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254 CHAPTER 6. EXPECTED VALUE AND VARIANCE (d) In Example 6.11 we stated that 1 + 1 2 + 1 3 + · · · + 1 n ∼log n + .5772 + 1 2n . Use this to estimate the expression in (c). Compare these estimates with the exact values and also with your estimates obtained by simulation for the case n = 26. *31 (Feller14) A large numb... | prob_Page_262_Chunk5101 |
6.1. EXPECTED VALUE 255 (b) Show that E(X) = n X j=0 P(X ≥j + 1) . (c) From these two facts, find an expression for E(X). This proof is due to Harris Schultz.15 *34 (Banach’s Matchbox16) A man carries in each of his two front pockets a box of matches originally containing N matches. Whenever he needs a match, he chooses... | prob_Page_263_Chunk5102 |
256 CHAPTER 6. EXPECTED VALUE AND VARIANCE 37 The reader is referred to Example 6.13 for an explanation of the various op- tions available in Monte Carlo roulette. (a) Compute the expected winnings of a 1 franc bet on red under option (a). (b) Repeat part (a) for option (b). (c) Compare the expected winnings for all th... | prob_Page_264_Chunk5103 |
6.2. VARIANCE OF DISCRETE RANDOM VARIABLES 257 that was proved in Exercise 33, show that E(T) = ∞ X s=0 1 − 2s n n! 2sn . Show that for n = 52, this expression is approximately equal to 11.7. (As was stated in Chapter 3, this means that on the average, almost 12 riffle shuffles of a 52-card deck are required in orde... | prob_Page_265_Chunk5104 |
258 CHAPTER 6. EXPECTED VALUE AND VARIANCE x m(x) (x −7/2)2 1 1/6 25/4 2 1/6 9/4 3 1/6 1/4 4 1/6 1/4 5 1/6 9/4 6 1/6 25/4 Table 6.6: Variance calculation. From this table we find E((X −µ)2) is V (X) = 1 6 25 4 + 9 4 + 1 4 + 1 4 + 9 4 + 25 4 = 35 12 , and the standard deviation D(X) = p 35/12 ≈1.707. 2 Calculation of ... | prob_Page_266_Chunk5105 |
6.2. VARIANCE OF DISCRETE RANDOM VARIABLES 259 Properties of Variance The variance has properties very different from those of the expectation. If c is any constant, E(cX) = cE(X) and E(X + c) = E(X) + c. These two statements imply that the expectation is a linear function. However, the variance is not linear, as seen i... | prob_Page_267_Chunk5106 |
260 CHAPTER 6. EXPECTED VALUE AND VARIANCE It is easy to extend this proof, by mathematical induction, to show that the variance of the sum of any number of mutually independent random variables is the sum of the individual variances. Thus we have the following theorem. Theorem 6.9 Let X1, X2, . . . , Xn be an independ... | prob_Page_268_Chunk5107 |
6.2. VARIANCE OF DISCRETE RANDOM VARIABLES 261 1 2 3 4 5 6 0 0.1 0.2 0.3 0.4 0.5 0.6 2 2.5 3 3.5 4 4.5 5 0 0.5 1 1.5 2 n = 10 n = 100 Figure 6.7: Empirical distribution of An. The last equation in the above theorem implies that in an independent trials process, if the individual summands have finite variance, then the s... | prob_Page_269_Chunk5108 |
262 CHAPTER 6. EXPECTED VALUE AND VARIANCE This suggests that the frequency interpretation of probability is a correct one. We shall make this more precise in Chapter 8. Example 6.19 Let T denote the number of trials until the first success in a Bernoulli trials process. Then T is geometrically distributed. What is the ... | prob_Page_270_Chunk5109 |
6.2. VARIANCE OF DISCRETE RANDOM VARIABLES 263 Poisson Distribution Just as in the case of expected values, it is easy to guess the variance of the Poisson distribution with parameter λ. We recall that the variance of a binomial distribution with parameters n and p equals npq. We also recall that the Poisson distributi... | prob_Page_271_Chunk5110 |
264 CHAPTER 6. EXPECTED VALUE AND VARIANCE (c) It is decided to report the temperature readings on a Celsius scale, that is, C = (5/9)(F −32). What is the expected value and variance for the readings now? 6 Write a computer program to calculate the mean and variance of a distribution which you specify as data. Use the ... | prob_Page_272_Chunk5111 |
6.2. VARIANCE OF DISCRETE RANDOM VARIABLES 265 15 Suppose that n people have their hats returned at random. Let Xi = 1 if the ith person gets his or her own hat back and 0 otherwise. Let Sn = Pn i=1 Xi. Then Sn is the total number of people who get their own hats back. Show that (a) E(X2 i ) = 1/n. (b) E(Xi · Xj) = 1/n... | prob_Page_273_Chunk5112 |
266 CHAPTER 6. EXPECTED VALUE AND VARIANCE (b) E | prob_Page_274_Chunk5113 |
6.2. VARIANCE OF DISCRETE RANDOM VARIABLES 267 21 Let X be a random variable with E(X) = µ and V (X) = σ2. Show that the function f(x) defined by f(x) = X ω (X(ω) −x)2p(ω) has its minimum value when x = µ. 22 Let X and Y be two random variables defined on the finite sample space Ω. Assume that X, Y , X + Y , and X −Y all ... | prob_Page_275_Chunk5114 |
268 CHAPTER 6. EXPECTED VALUE AND VARIANCE (a) What is the common distribution, expected value, and variance for Xj? (b) Let Tn = X1 + X2 + · · · + Xn. Then Tn is the time until the nth success. Find E(Tn) and V (Tn). (c) Use the results of (b) to find the expected value and variance for the number of tosses of a coin u... | prob_Page_276_Chunk5115 |
6.3. CONTINUOUS RANDOM VARIABLES 269 Theorem 6.10 If X and Y are real-valued random variables and c is any constant, then E(X + Y ) = E(X) + E(Y ) , E(cX) = cE(X) . The proof is very similar to the proof of Theorem 6.2, and we omit it. 2 More generally, if X1, X2, . . . , Xn are n real-valued random variables, and c1, ... | prob_Page_277_Chunk5116 |
270 CHAPTER 6. EXPECTED VALUE AND VARIANCE = Z 1 0 2z(1 −z) dz = h z2 −2 3z3i1 0 = 1 3 . 2 Expectation of a Function of a Random Variable Suppose that X is a real-valued random variable and φ(x) is a continuous function from R to R. The following theorem is the continuous analogue of Theorem 6.1. Theorem 6.11 If X is a... | prob_Page_278_Chunk5117 |
6.3. CONTINUOUS RANDOM VARIABLES 271 In the same way, one can show that if X1, X2, . . . , Xn are n mutually indepen- dent real-valued random variables, then E(X1X2 · · · Xn) = E(X1) E(X2) · · · E(Xn) . Example 6.23 Let Z = (X, Y ) be a point chosen at random in the unit square. Let A = X2 and B = Y 2. Then Theorem 4.3... | prob_Page_279_Chunk5118 |
272 CHAPTER 6. EXPECTED VALUE AND VARIANCE The next result follows easily from Theorem 6.1. There is another way to calculate the variance of a continuous random variable, which is usually slightly easier. It is given in Theorem 6.15. Theorem 6.13 If X is a real-valued random variable with E(X) = µ, then σ2 = Z ∞ −∞ (x... | prob_Page_280_Chunk5119 |
6.3. CONTINUOUS RANDOM VARIABLES 273 Example 6.26 Let X be an exponentially distributed random variable with pa- rameter λ. Then the density function of X is fX(x) = λe−λx . From the definition of expectation and integration by parts, we have E(X) = Z ∞ 0 xfX(x) dx = λ Z ∞ 0 xe−λx dx = −xe−λx ∞ 0 + Z ∞ 0 e−λx dx = 0 + e... | prob_Page_281_Chunk5120 |
274 CHAPTER 6. EXPECTED VALUE AND VARIANCE which one can easily show is finite. Thus, the expected value of Z is 0. To calculate the variance of Z, we begin by applying Theorem 6.15: V (Z) = Z +∞ −∞ x2fZ(x) dx −µ2 . If we write x2 as x · x, and integrate by parts, we obtain 1 √ 2π (−xe−x2/2) +∞ −∞ + 1 √ 2π Z +∞ −∞ e−x2/... | prob_Page_282_Chunk5121 |
6.3. CONTINUOUS RANDOM VARIABLES 275 Corollary 6.1 If X1, X2, . . . , Xn is an independent trials process of real-valued random variables, with E(Xi) = µ and V (Xi) = σ2, and if Sn = X1 + X2 + · · · + Xn , An = Sn n , then E(Sn) = nµ , E(An) = µ , V (Sn) = nσ2 , V (An) = σ2 n . It follows that if we set S∗ n = Sn −nµ √... | prob_Page_283_Chunk5122 |
276 CHAPTER 6. EXPECTED VALUE AND VARIANCE We expect on grounds of our everyday experience with queues that if the service rate is greater than the arrival rate, then the average queue size will tend to stabilize, but if the service rate is less than the arrival rate, then the queue will tend to increase in length with... | prob_Page_284_Chunk5123 |
6.3. CONTINUOUS RANDOM VARIABLES 277 0 10 20 30 40 50 0 0.02 0.04 0.06 0.08 Figure 6.8: Distribution of queue lengths. We note that the distribution appears to be a geometric distribution. In the study of queueing theory it is shown that the distribution for the queue length in equilibrium is indeed a geometric distrib... | prob_Page_285_Chunk5124 |
278 CHAPTER 6. EXPECTED VALUE AND VARIANCE 0 10 20 30 40 50 0 0.02 0.04 0.06 0.08 Figure 6.9: Distribution of queue waiting times. Exercises 1 Let X be a random variable with range [−1, 1] and let fX(x) be the density function of X. Find µ(X) and σ2(X) if, for |x| < 1, (a) fX(x) = 1/2. (b) fX(x) = |x|. (c) fX(x) = 1 −|... | prob_Page_286_Chunk5125 |
6.3. CONTINUOUS RANDOM VARIABLES 279 (d) Show that σ2(X) = (2/3)b −(4/9)a2 = 1/3 −(4/9)a2. 5 Let X be a random variable with range [−1, 1] and density function fX(x) = ax2 + bx + c if |x| < 1 and 0 otherwise. (a) Show that 2a/3 + 2c = 1 (see Exercise 4). (b) Show that 2b/3 = µ(X). (c) Show that 2a/5 + 2c/3 = σ2(X). (d)... | prob_Page_287_Chunk5126 |
280 CHAPTER 6. EXPECTED VALUE AND VARIANCE 11 The PilsdorffBeer Company runs a fleet of trucks along the 100 mile road from Hangtown to Dry Gulch. The trucks are old, and are apt to break down at any point along the road with equal probability. Where should the company locate a garage so as to minimize the expected dista... | prob_Page_288_Chunk5127 |
6.3. CONTINUOUS RANDOM VARIABLES 281 This result is correct but quite difficult to prove. Write a program that will allow you to specify the density fX, and the time t, and simulate this experi- ment to find N(t)/t. Have your program repeat the experiment 500 times and plot a bar graph for the random outcomes of N(t)/t. F... | prob_Page_289_Chunk5128 |
282 CHAPTER 6. EXPECTED VALUE AND VARIANCE *20 When studying certain physiological data, such as heights of fathers and sons, it is often natural to assume that these data (e.g., the heights of the fathers and the heights of the sons) are described by random variables with normal densities. These random variables, howe... | prob_Page_290_Chunk5129 |
6.3. CONTINUOUS RANDOM VARIABLES 283 (b) Using Exercises 17 and 18, show that the correlation of X and Y is ρ. (c) In Exercise 20, the joint density function fX,Y (x, y) for the random vari- able (X, Y ) is given. Now suppose that we want to know the set of points (x, y) in the xy-plane such that fX,Y (x, y) = C for so... | prob_Page_291_Chunk5130 |
284 CHAPTER 6. EXPECTED VALUE AND VARIANCE Plot the equal density ellipses as defined in Exercise 23 for r = 4, 6, and 8, and on the same graph print the values that appear in the table at the appropriate points. For example, print 12 at the point (70.5, 68.2), indicating that there were 12 cases where the parent’s heig... | prob_Page_292_Chunk5131 |
Chapter 7 Sums of Independent Random Variables 7.1 Sums of Discrete Random Variables In this chapter we turn to the important question of determining the distribution of a sum of independent random variables in terms of the distributions of the individual constituents. In this section we consider only sums of discrete ... | prob_Page_293_Chunk5132 |
286 CHAPTER 7. SUMS OF RANDOM VARIABLES Definition 7.1 Let X and Y be two independent integer-valued random variables, with distribution functions m1(x) and m2(x) respectively. Then the convolution of m1(x) and m2(x) is the distribution function m3 = m1 ∗m2 given by m3(j) = X k m1(k) · m2(j −k) , for j = . . . , −2, −1,... | prob_Page_294_Chunk5133 |
7.1. SUMS OF DISCRETE RANDOM VARIABLES 287 = 1 36 · 1 6 = 1 216 , P(S3 = 4) = P(S2 = 3)P(X3 = 1) + P(S2 = 2)P(X3 = 2) = 2 36 · 1 6 + 1 36 · 1 6 = 3 216 , and so forth. This is clearly a tedious job, and a program should be written to carry out this calculation. To do this we first write a program to form the convolution... | prob_Page_295_Chunk5134 |
288 CHAPTER 7. SUMS OF RANDOM VARIABLES 20 40 60 80 100 120 140 0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 20 40 60 80 100 120 140 0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 20 40 60 80 100 120 140 0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 n = 10 n = 20 n = 30 Figure 7.1: Density of Sn for rolling a die n times. | prob_Page_296_Chunk5135 |
7.1. SUMS OF DISCRETE RANDOM VARIABLES 289 For certain special distributions it is possible to find an expression for the dis- tribution that results from convoluting the distribution with itself n times. The convolution of two binomial distributions, one with parameters m and p and the other with parameters n and p, is... | prob_Page_297_Chunk5136 |
290 CHAPTER 7. SUMS OF RANDOM VARIABLES 6 People arrive at a queue according to the following scheme: During each minute of time either 0 or 1 person arrives. The probability that 1 person arrives is p and that no person arrives is q = 1 −p. Let Cr be the number of customers arriving in the first r minutes. Consider a B... | prob_Page_298_Chunk5137 |
7.2. SUMS OF CONTINUOUS RANDOM VARIABLES 291 a and b is the equiprobable distribution on the set 0, 1, 2, . . . , n −1. If n is prime this is not possible, but the proof is not so easy. (Assume that neither a nor b is concentrated at 0.) 11 Assume that you are playing craps with dice that are loaded in the following wa... | prob_Page_299_Chunk5138 |
292 CHAPTER 7. SUMS OF RANDOM VARIABLES Sum of Two Independent Uniform Random Variables Example 7.3 Suppose we choose independently two numbers at random from the interval [0, 1] with uniform probability density. What is the density of their sum? Let X and Y be random variables describing our choices and Z = X + Y thei... | prob_Page_300_Chunk5139 |
7.2. SUMS OF CONTINUOUS RANDOM VARIABLES 293 0.5 1 1.5 2 0.2 0.4 0.6 0.8 1 Figure 7.2: Convolution of two uniform densities. 1 2 3 4 5 6 0.05 0.1 0.15 0.2 0.25 0.3 0.35 Figure 7.3: Convolution of two exponential densities with λ = 1. and so, if z > 0, fZ(z) = Z +∞ −∞ fX(z −y)fY (y) dy = Z z 0 λe−λ(z−y)λe−λy dy = Z z 0 ... | prob_Page_301_Chunk5140 |
294 CHAPTER 7. SUMS OF RANDOM VARIABLES Sum of Two Independent Normal Random Variables Example 7.5 It is an interesting and important fact that the convolution of two normal densities with means µ1 and µ2 and variances σ1 and σ2 is again a normal density, with mean µ1 + µ2 and variance σ2 1 + σ2 2. We will show this in... | prob_Page_302_Chunk5141 |
7.2. SUMS OF CONTINUOUS RANDOM VARIABLES 295 This integral requires some effort, and we give here only the result (see Section 10.3, or Dwass3): fZ(z) = 2 π(4 + z2) . Now, suppose that we ask for the density function of the average A = (1/2)(X + Y ) of X and Y . Then A = (1/2)Z. Exercise 5.2.19 shows that if U and V are... | prob_Page_303_Chunk5142 |
296 CHAPTER 7. SUMS OF RANDOM VARIABLES This is a gamma density with λ = 1/2, β = 1/2 (see Example 7.4). Now let R2 = X2 + Y 2. Then fR2(r) = Z +∞ −∞ fX2(r −s)fY 2(s) ds = 1 4π Z +∞ −∞ e−(r−s)/2 r −s 2 −1/2 e−s s 2 −1/2 ds , = 1 2e−r2/2, if r ≥0, 0, otherwise. Hence, R2 has a gamma density with λ = 1/2, β = 1. We can... | prob_Page_304_Chunk5143 |
7.2. SUMS OF CONTINUOUS RANDOM VARIABLES 297 Outcome Observed Frequency 1 15 2 8 3 7 4 5 5 7 6 18 Table 7.1: Observed data. for moderate or large values of n, the quantity V is approximately chi-squared distributed, with ν−1 degrees of freedom, where ν represents the number of possible outcomes. The proof of this is be... | prob_Page_305_Chunk5144 |
298 CHAPTER 7. SUMS OF RANDOM VARIABLES 5 10 15 20 0.025 0.05 0.075 0.1 0.125 0.15 Figure 7.4: Chi-squared density with 5 degrees of freedom. 0 5 10 15 20 25 30 0 0.025 0.05 0.075 0.1 0.125 0.15 1000 experiments 60 rolls per experiment Figure 7.5: Rolling a fair die. | prob_Page_306_Chunk5145 |
7.2. SUMS OF CONTINUOUS RANDOM VARIABLES 299 1 2 3 4 5 6 7 8 0 0.2 0.4 0.6 0.8 1 n = 2 n = 4 n = 6 n = 8 n = 10 Figure 7.6: Convolution of n uniform densities. Independent Trials We now consider briefly the distribution of the sum of n independent random vari- ables, all having the same density function. If X1, X2, . . ... | prob_Page_307_Chunk5146 |
300 CHAPTER 7. SUMS OF RANDOM VARIABLES -15 -10 -5 5 10 15 0.025 0.05 0.075 0.1 0.125 0.15 0.175 n = 5 n = 10 n = 15 n = 20 n = 25 Figure 7.7: Convolution of n standard normal densities. and fSn(x) = 1 √ 2πne−x2/2n . Here the density fSn for n = 5, 10, 15, 20, 25 is shown in Figure 7.7. If the Xi are all exponentially ... | prob_Page_308_Chunk5147 |
7.2. SUMS OF CONTINUOUS RANDOM VARIABLES 301 5 10 15 20 0.05 0.1 0.15 0.2 0.25 0.3 0.35 n = 2 n = 4 n = 6 n = 8 n = 10 Figure 7.8: Convolution of n exponential densities with λ = 1. (a) fX(x) = fY (x) = 1/2, if −1 ≤x ≤+1, 0, otherwise. (b) fX(x) = fY (x) = 1/2, if 3 ≤x ≤5, 0, otherwise. (c) fX(x) = 1/2, if −1 ≤x ... | prob_Page_309_Chunk5148 |
302 CHAPTER 7. SUMS OF RANDOM VARIABLES fY (x) = x/2, if 0 < x < 2, 0, otherwise. (d) What can you say about the set E = { z : fZ(z) > 0 } in each case? 4 Let X, Y , and Z be independent random variables with fX(x) = fY (x) = fZ(x) = 1, if 0 < x < 1, 0, otherwise. Suppose that W = X + Y + Z. Find fW directly, and c... | prob_Page_310_Chunk5149 |
7.2. SUMS OF CONTINUOUS RANDOM VARIABLES 303 10 Let X1, X2, . . . , Xn be n independent random variables each of which has an exponential density with mean µ. Let M be the minimum value of the Xj. Show that the density for M is exponential with mean µ/n. Hint: Use cumulative distribution functions. 11 A company buys 10... | prob_Page_311_Chunk5150 |
304 CHAPTER 7. SUMS OF RANDOM VARIABLES 16 Verify your answers in Exercise 2(a) by computer simulation: Choose X and Y from [−1, 1] with uniform density and calculate Z = X + Y . Repeat this experiment 500 times, recording the outcomes in a bar graph on [−2, 2] with 40 bars. Does the density fZ calculated in Exercise 2... | prob_Page_312_Chunk5151 |
Chapter 8 Law of Large Numbers 8.1 Law of Large Numbers for Discrete Random Variables We are now in a position to prove our first fundamental theorem of probability. We have seen that an intuitive way to view the probability of a certain outcome is as the frequency with which that outcome occurs in the long run, when th... | prob_Page_313_Chunk5152 |
306 CHAPTER 8. LAW OF LARGE NUMBERS We know that V (X) = X x (x −µ)2m(x) , and this is clearly at least as large as X |x−µ|≥ϵ (x −µ)2m(x) , since all the summands are positive and we have restricted the range of summation in the second sum. But this last sum is at least X |x−µ|≥ϵ ϵ2m(x) = ϵ2 X |x−µ|≥ϵ m(x) = ϵ2P(|X −µ|... | prob_Page_314_Chunk5153 |
8.1. DISCRETE RANDOM VARIABLES 307 Law of Large Numbers Theorem 8.2 (Law of Large Numbers) Let X1, X2, . . . , Xn be an independent trials process, with finite expected value µ = E(Xj) and finite variance σ2 = V (Xj). Let Sn = X1 + X2 + · · · + Xn. Then for any ϵ > 0, P Sn n −µ ≥ϵ →0 as n →∞. Equivalently, P Sn n −... | prob_Page_315_Chunk5154 |
308 CHAPTER 8. LAW OF LARGE NUMBERS Consider the important special case of Bernoulli trials with probability p for success. Let Xj = 1 if the jth outcome is a success and 0 if it is a failure. Then Sn = X1 + X2 + · · ·+ Xn is the number of successes in n trials and µ = E(X1) = p. The Law of Large Numbers states that fo... | prob_Page_316_Chunk5155 |
8.1. DISCRETE RANDOM VARIABLES 309 0 0.2 0.4 0.6 0.8 1 0 0.02 0.04 0.06 0.08 0.1 0 0.2 0.4 0.6 0.8 1 0 0.02 0.04 0.06 0.08 0 0.2 0.4 0.6 0.8 1 0 0.02 0.04 0.06 0.08 0.1 0.12 0.14 0 0.2 0.4 0.6 0.8 1 0 0.02 0.04 0.06 0.08 0.1 0.12 0 0.2 0.4 0.6 0.8 1 0 0.05 0.1 0.15 0.2 0.25 0 0.2 0.4 0.6 0.8 1 0 0.025 0.05 0.075 0.1 0.... | prob_Page_317_Chunk5156 |
310 CHAPTER 8. LAW OF LARGE NUMBERS Example 8.3 Let X1, X2, . . . , Xn be a Bernoulli trials process with probability .3 for success and .7 for failure. Let Xj = 1 if the jth outcome is a success and 0 otherwise. Then, E(Xj) = .3 and V (Xj) = (.3)(.7) = .21. If An = Sn n = X1 + X2 + · · · + Xn n is the average of the X... | prob_Page_318_Chunk5157 |
8.1. DISCRETE RANDOM VARIABLES 311 that occurs with probability p but he does not know p. He wants to estimate p by the fraction ¯p of the times the event occurs when the experiment is repeated a number of times. He discusses in detail the problem of estimating, by this method, the proportion of white balls in an urn t... | prob_Page_319_Chunk5158 |
312 CHAPTER 8. LAW OF LARGE NUMBERS the world would be perceived to happen in fixed ratios and according to a constant law of alternation, so that even in the most accidental and fortuitous occurrences we would be bound to recognize, as it were, a certain necessity and, so to speak, a certain fate. I do now know whether... | prob_Page_320_Chunk5159 |
8.1. DISCRETE RANDOM VARIABLES 313 8 A fair coin is tossed a large number of times. Does the Law of Large Numbers assure us that, if n is large enough, with probability > .99 the number of heads that turn up will not deviate from n/2 by more than 100? 9 In Exercise 6.2.15, you showed that, for the hat check problem, th... | prob_Page_321_Chunk5160 |
314 CHAPTER 8. LAW OF LARGE NUMBERS *15 We have proved a theorem often called the “Weak Law of Large Numbers.” Most people’s intuition and our computer simulations suggest that, if we toss a coin a sequence of times, the proportion of heads will really approach 1/2; that is, if Sn is the number of heads in n times, the... | prob_Page_322_Chunk5161 |
8.1. DISCRETE RANDOM VARIABLES 315 (d) Find the variance of Sn. (e) Show that the sequence ⟨Xi⟩satisfies the Weak Law of Large Numbers, i.e. prove that for any ϵ > 0 P Sn n ≥ϵ →0 , as n tends to infinity. We now show that {Xi} does not satisfy the Strong Law of Large Num- bers. Suppose that Sn/n →0. Then because Xn n... | prob_Page_323_Chunk5162 |
316 CHAPTER 8. LAW OF LARGE NUMBERS 8.2 Law of Large Numbers for Continuous Ran- dom Variables In the previous section we discussed in some detail the Law of Large Numbers for discrete probability distributions. This law has a natural analogue for continuous probability distributions, which we consider somewhat more br... | prob_Page_324_Chunk5163 |
8.2. CONTINUOUS RANDOM VARIABLES 317 Note that this theorem is not necessarily true if σ2 is infinite (see Example 8.8). As in the discrete case, the Law of Large Numbers says that the average value of n independent trials tends to the expected value as n →∞, in the precise sense that, given ϵ > 0, the probability that ... | prob_Page_325_Chunk5164 |
318 CHAPTER 8. LAW OF LARGE NUMBERS n=2 n=5 n=10 n=20 n=30 n=50 Figure 8.2: Illustration of Law of Large Numbers — uniform case. Normal Case Example 8.6 Suppose we choose n real numbers at random, using a normal dis- tribution with mean 0 and variance 1. Then µ = E(Xi) = 0 , σ2 = V (Xi) = 1 . Hence, E Sn n = 0 , V ... | prob_Page_326_Chunk5165 |
8.2. CONTINUOUS RANDOM VARIABLES 319 n P(|Sn/n| ≥.1) Chebyshev 100 .31731 1.00000 200 .15730 .50000 300 .08326 .33333 400 .04550 .25000 500 .02535 .20000 600 .01431 .16667 700 .00815 .14286 800 .00468 .12500 900 .00270 .11111 1000 .00157 .10000 Table 8.1: Chebyshev estimates. Monte Carlo Method Here is a somewhat more ... | prob_Page_327_Chunk5166 |
320 CHAPTER 8. LAW OF LARGE NUMBERS Y X Y = g (x) 0 1 1 Figure 8.3: Area problem. The Law of Large Numbers requires that the variance σ2 of the original under- lying density be finite: σ2 < ∞. In cases where this fails to hold, the Law of Large Numbers may fail, too. An example follows. Cauchy Case Example 8.8 Suppose w... | prob_Page_328_Chunk5167 |
8.2. CONTINUOUS RANDOM VARIABLES 321 (a) P(|X −10| ≥2). (b) P(|X −10| ≥5). (c) P(|X −10| ≥9). (d) P(|X −10| ≥20). 2 Let X be a continuous random variable with values unformly distributed over the interval [0, 20]. (a) Find the mean and variance of X. (b) Calculate P(|X −10| ