Introductory Probability
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1 Introductory Probability Conditional Probability: Bayes Probabilities Nicholas Nguyen Department of Mathematics UK
2 Agenda Computing Bayes Probabilities Conditional Probability and Independence on the Real Line Announcement: The fth homework is due this Saturday. There is a quiz this Friday.
3 Reminder The conditional probability that F occurs given that E occurs is P(F E) = P(E F ), P(E) so P(E F ) = P(F E) P(E).
4 Three Dice We have a choice of 3 fair dice, with 4, 8, and 10 sides. A two-stage experiment consists of picking a die, and rolling it. Let D record the die chosen (4, 8, or 10 sided) and R record the roll (1-10). Suppose P(D = 4) = P(D = 8) = P(D = 10) = 1/3, and given an n-sided die, for any natural number j, { 1/n, j n P(R = j D = n) = 0, j > n. Let us nd P(D = 8 R = 5) = P(D = 8 and R = 5). P(R = 5)
5 Three Dice We have a choice of 3 fair dice, with 4, 8, and 10 sides. A two-stage experiment consists of picking a die, and rolling it. Let D record the die chosen (4, 8, or 10 sided) and R record the roll (1-10). Suppose P(D = 4) = P(D = 8) = P(D = 10) = 1/3, and given an n-sided die, for any natural number j, { 1/n, j n P(R = j D = n) = 0, j > n. Let us nd P(D = 8 R = 5) = P(D = 8 and R = 5). P(R = 5)
6 Three Dice We have a choice of 3 fair dice, with 4, 8, and 10 sides. A two-stage experiment consists of picking a die, and rolling it. Let D record the die chosen (4, 8, or 10 sided) and R record the roll (1-10). Suppose P(D = 4) = P(D = 8) = P(D = 10) = 1/3, and given an n-sided die, for any natural number j, { 1/n, j n P(R = j D = n) = 0, j > n. Let us nd P(D = 8 R = 5)= P(D = 8 and R = 5). P(R = 5)
7 Three Dice We have a choice of 3 fair dice, with 4, 8, and 10 sides. A two-stage experiment consists of picking a die, and rolling it. Let D record the die chosen (4, 8, or 10 sided) and R record the roll (1-10). Suppose P(D = 4) = P(D = 8) = P(D = 10) = 1/3, and given an n-sided die, for any natural number j, { 1/n, j n P(R = j D = n) = 0, j > n. Let us nd P(D = 8 R = 5) = P(D = 8 and R = 5). P(R = 5)
8 We have a choice of 3 fair dice, with 4, 8, and 10 sides. A two-stage experiment consists of picking a die, and rolling it. Let D record the die chosen (4, 8, or 10 sided) and R record the roll (1-10). Suppose P(D = 4) = P(D = 8) = P(D = 10) = 1/3, and given an n-sided die, for any natural number j, { 1/n, j n P(R = j D = n) = 0, j > n. Let us nd P(D = 8 R = 5) = P(D = 8 and R = 5). P(R = 5) Find P(R = 5) using cases based on the die picked: add up probabilities for picking die D AND rolling R = 5.
9 Cases Details If we roll a 5, then either: we picked the 4-sided die and rolled a 5 on it, or we picked the 8-sided die and rolled a 5 on it, or we picked the 10-sided die and rolled a 5 on it. Hence P(R = 5) = P(D = 4 and R = 5) + P(D = 8 and R = 5) + P(D = 10 and R = 5) = P(D = n and R = 5) n=4,8,10
10 Let D record the die chosen (4, 8, or 10 sided) and R record the roll (1-10). Suppose P(D = 4) = P(D = 8) = P(D = 10) = 1/3, and given an n-sided die, for any natural number j, { 1/n, j n P(R = j D = n) = 0, j > n. Let us nd P(D = 8 R = 5) = P(D = 8 and R = 5). P(R = 5) Find P(R = 5) using cases based on the die picked: add up probabilities for picking die D AND rolling R = 5. Start with D = 8: P(D = 8 and R = 5) = P(R = 5 D = 8) P(D = 8)
11 P(D = 4) = P(D = 8) = P(D = 10) = 1/3, and given an n-sided die, for any natural number j, { 1/n, j n P(R = j D = n) = 0, j > n. Let us nd P(D = 8 R = 5) = P(D = 8 and R = 5). P(R = 5) Find P(R = 5) using cases based on the die picked: add up probabilities for picking die D AND rolling R = 5. Start with D = 8: P(D = 8 and R = 5) = P(R = 5 D = 8) P(D = 8) = (1/8)(1/3) = 1/24.
12 Let us nd P(D = 8 R = 5) = P(D = 8 and R = 5). P(R = 5) Find P(R = 5) using cases based on the die picked: add up probabilities for picking die D AND rolling R = 5. Start with D = 8: P(D = 8 and R = 5) = P(R = 5 D = 8) P(D = 8) = (1/8)(1/3) = 1/24. Similarly, P(D = 4 and R = 5) = 0. P(D = 10 and R = 5) = 1/30.
13 Intersection Details The 4-sided die cannot roll a 5, so P(R = 5 D = 4) = 0, not 1/4. Hence P(D = 4 and R = 5) P(D = 4) = P(R = 5 D = 4), so P(D = 4 and R = 5) = P(R = 5 D = 4) P(D = 4) = (0)(1/3) = 0. Each number on the 10-sided die is equally likely - P(R = 5 D = 10) = 1/10 in particular. P(D = 10 and R = 5) P(D = 10) = P(R = 5 D = 10), so P(D = 10 and R = 5) = P(R = 5 D = 10) P(D = 10) = (1/10)(1/3) = 1/30.
14 Find P(R = 5) using cases based on the die picked: add up probabilities for picking die D AND rolling R = 5. Start with D = 8: P(D = 8 and R = 5) = P(R = 5 D = 8) P(D = 8) = (1/8)(1/3) = 1/24. Similarly, P(D = 4 and R = 5) = 0. Hence P(R = 5) is the sum P(D = 10 and R = 5) = 1/30. P(D = n and R = 5) = = 9 120, n=4,8,10
15 P(D = 8 and R = 5) = P(R = 5 D = 8) P(D = 8) Similarly, Hence P(R = 5) is the sum so = (1/8)(1/3) = 1/24. P(D = 4 and R = 5) = 0. P(D = 10 and R = 5) = 1/30. P(D = n and R = 5) = = 9 120, n=4,8,10 P(D = 8 R = 5) = P(D = 8 and R = 5) P(R = 5) = 1/24 9/120 = 5/9.
16 Similarly, P(D = 4 and R = 5) = 0. Hence P(R = 5) is the sum so P(D = 10 and R = 5) = 1/30. P(D = n and R = 5) = = 9 120, n=4,8,10 P(D = 8 R = 5) = Similarly, P(D = 8 and R = 5) P(R = 5) P(D = 4 R = 5) = 0 9/120 = 0, P(D = 10 R = 5) = 1/30 9/120 = 4/9. = 1/24 9/120 = 5/9.
17 Bayes' Formula (Variable Form) Let X 1 record the outcome of the rst stage, and suppose it can be one of these m outcomes: h 1,h 2,...,h m. Let X 2 record the outcome of the second stage. Suppose that X 2 = ω for some outcome. Then P(X 1 = h i X 2 = ω) = P(X 1 = h i and X 2 = ω) P(X 2 = ω) P(X 1 = h i and X 2 = ω) = m P(X 1 = h k and X 2 = ω) k=1 P(X 1 = h i and X 2 = ω) = m P(X 1 = h k ) P(X 2 = ω X 1 = h k ) k=1 P(X 1 = h i ) P(X 2 = ω X 1 = h i ) = m. P(X 1 = h k ) P(X 2 = ω X 1 = h k ) k=1
18 Continuous Conditional Probability Let Ω be a continuous sample space, and E be an event with P(E) 0. For any event F of Ω, the conditional probability that F occurs given that E occurs is P(F E) = P(F E). P(E) For any x in Ω, if f is the density function on Ω, then the conditional density function on E is { f (x)/p(e), x in E f (x E) = 0 x not in E.
19 Continuous Conditional Probability Let Ω be a continuous sample space, and E be an event with P(E) 0. For any event F of Ω, the conditional probability that F occurs given that E occurs is P(F E) = P(F E). P(E) For any x in Ω, if f is the density function on Ω, then the conditional density function on E is { f (x)/p(e), x in E f (x E) = 0 x not in E.
20 Uniform Density Functions Let Ω = [ω 1,ω 2 ] be an interval in R, and suppose X is a continuous random variable with a uniform density function (uniform probability). Then for any real numbers a and b with ω 1 a b ω 2, P(a X b) = P([a,b]) = Let Ω = [0,1] and F = [0,0.3]. Then Let E = [0.1,0.6]. Then P(F ) = = P(E) = = b a length of [a,b] = ω 2 ω 1 length of Ω.
21 Uniform Density Functions Let Ω = [ω 1,ω 2 ] be an interval in R, and suppose X is a continuous random variable with a uniform density function (uniform probability). Then for any real numbers a and b with ω 1 a b ω 2, P(a X b) = P([a,b]) = Let Ω = [0,1] and F = [0,0.3]. Then Let E = [0.1,0.6]. Then P(F ) = = P(E) = = b a length of [a,b] = ω 2 ω 1 length of Ω.
