Chapter 4 Random Variables - 1. Outline Random variables Discrete random variables Expected value 2.

Chapter 4 Random Variables - 1

Outline

• Random variables

• Discrete random variables

• Expected value

2

Sample Space and Events• Genes relating to albinism are denoted by A and a. Only those

people who receive the a gene from both parents will be albino. Persons having the gene pair A, a are normal in appearance and, because they can pass on the trait to their offspring, are called carriers. Suppose that a normal couple has two children, exactly one of whom is an albino. Suppose that the nonalbino child mates with a person who is known to be a carrier for albinism.(a) What is the probability that their first offspring is an albino?(b) What is the conditional probability that their second offspring is an albino given that their firstborn is not?

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Sample space

• If we measure the appearance (phenotype), then the sample space has two outcomes {albino, nonalbino}.

• If we measure the genes (genotype), then the sample space has three outcomes {(A,A), (A,a), (a,a)}.

• If we measure the phenotypes of two chirldren, then the sample space has four outcomes {(a,n), (a,a), (n,a), (n,n)}.

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Events• We may be only interested in, for example, whether a

person is albino or nonalbino even we measured the genotypes. When we know a person is nonalbino, we may be interested in whether he/she is a carrier. In these situations, we need to define events to help us to solve our problems.– Event E: the child is nonalbino, E = {(A,A),(A,a)}– Event F: the child is albino, F = {(a,a)}– Event G: the child is a carrier, G = {(A,a)}– Event Hi: the i-th offspring is an albino, H1: the 1st

offspring is an albino– (a) P(H1) = ?– (b) P(H2|H1

c) = ? 5

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S E

(A,A)(A,a) (a,a)

Albino, Nonalbino

Random Variables

• Some times events are not very convenient to use.– In an exam with ten problems of ten points each, the

student may be only interested in how many problems he/she answered correctly.

– In a clinical trial of certain new treatment, the proportion of patients cured.

• We are interested in some functions of the outcome of an experiment.

• Random variables– Real valued functions defined on sample space.

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Flip a coin three times and record the flips. Then S = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}. Define a function X on S by X(s) = “the # of heads in the three flips.” So X(HHH) = 3, X(HTT) = 1, X(THT) = 1, and X(TTT) = 0. The range of the random variable X is {0, 1, 2, 3}. P(X=0) + P(X=1) + P(X=2) + P(X=3) = 1/8 + 3/8 + 3/8 + 1/8 = 1

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9

S RV

HHH, HHT, HTH, THH,HTT, THT, TTH, TTT

0, 1, 2, 3

• Roll a red die and a clear die and record the two rolls. Sample space is S = {(1,1), (1,2),…,(6,6)} with 36 elements. There are many random variables definable on S.– Define X(r,c) = r+c. That is, X of a pair is the sum of the

numbers. So X(1,5) = 6 and X(6,6) = 12. Then the range of X is {2,3,…,12}.

– Define Y(r,c) = max{r,c}. So Y(1,5) = 5 and Y(6,6) = 6. Then the range of Y is {1,2,3,4,5,6}.

– Define Z(r,c) = r-c. So Z(1,5) = -4 and Z(6,6)=0. Then the range of Z is {-5, -4,…, 4, 5}.

– Define W(r,c) = r/c. So W(1,5) = 1/5 and W(6,6) = 1. Then the range of W is {1, 2, 3, 4, 5, 6, 1/2, 3/2, 5/2, 1/3, 2/3, 4/3, 5/3, 1/4, 3/4, 5/4, 1/5, 2/5, 3/5, 4/5, 6/5, 1/6, 5/6}.

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• Ex 1d. Three balls are randomly chosen from an urn containing 3 white, 3 red, and 5 black balls. Suppose that we win $1 for each white ball selected and lose $1 for each red selected. If we let X denote our total winning from the experiment, then X is a random variable taking on the possible values 0, ±1, ±2, ±3 with respective probabilities

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Benefits of Using Random Variables

• The concept of random variables helps us to formulate our problems better.

