Download presentation

Presentation is loading. Please wait.

Published byLucia Wallman Modified over 2 years ago

1
Lecture 9 5.3 Discrete Probability

2
5.3 Bayes’ Theorem We have seen that the following holds: We can write one conditional probability in terms of the other: Bayes’ Theorem

3
5.3 Example: What is the probability that a family with 2 kids has two boys, given that they have at least one boy? (all possibilities are equally likely). S: all possibilities: {BB, GB, BG, GG}. E: family has two boys: {BB}. F: family has at least one boy: {BB, GB, BG}. E F = {BB} p(E|F) = (1/4) / (3/4) = 1/3 E BB F GB BG GG Now we compute the probability of P(F|E), what is the probability that a family with two boys has at least one boy ? P(F|E) = P(E|F) P(F) / P(E) = 1/3 * ¾ / ¼ = 1

4
5.3 Expected Values The definition of an expected value of a random variable is: This equivalent to: Example: What is the expected number of heads if we toss a fair coin n times? We know that the distribution for this experiment is the Binomial distribution:

5
5.3 Therefore we need to compute:

6
5.3 Expectation are linear: Theorem: E(X1+X2) = E(X1) + E(X2) E(aX + b) = aE(X) + b Examples: 1) Expected value for the sum of the values when a pair of dice is rolled: X1 = value of first die, X2 value of second die: E(X1+X2) = E(X1) + E(X2) = 2 * (1+2+3+4+5+6)/6 = 7. 2) Expected number of heads when a fair coin is tossed n times (see example previous slide) Xi is the outcome coin toss i. Each has a probability of p of coming up heads. linearity: E(X1+...+Xn) = E(X1) +... + E(xn) = n p.

7
5.3 More examples: A person checking out coats mixed the labels up randomly. When someone collects his coat, he checks out a coat randomly from the remaining coats. What is the expected number of correctly returned coats? There are n coats checked in. Xi = 1 of correctly returned, and 0 if wrongly returned. Since the labels are randomly permuted, E(Xi) = 1/n E(X1+...Xn) = n 1/n = 1 (independent of the number of checked in coats)

8
5.3 Geometric distribution Q: What is the distribution of waiting times until a tail comes up, when we toss a fair coin? A: Possible outcomes: T, HT, HHT, HHHT, HHHHT,.... (infinitely many possibilities) P(T) = p, P(HT) = (1-p) p, P(HHT) = (1-p)^2 p,.... geometric distribution Normalization: (matlab) X(s) = number of tosses before success.

9
5.3 Geometric Distr. Here is how you can compute the expected value of the waiting time:

10
5.3 Independence Definition: Two random variables X(s) and Y(s) on a sample space S are independent if the following holds: Examples 1) Pair of dice is rolled. X1 is value first die, X2 value second die. Are these independent? P(x1=r1) = 1/6 P(X2=r2)=1/6 P(X1=r1 AND X2=r2)=1/36 = P(X1=r1) P(X2=r2): YES independent. 2) Are X1 and X=X1+X2 independent? P(X=12) =1/36 P(X1=1)=1/6 P(X=12 AND X1=1)=0 which is not the product: P(X=12) P(X1=1)

11
5.3 Independence Theorem: If two random variables X and Y are independent over a sample space S then: E(XY)=E(X) E(Y). (proof, read book) Note1: The reverse is not true: Two random variables do not have to be independent for E(XY)=E(X)E(Y) to hold. Note2: If 2 random variables are not independent, it follows that E(XY) does not have to be equal to E(X)E(Y), although it might still happen. Example: X counts number of heads when a coin is tossed twice: P(X=0) =1/4 (TT) P(X=1)=1/2 (HT,TH) P(X=2) =1/4 (HH). E(X) = 1x½+2x1/4=1. Y counts the number of tails: E(Y)=1 as well (symmetry, switch role H,T). However, P(XY=0) = 1/2 (HH,TT) P(XY=1) =1/2 (HT,TH) E(XY) = 0x1/2 + 1x1/2=1/2

12
5.3 Variance The average of a random variable tells us noting about the spread of a probability distribution. (matlab demo) Thus we introduce the variance of a probability distribution: Definition: The variance of a random variable X over a sample space S is given by: variance standard deviation (this is the width of the distribution)

13
5.3 Variance Theorem: For independent random variables the variances add: (proof in book) Example: 1) We toss 2 coins, Xi(H)=1, Xi(T)=0. What is the STD of X=X1+X2? X1 and X2 are independent. V(X1+X2)=V(X1)+V(X2)=2V(X1) E(X1)=1/2 V(X1) = (0-1/2)^2 x ½ + (1-1/2)^2 x ½ =1/4 V(X) = ½ STD(X)=sqrt(1/2).

14
5.3 Variance What is the variance of the number of successes when n independent Bernoulli trials are performed. V(X) = V(X1+...+Xn)=nV(X1) V(X1) = (0-p)^2 x (1-p) + (1-p)^2 x p = p^2(1-p) + p(1-p)^2=p(1-p) V(X)=np(1-p) (matlab demo)

Similar presentations

Presentation is loading. Please wait....

OK

Probability Distributions

Probability Distributions

© 2017 SlidePlayer.com Inc.

All rights reserved.

Ads by Google

Ppt on food subsidy in india Ppt on types of web browser Ppt on intellectual property in vlsi Ppt on charge coupled device detector Ppt on amplitude shift keying pdf Ppt on do's and don'ts of group discussion ideas Ppt on global marketing strategies Project ppt on student information system Ppt on law against child marriage in nigeria Ppt online viewer for word