45 | 475 |
| Total | 900 | 100 | 1,000 |
- Randomly selecting a person in this region, what is the probability this person is:
- Woman
- Unemployed
- Unemployed woman
Example
Let’s define the events:
\(W=\{Woman\}\), \(M=\{Man\}\), \(U=\{Unemployed\}\)
- Woman: \(P(W)=\frac{525}{1000}=0.525\)
- Unemployed: \(P(U)=\frac{100}{1000} = 0.1\)
- Unemployed woman: \(P(W\cap U)=\frac{55}{1000}=0.055\)
Example
- A citizen is randomly chosen from the population, and it happens to be a woman. What is the probability she is unemployed?
\(P(U|W)=\frac{P(U\cap W)}{P(W)}=\frac{0.055}{0.525}=0.105\)
- A citizen is randomly chosen from the population, and it happens to be unemployed. What is the probability this citizen is a woman?
\(P(W|U)=\frac{P(W\cap U)}{P(U)}=\frac{0.055}{0.1}=0.55\)
Independent Events
Independent Events
From the definition of independence, we can obtain several properties. Let \(A\) and \(B\) independent events with \(P(A)P(B)>0\):
- \(P(A|B)=P(A)\) and \(P(B|A)=P(B)\) (remember conditional example with replacement)
- \(A^c\) and \(B\) are independent, as well as \(A\) and \(B^c\), and even \(A^c\) and \(B^c\).
- If \(A\) and \(B\) are incompatible, they cannot be independent. \(P(A\cap B)=P(\emptyset)=0\neq P(A)P(B)\)
- Any event is independent of \(\Omega\) and \(\emptyset\).
Example
Are \(W\) and \(U\) from the previous example independent?
\(P(W)=0.525\), \(P(U)=0.1\), \(P(W|U)=0.55\), \(P(U|W)=0.105\).
Note that \(P(W)\neq P(W|U)\) and \(P(U)\neq P(U|W)\). Therefore, they cannot be independent.
Example
Consider a die :game_die: that is thrown twice. Consider the following two events:
\(A=\{\text{The die shows an odd number the first time}\}\) \(B=\{\text{The die shows a number }>4\text{ the second time}\}\)
Are \(A\) and \(B\) independent events?
Example
In this case, \(\Omega=\{(x,y)\in \mathbb{N}^2| x,y \leq 6\}\) with \(\# \Omega = 6^2=36\)
\(P(A)=\frac{18}{36}=\frac{1}{2}\)
\(P(B)=\frac{12}{36}=\frac{1}{3}\)
\(P(A\cap B)=\frac{1}{6}=P(A)P(B)\)
\(P(A|B)=\frac{P(A\cap B)}{P(B)}=\frac{1}{2}=P(A)\)
\(P(B|A)=\frac{P(B\cap A)}{P(A)}=\frac{1}{3}=P(B)\)
They are independent events!
Remark
Two events being independent is not the same that they being incompatible:
| \(A\) and \(B\) independent | \(A\) and \(B\) incompatible |
|---|---|
| \(P(A\cap B)=P(A)P(B)\) | \(P(A\cap B)=0\) |
| \(P(A|B)=P(A)\) and \(P(B|A)=P(B)\) | \(P(A|B)=0\) and \(P(B|A)=0\) |
Example
Let \(A\) and \(B\) be two events such that: \(P(A)=0.6\), \(P(B)=t\), and \(P(A\cup B)=0.8\)
Find \(t\) such that \(A\) and \(B\) are:
- Mutually exclusive or incompatible.
- Independent.
Example
- In this case, what we need is that \(P(A\cap B)=0\).
From probability theory, we have \[P(A\cup B)=P(A)+P(B)-P(A\cap B)\] and therefore we get: \[0.8 = 0.6+t\Rightarrow t=0.2\]
Example
- To make \(A\) and \(B\) independent, we need that \(P(A\cap B)=P(A)P(B)=0.6t\):
Using the same identity we just used: \[0.8=0.6+t-0.6t\Rightarrow t= 0.5\]
Law of Total Probability
Example
Consider a financial institution that sells two products, \(\alpha\) and \(\beta\), with very high yields. It is known that, among its clients, 10% invest a share of their wealth in \(\alpha\) and the rest in \(\beta\). From those who invest in \(\alpha\), 70% manage to get returns above the market. From among those who do not invest in \(\alpha\), 55% get returns above the market. Randomly choosing a client of this firm, find the probability this customer gets a return above the market.
Example
Let’s define the events:
- \(A_1\) the client invest in \(\alpha\)
- \(A_2\) the client invest in \(\beta\)
- \(B\) has returns above the market.
Matching with the available data we obtain:
\(P(A_1)=0.1\), \(P(A_2)=0.9\), \(P(B|A_1)=0.7\), and \(P(B|A_2)=0.55\).