• It allows us to apply all our knowledge of functions to the study of probability.

• Also, it strips away real world differences to reveal situations that are probabilistically identical. – For instance, if X is the number of heads in three flips of a

coin and Y is the number of girls in a family with three children, then X and Y are probabilistically identical (assuming the probability is 1/2 of each child being a girl) even though the physical situations they model are quite different.

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Discrete Random Variables• A random variable is discrete if it can take on

at most a countable (or countably infinite) number of possible values.

• Countably infinite– it is possible to make a list of the elements even

though they are infinite. For instance• the set of positive even numbers is countably infinite: 2,

4, 6, 8,…. • The set of positive numbers of no more than two

decimal digits is countably infinite: 0.01, 0.02,…, 0.99, 1.00, 1.01, ….

– An interval like [0,1] is uncountably infinite; it is impossible to make a list – even an infinite list – of the real numbers between zero and one.

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Discrete Random Variables• ExampleFlip a coin repeatedly until it shows heads. Record the flips in order. So S = {H, TH, TTH, TTTH, TTTTH,…}. Define a function X on S by X(s) = “the # of flips in the outcomes.” So X(H) = 1, X(TH) = 2, and X(TTTTH) =5. The range of X is {1, 2, 3, 4, …} = “the set of positive integers.” This set is countably infinite, so X is a discrete random variable.

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Probability Density Functions (pdf) • Given a discrete random variable X with sample

space S, there is an associated probability density function (probability mass function) p defined by p(a) = P{X=a}. That is, the pdf tells us the probability of each particular value of x occurring. If x is not in the range of p, then p(x)=0.

• The sum of the values of p(x) for all x’s in the range of X must be 1.

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Probability Density Functions (pdf) • Flip a coin and record the result. S = {H,T}.

Define a random variable X on S by X(H) = 1 and X(T) = 0. The range of X is {0,1}. The pdf for X is given by p(0) = P(X=0) = P({T}) = 1/2 and p(1) = P(X=1) = P({H}) = 1/2.

• Flip a coin three times and record the results. S = {TTT, TTH, THT, HTT, HHT, HTH, THH, HHH,}. Define X(s) to be the number of heads in the three flips. The range of X is {0,1,2,3}. The values of the pdf for this random variable are p(0) = 1/8, p(1) = 3/8, p(2) = 3/8, p(3) = 1/8. For instance, p(1) = P(X=1) = P({TTH, THT, HTT}) = 3/8.

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18

S RV

HHH, HHT, HTH, THH,HTT, THT, TTH, TTT

0, 1, 2, 3

PDF

0.125, 0.375

Probability Density Functions (pdf) • Graph the pdf for

flipping a coin three times using a “discrete density graph” or a histogram. We can also display them using tables.

PDF for # of Heads in 3 f lips of a Coin

1/8

3/8 3/8

1/8

0

0.25

0.5

0.75

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0 1 2 3 4

# of Heads

Prob

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ty

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Probability Density Functions (pdf) PDF for # of Heads in 3 f lips of a Coin

1/8

3/8 3/8

1/8

0

0.25

0.5

0.75

1

# of Heads

Prob

abili

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PDF for # of Heads in 3 f lips of aCoin

1/8 3/8 3/8 1/8

0 1 2 3

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Cumulative Distribution Functions (cdf)

• In many practical situations, we are interested in whether the value of a random variable is smaller (or greater) than a certain number.– The hurricane is less than certain strength, the earthquake is less

than certain level, the risk of losing money in an investment is less than certain percentage, etc…

• Cumulative distribution function is defined as F(x) = P(X<=x)

• Every cdf is an increasing function. Its limit at negative infinity (to the left) is 0 and its limit at positive infinity (to the right) is 1.

• Once you know the cdf, you can easily find almost any probability that interests you.

