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154 CHAPTER 4. CONDITIONAL PROBABILITY man with mustache 1/4 girl with blond hair 1/3 girl with ponytail 1/10 black man with beard 1/10 interracial couple in a car 1/1000 partly yellow car 1/10 Table 4.5: Collins case probabilities. If you were the lawyer for the Collins couple how would you have countered the above ar... | prob_Page_162_Chunk5001 |
4.1. DISCRETE CONDITIONAL PROBABILITY 155 Woodstock Tunbridge p q C D T W .8 .9 .9 .8 .95 W T p q (a) (b) (c) Figure 4.6: From Woodstock to Tunbridge. (c) Now suppose W and T are connected like the bottom graph in Figure 4.6. Find the probability of Helen’s getting from W to T. Hint: If the road from C to D is impassab... | prob_Page_163_Chunk5002 |
156 CHAPTER 4. CONDITIONAL PROBABILITY Y -1 0 1 2 X -1 0 1/36 1/6 1/12 0 1/18 0 1/18 0 1 0 1/36 1/6 1/12 2 1/12 0 1/12 1/6 Table 4.6: Joint distribution. 36 A die is thrown twice. Let X1 and X2 denote the outcomes. Define X = min(X1, X2). Find the distribution of X. *37 Given that P(X = a) = r, P(max(X, Y ) = a) = s, an... | prob_Page_164_Chunk5003 |
4.1. DISCRETE CONDITIONAL PROBABILITY 157 (a) P(r, N) = 0 if r < N, (b) P(N, s) = 1 if s < N, (c) P(r, s) = pP(r + 1, s) + qP(r, s + 1) if r < N and s < N; and (1), (2), and (3) determine P(r, s) for r ≤N and s ≤N. Pascal used these facts to find P(r, s) by working backward: He first obtained P(N −1, j) for j = N −1, N −... | prob_Page_165_Chunk5004 |
158 CHAPTER 4. CONDITIONAL PROBABILITY is now the person, in the extended game, who wins the most points. Show that playing these additional points has not changed the winner. (c) Show that (a) and (b) prove that you have the same probability of win- ning the game under either convention. 45 In the previous problem, as... | prob_Page_166_Chunk5005 |
4.1. DISCRETE CONDITIONAL PROBABILITY 159 49 You are given two urns each containing two biased coins. The coins in urn I come up heads with probability p1, and the coins in urn II come up heads with probability p2 ̸= p1. You are given a choice of (a) choosing an urn at random and tossing the two coins in this urn or (b... | prob_Page_167_Chunk5006 |
160 CHAPTER 4. CONDITIONAL PROBABILITY 54 (Suggested by Eisenberg and Ghosh12) A deck of playing cards can be de- scribed as a Cartesian product Deck = Suit × Rank , where Suit = {♣, ♦, ♥, ♠} and Rank = {2, 3, . . ., 10, J, Q, K, A}. This just means that every card may be thought of as an ordered pair like (♦, 2). By a... | prob_Page_168_Chunk5007 |
4.1. DISCRETE CONDITIONAL PROBABILITY 161 57 Prove that A neither attracts nor repels B if and only if A and B are inde- pendent. 58 Prove that A and B are mutually attractive if and only if P(B|A) > P(B| ˜A). 59 Prove that if A attracts B, then A repels ˜B. 60 Prove that if A attracts both B and C, and A repels B ∩C, ... | prob_Page_169_Chunk5008 |
162 CHAPTER 4. CONDITIONAL PROBABILITY (b) You have observed the show for a long time and found that the car is put behind door A 45% of the time, behind door B 40% of the time and behind door C 15% of the time. Assume that everything else about the show is the same. Again you pick door A. Monty opens a door with a goa... | prob_Page_170_Chunk5009 |
4.2. CONTINUOUS CONDITIONAL PROBABILITY 163 f(x|E) = 2, if 0 ≤x < 1/2, 0, if 1/2 ≤x < 1. Thus the conditional density function is nonzero only on [0, 1/2], and is uniform there. 2 Example 4.19 In the dart game (cf. Example 2.8), suppose we know that the dart lands in the upper half of the target. What is the probabil... | prob_Page_171_Chunk5010 |
164 CHAPTER 4. CONDITIONAL PROBABILITY This tells us the rather surprising fact that the probability that we have to wait s seconds more for an emission, given that there has been no emission in r seconds, is independent of the time r. This property (called the memoryless property) was introduced in Example 2.17. When ... | prob_Page_172_Chunk5011 |
4.2. CONTINUOUS CONDITIONAL PROBABILITY 165 Joint Density and Cumulative Distribution Functions In a manner analogous with discrete random variables, we can define joint density functions and cumulative distribution functions for multi-dimensional continuous random variables. Definition 4.6 Let X1, X2, . . . , Xn be cont... | prob_Page_173_Chunk5012 |