From the Law of Total Probability:
Example
\[P(B)=\sum_{i=1}^2 P(A_i\cap B)\] \[P(B)=P(B|A_1)P(A_1)+P(B|A_2)P(A_2)\] \[P(B)=0.7\times 0.1 + 0.55\times 0.9 = 0.565\]
Bayes Theorem
Note that this is a consequence of the Law of Total Probability.
Bayes Theorem
On the other side, \(\sum_i P(A_i)=1\) and \(\sum_{i} P(A_i|B)=1\)
Bayes Theorem has been widely used in economics, in biomedical sciences, and social sciences when looking for causality.
If event \(B\) represents consequences and event \(A_i\) probable cause, Bayes Theorem allows to assess the probability of this cause \((P(A_i))\).
Example
Let’s go back to the previous example, about our investors.
Let’s compute the probability that the client invested his money on product \(\beta\), but given that the client had returns above the market (event \(B\)).
Example
If the customer invested in \(\beta\), then the event we are trying to is \(A_2\), but conditional on event \(B\), \(P(A_2|B)\):
\[P(A_2|B)=\frac{P(A_2\cap B)}{P(B)}=\frac{P(A_2)\times P(B|A_2)}{\sum_i P(A_i)\times P(B|A_i)}\]
We knew from the previous exercise that \(P(B)=0.565\), and therefore we obtain:
\[P(A_2|B)=\frac{0.9\times 0.55}{0.565}=0.876\]
Example
How do we interpret this?
The probability that the client invested in \(\beta\), given that he had a return above the market, is 0.876.
Example
All these computations can be very easy with the help of the following table:
| \(A_i\) | \(P(A_i)\) | \(P(B|A_i)\) | \(P(A_i)P(B|A_i)\) | \(P(A_i|B)\) |
|---|---|---|---|---|
| \(A_1\) | 0.1 | 0.7 | 0.07 | 0.124 |
| \(A_2\) | 0.9 | 0.55 | 0.495 | 0.876 |
| 1 | 0.565 | 1 |
A test that is right 99% of the time
A screening test for a rare condition. Move the sliders and read off the only number a patient cares about: given a positive test, what is the chance of actually being ill?
What the chart showed
At 99% sensitivity and 99% specificity, a positive result on a condition affecting 1 in 100 people leaves you only about 50% likely to be ill. Push prevalence down to 1 in 1000 and the answer collapses to roughly 9%.
The reason is in the two blocks of the bar: the healthy group is so much larger that its 1% error rate still produces more positives than the ill group produces in total.
\(P(\text{ill}\mid +)\) is not \(P(+\mid \text{ill})\). Swapping them is the single most common probability mistake outside this classroom, and Bayes is what keeps them apart.
Example
Let’s verify now if the event \(A_1\) and \(B^c\) are independent or not!
According to the definition of independence: \(P(A_1\cap B^c)=P(A_1)P(B^c)\)
- \(P(A_1\cap B^c)=P(A_1)\times P(B^c|A_1)=0.1\times 0.3=0.03\)
- \(P(A_1)\times P(B^c)=0.1\times(1-0.565)=0.0435\)
- Then: \[P(A_1\cap B^c)=P(A_1)\times P(B^c)\Leftrightarrow 0.03\neq 0.0435\]
Then \(A_1\) and \(B^c\) are not independent.
❓ Axioms, Independence and Bayes · Question 1
\(A\) and \(B\) are incompatible, with \(P(A)>0\) and \(P(B)>0\). Then they are:
A. independent
B. independent only if disjoint
C. never independent
D. impossible to classify
✅ C. Incompatible means \(P(A\cap B)=0\), but independence needs \(P(A\cap B)=P(A)P(B)>0\). The two cannot hold together.
❓ Axioms, Independence and Bayes · Question 2
\(P(A)=0.5\), \(P(B)=0.4\) and \(P(A\cap B)=0.2\). Then \(A\) and \(B\) are:
A. incompatible
B. neither
C. independent
D. both
✅ C. \(P(A)P(B)=0.5\times 0.4=0.2=P(A\cap B)\) ✅
✏️ Axioms, Independence and Bayes · Question 3
A disease affects 2% of the population. A test is positive for 95% of ill people, and also for 5% of healthy people.
A randomly chosen person tests positive. What is the probability that this person is ill?
✅ Axioms, Independence and Bayes · Solution
Let \(D\) be “ill” and \(+\) be “tests positive”. By the Law of Total Probability:
\[P(+)=0.95\times 0.02 + 0.05\times 0.98 = 0.019+0.049=0.068\]
\[P(D|+)=\frac{P(+|D)P(D)}{P(+)}=\frac{0.019}{0.068}\approx 0.279\]
Only about 28%, even though the test is right 95% of the time. The disease is rare, so most positives come from the large healthy group.