• If the random variable X is discrete, then the cdf is a step function.

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Cumulative Distribution Functions Cumulative distribution function for the random variable X that counts the heads in three coin flips. Note that it jumps at every value in the range of X.

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CDF for # of Heads in 3 f lips of a Coin

1/8

4/8

7/8

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0.25

0.5

0.75

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-2 -1 0 1 2 3 4

# of Heads

Prob

abili

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CDF ExampleFlip a coin repeatedly until it shows heads. Record the flips in order. So S = {H, TH, TTH, TTTH, TTTTH,…}. X: the number of flips in the outcomes. So X(H) = 1, X(TH) = 2, and X(TTTTH) =5. The range of X is {1, 2, 3, 4, …} = “the set of positive integers.” What does the CDF of X look like?

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Play or Not to Play?

• Suppose I want to play a game with you. In each round, I toss two coins and pay you $3 if there are two heads, $2.5 if there are two tails, and $2 if there are one head and one tail. We play this for 100 rounds.

• If I charge you $2.4 for each round. Do you want to play the game or not?

• What you really care is the average of the money you make in each round.

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Expected Value• Expected value of X is a weighted average of the

possible values that X can take on, each value being weighted by the probability that X assumes it.

• E[X] = Σxp(x)• Expected value is what one should expect if the

experiment is repeated many times.• p(0) = 1/2, p(1) = 1/2

– E[X] = 0(1/2) + 1 (1/2) = 1/2• p(0) = 1/3, p(1) = 2/3

– E[X] = 0(1/3) + 1(2/3) = 2/325

Play or Not to Play• X: amount of dollars one makes in a single

round.– A random variable of the outcome of tossing two

coins.• x = X(two heads) = 3, x = X(one head, one tail)

= 2, x = X(two tails) = 2.5.• E[X] = 3×p(x=3) + 2 ×p(x=2) + 2.5×p(x=2.5)

= 3×(1/4) + 2 × (1/2) + 2.5×(1/4) = 2.375You are charged $2.4 for each round...

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• Ex 3b. I is an indicator variable for the event A if

Find E[I]p(1) = P(A), p(0) = 1-P(A), we haveE[I] = 1×P(A) + 0×(1-P(A))The expected value of the indicator variable for the event A is equal to the probability that A occurs.

occurs if 0occurs if 1

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• Ex 3d. A school class of 120 students are driven in 3 buses to a symphonic performance. There are 36 students in one of the buses, 40 in another, and 44 in the third bus. When the buses arrive, one of the 120 students is randomly chosen. Let X denote the number of students on the bus of that randomly chosen student, find E[X].

• The range of X is {36, 40, 44}.• P(X=36) = 36/120, P(X=40) = 40/120, P(X=44) =

44/120• E[X] = 36(36/120) + 40(40/120) + 44(44/120) =

40.266728

Two Envelop Problem

• Let's say you are given two indistinguishable envelopes, each of which contains a positive sum of money. One envelope contains twice as much as the other. You may pick one envelope and keep whatever amount it contains. You pick one envelope at random but before you open it you're offered the possibility to take the other envelope instead.

• What will you do?

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The Switching Argument• Let’s denote by A the amount in your selected envelope.• The probability that A is the smaller amount is 1/2, and that it's

the larger also 1/2 • The other envelope may contain either 2A or A/2

– If A is the smaller amount the other envelope contains 2A – If A is the larger amount the other envelope contains A/2– The other envelope contains 2A with probability 1/2 and A/2 with

probability 1/2

• So the expected value of the money in the other envelope is:

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The Contradiction

• Since the expected value in the other envelop is larger than what is in the envelop you selected, you should switch.

• After the switch one can reason in exactly the same manner as above – The most rational thing to do is to switch back again – To be rational one will thus end up switching envelopes

indefinitely • As it seems more rational to open just any envelope

than to switch indefinitely we have a contradiction! 31