22 Uniform Density Functions Let Ω = [ω 1,ω 2 ] be an interval in R, and suppose X is a continuous random variable with a uniform density function (uniform probability). Then for any real numbers a and b with ω 1 a b ω 2, P(a X b) = P([a,b]) = Let Ω = [0,1] and F = [0,0.3]. Then Let E = [0.1,0.6]. Then P(F ) = = P(E) = = b a length of [a,b] = ω 2 ω 1 length of Ω.
23 P(a X b) = P([a,b]) = b a length of [a,b] = ω 2 ω 1 length of Ω. Let Ω = [0,1] and F = [0,0.3]. Then Let E = [0.1,0.6]. Then Notice so P(F ) = = P(E) = = F E = [0.1,0.3], P(F E) = =
24 Let E = [0.1,0.6]. Then P(E) = = Notice so Hence F E = [0.1,0.3], P(F E) = = P(F E) = P(F E) P(E) = = 2 5.
25 Proportions of Lengths Notice F (in gold) takes up 3/10ths of the length of Ω: Here, E takes up 5/10ths of the length of Ω. Here, F E (green) takes up 2/5ths of the area of E 2 (blue and green):
26 Independent Events Let E and F be two events. Then they are independent if either: P(E F ) = P(E) or P(F E) = P(F ). One or both events has probability zero. Two events E and F are independent if and only if P(E F ) = P(E)P(F ).
27 Not Independent The events F and E are not independent: 0.3 = P(F ) P(F E) = 2/5 = 0.4. Also, P(F ) P(E) = = 0.15 P(F E) = 0.2 P(F ) P(E) P(F E).
28 Not Independent The events F and E are not independent: 0.3 = P(F ) P(F E) = 2/5 = 0.4. Also, P(F ) P(E) = = 0.15 P(F E) = 0.2 P(F ) P(E) P(F E).
29 Independent Example Let E 1 = [0,0.6] and F 1 = [0.3,0.8]. Then Then P(E 1 ) = 0.6, P(F 1 ) = 0.5, F 1 E 1 = [0.3,0.6], so P(F 1 E 1 ) = 0.3. P(F 1 E 1 ) = P(F 1 E 1 ) P(E 1 ) P(E 1 F 1 ) = P(F 1 E 1 ) P(F 1 ) so E 1 and F 1 are independent. Indeed, = = 0.5 = P(F 1), = = 0.6 = P(E 1), P(F 1 )P(E 1 ) = = 0.3 = P(F 1 E 1 ).
30 Independent Example Let E 1 = [0,0.6] and F 1 = [0.3,0.8]. Then Then P(E 1 ) = 0.6, P(F 1 ) = 0.5, F 1 E 1 = [0.3,0.6], so P(F 1 E 1 ) = 0.3. P(F 1 E 1 ) = P(F 1 E 1 ) P(E 1 ) P(E 1 F 1 ) = P(F 1 E 1 ) P(F 1 ) so E 1 and F 1 are independent. Indeed, = = 0.5 = P(F 1), = = 0.6 = P(E 1), P(F 1 )P(E 1 ) = = 0.3 = P(F 1 E 1 ).
31 Independent Example Let E 1 = [0,0.6] and F 1 = [0.3,0.8]. Then Then P(E 1 ) = 0.6, P(F 1 ) = 0.5, F 1 E 1 = [0.3,0.6], so P(F 1 E 1 ) = 0.3. P(F 1 E 1 ) = P(F 1 E 1 ) P(E 1 ) P(E 1 F 1 ) = P(F 1 E 1 ) P(F 1 ) so E 1 and F 1 are independent. Indeed, = = 0.5 = P(F 1), = = 0.6 = P(E 1), P(F 1 )P(E 1 ) = = 0.3 = P(F 1 E 1 ).
32 Conditional Density Functions The uniform density function on [0,1] is { { 1, 0 x 1 f (x) = 0, otherwise = 1/(length of [0,1]) 0 x 1 0, otherwise Since E 1 = [0.0,0.6], E 2 = [0.1,0.6], P(E 1 ) = = 0.6, and P(E ) = = 0.5, 1 0 we have { 1/0.6 = 1/(length of E 1 ), 0 x 0.6 f (x E 1 ) =, 0, otherwise { 1/0.5 = 1/(length of E 2 ), 0.1 x 0.6 f (x E 2 ) =. 0, otherwise
33 @Home: Reading Note: page numbers refer to printed version. Add 8 to get page numbers in a PDF reader. The book presents Bayes' formula in terms of events. You may want to review Chapter 2 before next class, since Section 4.2 is conditional probability with continuous random variables. You may want to brush up on multivariable functions (functions with more than one input).
34 Next Time Please read Section 4.2. We will continue this section on Friday. The fth homework is due this Saturday.
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