Probability theory. Bayes formula

Let some experiment be conducted.

$$$w_1, ..., w_N$- elementary events (elementary outcomes of an experiment).
$$$\ Omega = \ {w_i \} _ {i = 1} ^ N$ - the space of elementary events (the set of all possible elementary outcomes of the experiment).

Definition 1:

Set system $$$\ Sigma$is called a sigma algebra if the following properties are satisfied:

1. $$$\ Omega \ in \ Sigma;$
2. $$$A \ in \ Sigma \ Rightarrow \ overline {A} \ in \ Sigma;$
3. $$$A_1, A_2, ... \ in \ Sigma \ Rightarrow \ bigcup \ limits_ {i = 1} ^ \ infty A_i \ in \ Sigma.$

From properties 1 and 2 of Definition 1 it follows that $$$\ emptyset \ in \ Sigma$. From properties 2 and 3 of Definition 1 it follows that $$$\ bigcap \ limits_ {i = 1} ^ \ infty A_i \ in \ Sigma \ space ($because $$$A_i \ in \ Sigma \ Rightarrow_ {St. 3} \ overline {A_i} \ in \ Sigma \ Rightarrow_ {St. 3} \ bigcup \ limits_ {i = 1} ^ \ infty \ overline {A_i} \ in \ Sigma \ Rightarrow_ {sv.2} \\ \ Rightarrow_ {sv.2} \ overline {\ bigcup \ limits_ {i = 1} ^ \ infty \ overline {A_i}} \ in \ Sigma \ Rightarrow \ bigcap \ limits_ {i = 1} ^ \ infty A_i \ in \ Sigma).$

Definition 2:

• $$$A$- event $$$\ forall A \ in \ Sigma;$
• $$$P \ colon \ Sigma \ to \ mathbb R$- probabilistic measure (probability) if:
  
  1. $$$P (\ Sigma) = 1;$
  2. $$$\ forall A \ in \ Sigma \ space \ space P (A) \ geqslant 0;$
  3. $$$\ {A_i \} _ {i = 1} ^ \ infty, \ space A_i \ in \ Sigma, \ space A_i \ cap A_j = \ emptyset$ at $$$i \ not = j \ Rightarrow P (\ bigcup \ limits_ {i = 1} ^ \ infty A_i) = \ sum \ limits_ {i = 1} ^ \ infty P (A_i).$

Probability Properties:

1. $$$P (A) \ leqslant 1;$
2. $$$P (A) = 1-P (\ overline {A});$
3. $$$P (\ emptyset) = 0;$
4. $$$A \ subseteq B \ Rightarrow P (A) \ leqslant P (B);$
5. $$$P (A \ cup B) = P (A) + P (B) -P (A \ cap B);$
6. $$$\ forall \ {A_i \} _ {i = 1} ^ N \\ \ space \ space P (\ bigcup \ limits_ {i = 1} ^ N A_i) = \ sum \ limits_ {i = 1} ^ NP ( A_i) - \ sum \ limits_ {i <j} P (A_i \ cap A_j) + \ sum \ limits_ {i <j <k} P (A_i \ cap A_j \ cap A_k) -... + \\ + ( -1) ^ {n-1} P (A_1 \ cap A_2 \ cap ... \ cap A_n);$
7. $$$\ forall \ {A_i \} _ {i = 1} ^ \ infty \ colon (A_ {i + 1} \ subseteq A_i, \ space \ bigcap \ limits_ {i = 1} ^ \ infty A_i = \ emptyset) \ space \ space \ space \ lim \ limits_ {i \ to \ infty} P (A_i) = 0.$

Definition 3:

$$$(\ Omega, \ Sigma, P)$- probability space .

Definition 4:

$$$\ forall A, B \ in \ Sigma: P (B)> 0$
$$$\ qquad P (A | B) = \ frac {P (AB)} {P (B)}$- conditional probability of an event $$$A$subject to the event $$$B$.

Definition 5:

Let for $$$\ {A_i \} _ {i = 1} ^ N$ where $$$\ forall i \ in \ overline {1, N} A_i \ in \ Sigma$, performed $$$\ forall i, j \ in \ overline {1, N} \ space A_i \ cap A_j = \ emptyset$and $$$\ bigcup \ limits_ {i = 1} ^ N A_i = \ Omega$. Then $$$\ {A_i \} _ {i = 1} ^ N$ called a partition of the space of elementary events.

Theorem 1 (total probability formula):

$$$\ {A_i \} _ {i = 1} ^ N$ - partition of the space of elementary events, $$$\ forall i \ in \ overline {1, N} \ space P (A_i)> 0$.
Then $$$\ forall B \ in \ Sigma \ quad P (B) = \ sum \ limits_ {i = 1} ^ NP (B | A_i) P (A_i)$.

Theorem 2 (Bayes formula):

$$$\ {A_i \} _ {i = 1} ^ N$ - partition of the space of elementary events, $$$\ forall i \ in \ overline {1, N} \ space P (A_i)> 0$.

Then $$$\ forall B \ in \ Sigma \ colon P (B)> 0 \ quad P (A_i | B) = \ frac {P (B | A_i) P (A_i)} {\ sum \ limits_ {i = 1} ^ NP (B | A_i) P (A_i)} = \ frac {P (B | A_i) P (A_i)} {P (B)}$.

Using the Bayes formula, we can overestimate the a priori probabilities ( $$$P (A_i)$) based on observations ( $$$P (B | A_i)$), and get a whole new understanding of reality.

An example :

Suppose that there is a test that is applied to a person individually and determines whether he is infected with the “X” virus or not? We assume that the test was successful if it delivered the correct verdict for a particular person. It is known that this test has a probability of success of 0.95, and 0.05 is the probability of both errors of the first kind (false positive, i.e. the test passed a positive verdict, and the person is healthy), and errors of the second kind (false negative, i.e. the test passed a negative verdict, and the person is sick). For clarity, a positive verdict = test “said” that a person is infected with a virus. It is also known that 1% of the population is infected with this virus. Let some person get a positive verdict of the test. How likely is he really sick?

Denote: $$$t$- test result, $$$d$- the presence of the virus. Then according to the formula for total probability:

$$

$$P (t = 1) = P (t = 1 | d = 1) P (d = 1) + P (t = 1 | d = 0) P (d = 0).$$

By Bayes theorem:

$$

$$P (d = 1 | t = 1) = \ frac {P (t = 1 | d = 1) P (d = 1)} {P (t = 1 | d = 1) P (d = 1) + P (t = 1 | d = 0) P (d = 0)} = \\ = \ frac {0.95 \ times0.01} {0.95 \ times0.01 + 0.05 \ times0.99} = 0.16$$

It turns out that the probability of being infected with the "X" virus, subject to a positive test verdict, is 0.16. Why such a result? Initially, a person with a probability of 0.01 is infected with the “X” virus and even with a probability of 0.05 the test will fail. That is, in the case when only 1% of the population is infected with this virus, the probability of a test error of 0.05 has a significant impact on the likelihood that a person is really sick, provided that the test gives a positive result.

Bibliography:

• “Fundamentals of probability theory. Textbook ", M.E. Zhukovsky, I.V. Rodionov, Moscow Institute of Physics and Technology, MOSCOW, 2015;
• “Deep learning. Immersion in the world of neural networks ”, S. Nikulenko, A. Kadurin, E. Arkhangelskaya, PETER, 2018.