Practice
Practice
The exercises go from easy to hard:
- 🟢 warm-up: one formula, one step
- 🟡 standard: two or three steps
- 🟠 exam level: this is what you will find in the midterm
- 🔴 stretch: you need to combine several things
🟢 Practice · Exercise 1
Let \(A\) and \(B\) be two events with \(0<P(A)<1\) and \(0<P(B)<1\). You know that \(A\subset B\).
What is \(P[(A\cup B)\cap B^c]\)?
A. \(0\)
B. \(P(A)\)
C. \(P(B)\)
D. \(1\)
✅ A. If \(A\subset B\) then \(A\cup B=B\), and \(B\cap B^c=\emptyset\). Draw the Venn diagram if you doubt it.
🟢 Practice · Exercise 2
Let \(A\) and \(B\) be two events such that \(P(A)=0.75\), \(P(B)=0.5\) and \(P(A\cup B)=1\).
Find \(P(A|B)\).
\(P(A\cap B)=P(A)+P(B)-P(A\cup B)=0.75+0.5-1=0.25\)
\(P(A|B)=\frac{P(A\cap B)}{P(B)}=\frac{0.25}{0.5}=0.5\)
🟡 Practice · Exercise 3
Let \(P(A)=a\) and \(P(B)=b\), with \(0<a<1\) and \(0<b<1\). True or false, the probability that neither \(A\) nor \(B\) happen is:
(a) \(1-a-b+ab\), if \(A\) and \(B\) are independent.
(b) \(1\), if \(A\) and \(B\) are complementary.
(a) ✅ True. \(P(A^c\cap B^c)=P(A^c)P(B^c)=(1-a)(1-b)=1-a-b+ab\), since the complements of independent events are also independent.
(b) ❌ False. If they are complementary, one of them always happens, so the probability that none happens is \(0\).
🟡 Practice · Exercise 4
Let \(A,B,C\) be events such that:
\(A\cup B\cup C=\Omega\), \(P(A)=0.3\), \(P(C)=0.5\), \(P(B^c)=0.7\), \(A\cap B=\emptyset\) and \(B\cap C=\emptyset\).
Find \(P(A\cap C)\).
\(P(B)=1-0.7=0.3\). \(B\) does not intersect \(A\) nor \(C\), so all the intersections with \(B\) vanish:
\[1=P(A\cup B\cup C)=P(A)+P(B)+P(C)-P(A\cap C)\]
\[1=0.3+0.3+0.5-P(A\cap C)\Rightarrow P(A\cap C)=0.1\]
🟠 Practice · Exercise 5
\(A\) and \(B\) are independent, and \(A\) has a probability twice as large as \(B\). The probability that at least one of them happens is \(0.5\).
Find \(P(B)\).
Let \(p=P(B)\), so \(P(A)=2p\) and, by independence, \(P(A\cap B)=2p^2\):
\[0.5=2p+p-2p^2 \Leftrightarrow 2p^2-3p+0.5=0 \Leftrightarrow p=\frac{3\pm\sqrt{5}}{4}\]
\(\frac{3+\sqrt{5}}{4}\approx 1.31\) is not a probability! So \(P(B)=\frac{3-\sqrt{5}}{4}\approx 0.191\) and \(P(A)\approx 0.382\).
🟠 Practice · Exercise 6
5% of the students are excluded from continuous assessment 📚, and of these, 98% end up with a final grade above 13. Of those not excluded, 10% end up with a final grade below 13.
Pick a student randomly at the end of the semester.
(a) What is the probability of a grade above 13, given the student was in continuous assessment?
(b) Knowing the student got a grade above 13, what is the probability that the student was excluded?
✅ Practice · Exercise 6 · Solution
Let \(E\) be “excluded” and \(G\) “grade above 13”. \(P(E)=0.05\), \(P(G|E)=0.98\), \(P(G^c|E^c)=0.1\).
(a) \(P(G|E^c)=1-0.1=0.90\)
(b) By the Law of Total Probability, \(P(G)=0.05\times 0.98+0.95\times 0.90=0.049+0.855=0.904\)
\[P(E|G)=\frac{P(G|E)P(E)}{P(G)}=\frac{0.049}{0.904}\approx 0.0542\]
📝 Homework
Problem set 2.1, Questions 3 to 10. Questions 8 to 10 are Bayes, like Exercise 6, and we will come back to them before the midterm.
Bibliography
- Murteira, B.; Ribeiro C.; Silva, J. and Pimenta, C. (2010) Introdução à Estatística (2a Edição). McGraw-hill.
- Paulino C.D.; Branco J.A. (2005). Exercícios de Probabilidade e Estatística. Escolar Editora.
- Pedrosa, A.; Gama, S. (2004). Introdução Computacional à Probabilidade e Estatística. Porto Editora.