166 CHAPTER 4. CONDITIONAL PROBABILITY 1 1 r r 0 ω ω E 2 1 1 2 1 Figure 4.7: X1 and X2 are independent. Let’s look at some examples. Example 4.22 In this example, we define three random variables, X1, X2, and X3. We will show that X1 and X2 are independent, and that X1 and X3 are not independent. Choose a point ω = (ω1,... | prob_Page_174_Chunk5013 |
4.2. CONTINUOUS CONDITIONAL PROBABILITY 167 1 1 0 ω ω ω + ω = 1 1/2 1 2 1 2 Ε 2 Figure 4.8: X1 and X3 are not independent. = P(ω1 ≤1/2, ω1 + ω2 ≤1) = Area (E2) = 1 2 −1 8 = 3 8 . Now recalling that F3(r3) = 0, if r3 < 0, (1/2)r2 3, if 0 ≤r3 ≤1, 1 −(1/2)(2 −r3)2, if 1 ≤r3 ≤2, 1, if 2 < r3, (see Example 2.1... | prob_Page_175_Chunk5014 |
168 CHAPTER 4. CONDITIONAL PROBABILITY 0.2 0.4 0.6 0.8 1 0.5 1 1.5 2 2.5 3 α = β =.5 α = β =1 α = β = 2 0 Figure 4.9: Beta density for α = β = .5, 1, 2. Definition 4.8 A sequence X1, X2, . . . , Xn of random variables Xi that are mutually independent and have the same density is called an independent trials process. 2 A... | prob_Page_176_Chunk5015 |
4.2. CONTINUOUS CONDITIONAL PROBABILITY 169 one is, in effect, tossing a biased coin with probability x for heads. Before further experimentation, you do not know the value x but past experience might give some information about its possible values. It is natural to represent this information by sketching a density func... | prob_Page_177_Chunk5016 |
170 CHAPTER 4. CONDITIONAL PROBABILITY = xα+i−1(1 −x)β+j−1 B(α + i, β + j) , (4.5) that is, f(x|i) is another beta density. This says that if we observe i successes and j failures in n subjects, then the new density for the probability that the drug is effective is again a beta density but with parameters α + i, β + j. ... | prob_Page_178_Chunk5017 |
4.2. CONTINUOUS CONDITIONAL PROBABILITY 171 0.2 0.4 0.6 0.8 1 0.5 1 1.5 2 2.5 Machine Result 1 W 1 L 2 L 1 L 1 W 1 L 1 L 1 L 2 W 2 L 0 0 Figure 4.10: Play the best machine. choose the ith machine is p(i) = win(i) + 1 win(i) + lose(i) + 2 . Thus, if p(1) > p(2) you would play machine 1 and otherwise you would play machi... | prob_Page_179_Chunk5018 |
172 CHAPTER 4. CONDITIONAL PROBABILITY 0.2 0.4 0.6 0.8 1 0.5 1 1.5 2 Machine Result 1 W 1 W 1 L 2 L 1 W 1 W 1 L 2 L 1 L 2 W Figure 4.11: Play the winner. Neither of the strategies that we simulated is the best one in terms of maximizing our average winnings. This best strategy is very complicated but is reasonably ap- ... | prob_Page_180_Chunk5019 |
4.2. CONTINUOUS CONDITIONAL PROBABILITY 173 (a) Find the failure rate of this bulb (see Exercise 2.2.6). (b) Find the reliability of this bulb after 20 hours. (c) Given that it lasts 20 hours, find the probability that the bulb lasts another 20 hours. (d) Find the probability that the bulb burns out in the forty-first ho... | prob_Page_181_Chunk5020 |
174 CHAPTER 4. CONDITIONAL PROBABILITY (d) x + y < 1. 9 Suppose that X and Y are continuous random variables with density functions fX(x) and fY (y), respectively. Let f(x, y) denote the joint density function of (X, Y ). Show that Z ∞ −∞ f(x, y) dy = fX(x) , and Z ∞ −∞ f(x, y) dx = fY (y) . *10 In Exercise 2.2.12 you ... | prob_Page_182_Chunk5021 |
4.3. PARADOXES 175 13 Write a program to allow you to compare the strategies play-the-winner and play-the-best-machine for the two-armed bandit problem of Example 4.24. Have your program determine the initial payoffprobabilities for each machine by choosing a pair of random numbers between 0 and 1. Have your program car... | prob_Page_183_Chunk5022 |
176 CHAPTER 4. CONDITIONAL PROBABILITY First child Second child Conditional probability First child Second child Unconditional probability b g b g b g b 1/4 1/4 1/4 1/4 1/3 1/3 1/3 1/4 1/4 1/4 1/2 1/2 1/2 1/2 1/2 b g 1/2 1/2 1/2 g b Unconditional probability 1/2 1/2 1/2 Figure 4.12: Tree for Example 4.25. information. ... | prob_Page_184_Chunk5023 |
4.3. PARADOXES 177 Mr.Smith's children Walking with Mr.Smith Unconditional probability Mr.Smith's children Walking with Mr. Smith Unconditional probability b bb b g g g b 1/4 1/8 1/8 1/8 1/8 1/4 1/4 1/4 1/4 1/4 1 1/2 1/2 1/2 1/2 1 bg gb gg b bb b b 1/4 1/8 1/8 1/4 1/4 1/4 1 1/2 1/2 bg gb Conditional probability 1/2 1/4... | prob_Page_185_Chunk5024 |
178 CHAPTER 4. CONDITIONAL PROBABILITY she asks. “Yes,” she informs you with a smile. What is the probability that the other one is male? The reader is asked to decide whether the model which gives an answer of 1/3 is a reasonable one to use in this case. 2 In the preceding examples, the apparent paradoxes could easily... | prob_Page_186_Chunk5025 |
4.3. PARADOXES 179 $5 $10 $10 $5 1/2 1/2 1/2 1 1 1/2 In Ali's envelope In Baba's envelope Figure 4.14: John Finn’s version of Example 4.28. smaller amount is 1/2, and the probability that her envelope contains the larger amount is also 1/2. In fact, these conditional probabilities depend upon the distri- bution of the ... | prob_Page_187_Chunk5026 |
180 CHAPTER 4. CONDITIONAL PROBABILITY As above, we let X denote the smaller of the two amounts in the envelopes, and let px = P(X = x) . We are now in a position where we can calculate the long-term average winnings, if we switch. (This long-term average is an example of a probabilistic concept known as expectation, a... | prob_Page_188_Chunk5027 |
4.3. PARADOXES 181 it, revealing X, say. Is it possible to determine, with probability greater than 1/2, whether X is the smaller of the two dollar amounts? Even if we have no knowledge of the joint distribution of X and Y , the surprising answer is yes! Here’s how to do it. Toss a fair coin until the first time that he... | prob_Page_189_Chunk5028 |
182 CHAPTER 4. CONDITIONAL PROBABILITY (b) Suppose the person holding the hand is asked the more direct question “Do you have the ace of hearts?” and the answer is yes. What is the probability that he has a second ace? 4 Using the notation introduced in Example 4.29, show that in the example of Brams and Kilgour, if x ... | prob_Page_190_Chunk5029 |
Chapter 5 Important Distributions and Densities 5.1 Important Distributions In this chapter, we describe the discrete probability distributions and the continuous probability densities that occur most often in the analysis of experiments. We will also show how one simulates these distributions and densities on a comput... | prob_Page_191_Chunk5030 |
184 CHAPTER 5. DISTRIBUTIONS AND DENSITIES Binomial Distribution The binomial distribution with parameters n, p, and k was defined in Chapter 3. It is the distribution of the random variable which counts the number of heads which occur when a coin is tossed n times, assuming that on any one toss, the probability that a ... | prob_Page_192_Chunk5031 |
5.1. IMPORTANT DISTRIBUTIONS 185 5 10 15 20 0 0.2 0.4 p = .5 5 10 15 20 0 0.05 0.1 0.15 0.2 p = .2 Figure 5.1: Geometric distributions. The left-hand expression is just a geometric series with first term p and common ratio q, so its sum is p 1 −q which equals 1. In Figure 5.1 we have plotted this distribution using the ... | prob_Page_193_Chunk5032 |
186 CHAPTER 5. DISTRIBUTIONS AND DENSITIES Thus, Y is geometrically distributed with parameter p. To generate Y , all we have to do is solve Equation 5.1 for Y . We obtain Y = & log(1 −rnd) log q ' , where the notation ⌈x⌉means the least integer which is greater than or equal to x. Since log(1−rnd) and log(rnd) are ide... | prob_Page_194_Chunk5033 |
5.1. IMPORTANT DISTRIBUTIONS 187 let X represent the number of tosses. When k = 1, X is geometrically distributed. For a general k, we say that X has a negative binomial distribution. We now calculate the probability distribution of X. If X = x, then it must be true that there were exactly k −1 heads thrown in the first... | prob_Page_195_Chunk5034 |
188 CHAPTER 5. DISTRIBUTIONS AND DENSITIES 5 10 15 20 25 30 0 0.02 0.04 0.06 0.08 0.1 Figure 5.2: Negative binomial distribution with k = 2 and p = .25. Suppose that we have a situation in which a certain kind of occurrence happens at random over a period of time. For example, the occurrences that we are interested in ... | prob_Page_196_Chunk5035 |
5.1. IMPORTANT DISTRIBUTIONS 189 To decide upon the proper value of p, the probability of an occurrence in a given subinterval, we reason as follows. On the average, there are λt occurrences in a time interval of length t. If this time interval is divided into n subintervals, then we would expect, using the Bernoulli t... | prob_Page_197_Chunk5036 |
190 CHAPTER 5. DISTRIBUTIONS AND DENSITIES Poisson Binomial Poisson Binomial Poisson Binomial n = 100 n = 100 n = 1000 j λ = .1 p = .001 λ = 1 p = .01 λ = 10 p = .01 0 .9048 .9048 .3679 .3660 .0000 .0000 1 .0905 .0905 .3679 .3697 .0005 .0004 2 .0045 .0045 .1839 .1849 .0023 .0022 3 .0002 .0002 .0613 .0610 .0076 .0074 4 ... | prob_Page_198_Chunk5037 |
5.1. IMPORTANT DISTRIBUTIONS 191 Example 5.4 In his book,1 Feller discusses the statistics of flying bomb hits in the south of London during the Second World War. Assume that you live in a district of size 10 blocks by 10 blocks so that the total district is divided into 100 small squares. How likely is it that the squa... | prob_Page_199_Chunk5038 |
192 CHAPTER 5. DISTRIBUTIONS AND DENSITIES 0 2 4 6 8 10 0 0.05 0.1 0.15 0.2 Figure 5.3: Flying bomb hits. | prob_Page_200_Chunk5039 |
5.1. IMPORTANT DISTRIBUTIONS 193 Hypergeometric Distribution Suppose that we have a set of N balls, of which k are red and N −k are blue. We choose n of these balls, without replacement, and define X to be the number of red balls in our sample. The distribution of X is called the hypergeometric distribution. We note tha... | prob_Page_201_Chunk5040 |
194 CHAPTER 5. DISTRIBUTIONS AND DENSITIES Democrat Republican Female 24 4 28 Male 8 14 22 32 18 50 Table 5.2: Observed data. Democrat Republican Female s11 s12 t11 Male s21 s22 t12 t21 t22 n Table 5.3: General data table. nail down what is meant by “quite a bit,” we decide which possible data sets differ from the expec... | prob_Page_202_Chunk5041 |
5.1. IMPORTANT DISTRIBUTIONS 195 Democrat Republican Female 18 10 28 Male 14 8 22 32 18 50 Table 5.4: Expected data. of drawing exactly a yellow balls, i.e., what is the probability that s11 = a? It is | prob_Page_203_Chunk5042 |
196 CHAPTER 5. DISTRIBUTIONS AND DENSITIES 2 4 6 8 0 0.05 0.1 0.15 0.2 0.25 0.3 Figure 5.4: Leading digits in President Clinton’s tax returns. Theodore Hill2 gives a general description of the Benford distribution, when one considers the first d digits of integers in a data set. We will restrict our attention to the firs... | prob_Page_204_Chunk5043 |
5.1. IMPORTANT DISTRIBUTIONS 197 Exercises 1 For which of the following random variables would it be appropriate to assign a uniform distribution? (a) Let X represent the roll of one die. (b) Let X represent the number of heads obtained in three tosses of a coin. (c) A roulette wheel has 38 possible outcomes: 0, 00, an... | prob_Page_205_Chunk5044 |
198 CHAPTER 5. DISTRIBUTIONS AND DENSITIES (b) Find P(T > 3). (c) Find P(T > 6|T > 3). 8 If a coin is tossed a sequence of times, what is the probability that the first head will occur after the fifth toss, given that it has not occurred in the first two tosses? 9 A worker for the Department of Fish and Game is assigned t... | prob_Page_206_Chunk5045 |
5.1. IMPORTANT DISTRIBUTIONS 199 i.e., Y is Poisson with parameter λt. Hint: Suppose a Martian were to observe the police station. Let us also assume that the basic time interval used on Mars is exactly t Earth minutes. Finally, we will assume that the Martian understands the derivation of the Poisson distribution in t... | prob_Page_207_Chunk5046 |
200 CHAPTER 5. DISTRIBUTIONS AND DENSITIES 20 An advertiser drops 10,000 leaflets on a city which has 2000 blocks. Assume that each leaflet has an equal chance of landing on each block. What is the probability that a particular block will receive no leaflets? 21 In a class of 80 students, the professor calls on 1 student ... | prob_Page_208_Chunk5047 |
5.1. IMPORTANT DISTRIBUTIONS 201 Number of deaths Number of corps with x deaths in a given year 0 144 1 91 2 32 3 11 4 2 Table 5.5: Mule kicks. 26 Feller5 discusses the statistics of flying bomb hits in an area in the south of London during the Second World War. The area in question was divided into 24 × 24 = 576 small ... | prob_Page_209_Chunk5048 |
202 CHAPTER 5. DISTRIBUTIONS AND DENSITIES 32 It is often assumed that the auto traffic that arrives at the intersection during a unit time period has a Poisson distribution with expected value m. Assume that the number of cars X that arrive at an intersection from the north in unit time has a Poisson distribution with p... | prob_Page_210_Chunk5049 |
5.1. IMPORTANT DISTRIBUTIONS 203 tagged. Six months later 200 moose are captured and it is found that 8 of these were tagged. Estimate the number of moose on Isle Royale from these data, and then verify your guess by computer program (see Exercise 36). 38 A manufactured lot of buggy whips has 20 items, of which 5 are d... | prob_Page_211_Chunk5050 |
204 CHAPTER 5. DISTRIBUTIONS AND DENSITIES Dark Eyes Light Eyes Dark Hair 28 15 43 Light Hair 9 23 32 37 38 75 Table 5.6: Observed data. 0 10 20 30 40 0 500 1000 1500 2000 2500 3000 3500 Figure 5.5: Distribution of choices in the Powerball lottery. 45 (a) Compute the leading digits of the first 100 powers of 2, and see ... | prob_Page_212_Chunk5051 |
5.2. IMPORTANT DENSITIES 205 Integer Times Integer Times Integer Times Chosen Chosen Chosen 1 2646 2 2934 3 3352 4 3000 5 3357 6 2892 7 3657 8 3025 9 3362 10 2985 11 3138 12 3043 13 2690 14 2423 15 2556 16 2456 17 2479 18 2276 19 2304 20 1971 21 2543 22 2678 23 2729 24 2414 25 2616 26 2426 27 2381 28 2059 29 2039 30 22... | prob_Page_213_Chunk5052 |
206 CHAPTER 5. DISTRIBUTIONS AND DENSITIES 0 2 4 6 8 10 λ=1 λ=2 λ=1/2 Figure 5.6: Exponential densities. describe experiments involving a question of the form: How long until something happens? For example, the exponential density is often used to study the time between emissions of particles from a radioactive source.... | prob_Page_214_Chunk5053 |
5.2. IMPORTANT DENSITIES 207 = e−λ(r+s) e−λr = e−λs . There is a very important relationship between the exponential density and the Poisson distribution. We begin by defining X1, X2, . . . to be a sequence of independent exponentially distributed random variables with parameter λ. We might think of Xi as denoting the a... | prob_Page_215_Chunk5054 |
208 CHAPTER 5. DISTRIBUTIONS AND DENSITIES Using Corollary 5.2 (below), one can derive the above expression (see Exercise 3). We content ourselves for now with a short calculation that should convince the reader that the random variable Y has the required property. We have P(Y ≤y) = P −1 λ log(rnd) ≤y = P(log(rnd) ... | prob_Page_216_Chunk5055 |
5.2. IMPORTANT DENSITIES 209 2000 4000 6000 8000 10000 10 20 30 40 50 60 2000 4000 6000 8000 10000 200 400 600 800 1000 1200 λ = 1 λ = 1 µ = .9 µ = 1.1 Figure 5.7: Queue sizes. 0 10 20 30 40 50 0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 Figure 5.8: Waiting times. Then we plot N(t) as a function of t for different choices of t... | prob_Page_217_Chunk5056 |
210 CHAPTER 5. DISTRIBUTIONS AND DENSITIES Functions of a Random Variable Before continuing our list of important densities, we pause to consider random variables which are functions of other random variables. We will prove a general theorem that will allow us to derive expressions such as Equation 5.5. Theorem 5.1 Let... | prob_Page_218_Chunk5057 |
5.2. IMPORTANT DENSITIES 211 Proof. This result follows from Theorem 5.1 by using the Chain Rule. 2 If the function φ is neither strictly increasing nor strictly decreasing, then the situation is somewhat more complicated but can be treated by the same methods. For example, suppose that Y = X2, Then φ(x) = x2, and FY (... | prob_Page_219_Chunk5058 |
212 CHAPTER 5. DISTRIBUTIONS AND DENSITIES y = φ(x) x = FY (y) Y (y) Graph of F x y 1 0 Figure 5.9: Converting a uniform distribution FU into a prescribed distribution FY . Corollary 5.2 If F(y) is a given cumulative distribution function that is strictly increasing when 0 < F(y) < 1 and if U is a random variable with ... | prob_Page_220_Chunk5059 |
5.2. IMPORTANT DENSITIES 213 -4 -2 2 4 0.1 0.2 0.3 0.4 σ = 1 σ = 2 Figure 5.10: Normal density for two sets of parameter values. integral over the real line equals 1. The cumulative distribution function is given by the formula FX(x) = Z x −∞ 1 √ 2πσ e−(u−µ)2/2σ2 du . In Figure 5.10 we have included for comparison a pl... | prob_Page_221_Chunk5060 |
214 CHAPTER 5. DISTRIBUTIONS AND DENSITIES fX(x) = fZ x −µ σ · 1 σ = 1 √ 2πσ e−(x−µ)2/2σ2 . The reader will note that this last expression is the density function with parameters µ and σ, as claimed. We have seen above that it is possible to simulate a standard normal random variable Z. If we wish to simulate a norm... | prob_Page_222_Chunk5061 |
5.2. IMPORTANT DENSITIES 215 0 1 2 3 4 5 0 0.1 0.2 0.3 0.4 0.5 0.6 Figure 5.11: Distribution of dart distances in 1000 drops. This last expression can be evaluated by using tabulated values of the standard normal distribution function (see 11.5); when we use this table, we find that FZ(2) = .9772 and FZ(−2) = .0228. Thu... | prob_Page_223_Chunk5062 |
216 CHAPTER 5. DISTRIBUTIONS AND DENSITIES Female Male A 37 56 93 B 63 60 123 C 47 43 90 Below C 5 8 13 152 167 319 Table 5.8: Calculus class data. Female Male A 44.3 48.7 93 B 58.6 64.4 123 C 42.9 47.1 90 Below C 6.2 6.8 13 152 167 319 Table 5.9: Expected data. We have also plotted the theoretical density f(r) = re−r2... | prob_Page_224_Chunk5063 |
5.2. IMPORTANT DENSITIES 217 the traits are independent, we would still expect to see some differences between the numbers in corresponding boxes in the two tables. However, if the differences are large, then we might suspect that the two traits are not independent. In Ex- ample 5.6, we used the probability distribution ... | prob_Page_225_Chunk5064 |
218 CHAPTER 5. DISTRIBUTIONS AND DENSITIES 0 2 4 6 8 10 12 0 0.05 0.1 0.15 0.2 Figure 5.12: Chi-squared density with three degrees of freedom. As we stated above, if the value of the random variable χ2 is large, then we would tend not to believe that the two traits are independent. But how large is large? The actual va... | prob_Page_226_Chunk5065 |
5.2. IMPORTANT DENSITIES 219 Thus, for positive B, the cumulative distribution function of X is F(B) = 1 −π/2 −arctan(B) π . Therefore, the density function for positive B is f(B) = 1 π(1 + B2) . Since the physical situation is symmetric with respect to φ = 0, it is easy to see that the above expression for the density... | prob_Page_227_Chunk5066 |
220 CHAPTER 5. DISTRIBUTIONS AND DENSITIES 6 Check your results for Exercise 5 by simulation as described in Exercise 4. 7 Explain how you can generate a random variable whose cumulative distribu- tion function is F(x) = 0, if x < 0, x2, if 0 ≤x ≤1, 1, if x > 1. 8 Write a program to generate a sample of 1000 rand... | prob_Page_228_Chunk5067 |
5.2. IMPORTANT DENSITIES 221 16 Let X be a random variable with density function fX(x) = cx(1 −x), if 0 < x < 1, 0, otherwise. (a) What is the value of c? (b) What is the cumulative distribution function FX for X? (c) What is the probability that X < 1/4? 17 Let X be a random variable with cumulative distribution fun... | prob_Page_229_Chunk5068 |
222 CHAPTER 5. DISTRIBUTIONS AND DENSITIES Test Score Letter grade µ + σ < x A µ < x < µ + σ B µ −σ < x < µ C µ −2σ < x < µ −σ D x < µ −2σ F Table 5.10: Grading on the curve. 24 Let X be a random variable with density function fX. The mode of X is the value M for which f(M) is maximum. Then values of X near M are most ... | prob_Page_230_Chunk5069 |
5.2. IMPORTANT DENSITIES 223 30 Suppose that the number of years a car will run is exponentially distributed with parameter µ = 1/4. If Prosser buys a used car today, what is the probability that it will still run after 4 years? 31 Let U be a uniformly distributed random variable on [0, 1]. What is the probability that... | prob_Page_231_Chunk5070 |
224 CHAPTER 5. DISTRIBUTIONS AND DENSITIES 35 Consider the simple queueing process of Example 5.7. Suppose that you watch the size of the queue. If there are j people in the queue the next time the queue size changes it will either decrease to j −1 or increase to j + 1. Use the result of Exercise 34 to show that the pr... | prob_Page_232_Chunk5071 |
Chapter 6 Expected Value and Variance 6.1 Expected Value of Discrete Random Variables When a large collection of numbers is assembled, as in a census, we are usually interested not in the individual numbers, but rather in certain descriptive quantities such as the average or the median. In general, the same is true for... | prob_Page_233_Chunk5072 |
226 CHAPTER 6. EXPECTED VALUE AND VARIANCE n = 100 n = 10000 Winning Frequency Relative Frequency Relative Frequency Frequency 1 17 .17 1681 .1681 -2 17 .17 1678 .1678 3 16 .16 1626 .1626 -4 18 .18 1696 .1696 5 16 .16 1686 .1686 -6 16 .16 1633 .1633 Table 6.1: Frequencies for dice game. = 9 6 −12 6 = −3 6 = −.5 . This ... | prob_Page_234_Chunk5073 |
6.1. EXPECTED VALUE 227 (This is just the geometric distribution with parameter 1/2.) Thus, we have E(X) = ∞ X i=1 i 1 2i = ∞ X i=1 1 2i + ∞ X i=2 1 2i + · · · = 1 + 1 2 + 1 22 + · · · = 2 . 2 Example 6.3 (Example 6.2 continued) Suppose that we flip a coin until a head first appears, and if the number of tosses equals n,... | prob_Page_235_Chunk5074 |
228 CHAPTER 6. EXPECTED VALUE AND VARIANCE Example 6.4 Let T be the time for the first success in a Bernoulli trials process. Then we take as sample space Ωthe integers 1, 2, . . . and assign the geometric distribution m(j) = P(T = j) = qj−1p . Thus, E(T) = 1 · p + 2qp + 3q2p + · · · = p(1 + 2q + 3q2 + · · ·) . Now if |... | prob_Page_236_Chunk5075 |
6.1. EXPECTED VALUE 229 X Y HHH 1 HHT 2 HTH 3 HTT 2 THH 2 THT 3 TTH 2 TTT 1 Table 6.2: Tossing a coin three times. to be introduced in the next section, we shall be able to prove the Law of Large Numbers. This theorem will justify mathematically both our frequency concept of probability and the interpretation of expect... | prob_Page_237_Chunk5076 |
230 CHAPTER 6. EXPECTED VALUE AND VARIANCE Now suppose we didn’t group the values of X with a common Y -value, but instead, for each X-value x, we multiply the probability of x and the corresponding value of Y , and add the results. We obtain 1 1 8 + 2 1 8 + 3 1 8 + 2 1 8 + 2 1 8 + 3 1 8 + 2 1 8 + ... | prob_Page_238_Chunk5077 |
6.1. EXPECTED VALUE 231 Theorem 6.2 Let X and Y be random variables with finite expected values. Then E(X + Y ) = E(X) + E(Y ) , and if c is any constant, then E(cX) = cE(X) . Proof. Let the sample spaces of X and Y be denoted by ΩX and ΩY , and suppose that ΩX = {x1, x2, . . .} and ΩY = {y1, y2, . . .} . Then we can co... | prob_Page_239_Chunk5078 |
232 CHAPTER 6. EXPECTED VALUE AND VARIANCE X Y a b c 3 a c b 1 b a c 1 b c a 0 c a b 0 c b a 1 Table 6.3: Number of fixed points. It is easy to prove by mathematical induction that the expected value of the sum of any finite number of random variables is the sum of the expected values of the individual random variables. ... | prob_Page_240_Chunk5079 |
6.1. EXPECTED VALUE 233 Bernoulli Trials Theorem 6.3 Let Sn be the number of successes in n Bernoulli trials with prob- ability p for success on each trial. Then the expected number of successes is np. That is, E(Sn) = np . Proof. Let Xj be a random variable which has the value 1 if the jth outcome is a success and 0 i... | prob_Page_241_Chunk5080 |
234 CHAPTER 6. EXPECTED VALUE AND VARIANCE are the sample spaces of X and Y , respectively. Using Theorem 6.1, we have E(X · Y ) = X j X k xjykP(X = xj, Y = yk) . But if X and Y are independent, P(X = xj, Y = yk) = P(X = xj)P(Y = yk) . Thus, E(X · Y ) = X j X k xjykP(X = xj)P(Y = yk) = X j xjP(X = xj) X k ykP(Y ... | prob_Page_242_Chunk5081 |
6.1. EXPECTED VALUE 235 so E(Xj) = 1/j. Therefore, if Sn is the total number of records observed in the first n years, E(Sn) = 1 + 1 2 + 1 3 + · · · + 1 n . This is the famous divergent harmonic series. It is easy to show that E(Sn) ∼log n as n →∞. A more accurate approximation to E(Sn) is given by the expression log n ... | prob_Page_243_Chunk5082 |
236 CHAPTER 6. EXPECTED VALUE AND VARIANCE W L W L W L W L W L W L (2,3,12) L 10 9 8 6 5 4 (7,11) W 1/3 2/3 2/5 3/5 5/11 6/11 5/11 6/11 2/5 3/5 1/3 2/3 2/9 1/12 1/9 5/36 5/36 1/9 1/12 1/9 1/36 2/36 2/45 3/45 25/396 30/396 25/396 30/396 2/45 3/45 1/36 2/36 Figure 6.1: Tree measure for craps. | prob_Page_244_Chunk5083 |
6.1. EXPECTED VALUE 237 a dollar bet, E(X) = 1 244 495 + (−1) 251 495 = −7 495 ≈−.0141 . The game is unfavorable, but only slightly. The player’s expected gain in n plays is −n(.0141). If n is not large, this is a small expected loss for the player. The casino makes a large number of plays and so can afford a smal... | prob_Page_245_Chunk5084 |
238 CHAPTER 6. EXPECTED VALUE AND VARIANCE S W L E L L L L L L E P1 P1 P1 P2 P2 P2 Figure 6.2: Tree for 2-prison Monte Carlo roulette. It is interesting to compare the expected winnings of a 1 franc bet on red, under each of these three options. We leave the first two calculations as an exercise (see Exercise 37). Suppo... | prob_Page_246_Chunk5085 |
6.1. EXPECTED VALUE 239 P W L P P L 18/37 18/37 1/37 19/37 18/37 1 1 2 Figure 6.3: Your money is put in prison. It is interesting to note that the more romantic option (c) is less favorable than option (a) (see Exercise 37). If you bet 1 dollar on the number 17, then the distribution function for your winnings X is PX ... | prob_Page_247_Chunk5086 |
240 CHAPTER 6. EXPECTED VALUE AND VARIANCE Proof. We have X j E(X|Fj)P(Fj) = X j X k xkP(X = xk|Fj)P(Fj) = X j X k xkP(X = xk and Fj occurs) = X k X j xkP(X = xk and Fj occurs) = X k xkP(X = xk) = E(X) . 2 Example 6.14 (Example 6.12 continued) Let T be the number of rolls in a single play of craps. We can think of a si... | prob_Page_248_Chunk5087 |
6.1. EXPECTED VALUE 241 Martingales We can extend the notion of fairness to a player playing a sequence of games by using the concept of conditional expectation. Example 6.15 Let S1, S2, . . . , Sn be Peter’s accumulated fortune in playing heads or tails (see Example 1.4). Then E(Sn|Sn−1 = a, . . . , S1 = r) = 1 2(a + ... | prob_Page_249_Chunk5088 |
242 CHAPTER 6. EXPECTED VALUE AND VARIANCE 5 10 15 20 -1 -0.5 0.5 1 1.5 2 Figure 6.4: Mr. Ace’s system. of times, for n = 20, say, one finds that his expected winnings are very close to 0, but the probability that he is ahead after 20 days is significantly greater than 1/2. For small values of n, the exact distribution o... | prob_Page_250_Chunk5089 |
6.1. EXPECTED VALUE 243 -20 -15 -10 -5 0 5 10 0 0.05 0.1 0.15 0.2 Figure 6.5: Winnings distribution for n = 20. to separate without playing it, the first man must say: “I am certain to get 32 pistoles, even if I lose I still get them; but as for the other 32 pistoles, perhaps I will get them, perhaps you will get them, ... | prob_Page_251_Chunk5090 |
244 CHAPTER 6. EXPECTED VALUE AND VARIANCE for the modified game? If he wins he gets the total stake a + b and must pay B an amount b so ends up with a. If he loses he gets an amount b from player B. Thus player A wins a or b with equal chances and the value to him is (a + b)/2. Huygens illustrated this proof in terms o... | prob_Page_252_Chunk5091 |
6.1. EXPECTED VALUE 245 manipulation, Huygens ended up with values that agree with those given by our modern definition of expected value. One advantage of this method is that it gives a justification for the expected value in cases where it is not reasonable to assume that you can repeat the experiment a large number of... | prob_Page_253_Chunk5092 |
246 CHAPTER 6. EXPECTED VALUE AND VARIANCE God does not exist God exists p 1 −p believe −u v not believe 0 −x Table 6.4: Payoffs. Age Survivors 0 100 6 64 16 40 26 25 36 16 46 10 56 6 66 3 76 1 Table 6.5: Graunt’s mortality data. and choose the larger of the two. In general, the choice will depend upon the value of p. B... | prob_Page_254_Chunk5093 |
6.1. EXPECTED VALUE 247 A terminal annuity provides a fixed amount of money during a period of n years. To determine the price of a terminal annuity one needs only to know the appropriate interest rate. A life annuity provides a fixed amount during each year of the buyer’s life. The appropriate price for a life annuity i... | prob_Page_255_Chunk5094 |
248 CHAPTER 6. EXPECTED VALUE AND VARIANCE *7 Show that, if X and Y are random variables taking on only two values each, and if E(XY ) = E(X)E(Y ), then X and Y are independent. 8 A royal family has children until it has a boy or until it has three children, whichever comes first. Assume that each child is a boy with pr... | prob_Page_256_Chunk5095 |
6.1. EXPECTED VALUE 249 jth person gets his or her hat and 0 otherwise. Find E(Xj) and E(Xj · Xk) for j not equal to k. Are Xj and Xk independent? 15 A box contains two gold balls and three silver balls. You are allowed to choose successively balls from the box at random. You win 1 dollar each time you draw a gold ball... | prob_Page_257_Chunk5096 |
250 CHAPTER 6. EXPECTED VALUE AND VARIANCE (c) Assume that you pay 10 dollars for each play of the original game. Write a program to simulate 100 plays of the game and see how you do. (d) Now assume that the utility of n dollars is √n. Write an expression for the expected utility of the payment, and show that this expr... | prob_Page_258_Chunk5097 |
6.1. EXPECTED VALUE 251 (a) First, do a simulation of guessing without information, repeating the experiment at least 1000 times. Estimate the expected number of correct answers and compare your result with the theoretical expectation. (b) What is the best strategy for guessing with information? (c) Do a simulation of ... | prob_Page_259_Chunk5098 |
252 CHAPTER 6. EXPECTED VALUE AND VARIANCE 2 1 1 2 3 4 5 6 7 8 9 10 (10,2) L Figure 6.6: Random walk for ESP. the probability that the walk reaches (x, x) is | prob_Page_260_Chunk5099 |
6.1. EXPECTED VALUE 253 (a) Show that if n ≤2k −1, then there is a strategy that guarantees you will correctly guess the number in k tries. (b) Show that if n ≥2k −1, there is a strategy that assures you of identifying one of 2k −1 numbers and hence gives a probability of (2k −1)/n of winning. Why is this an optimal st... | prob_Page_261_Chunk5100 |
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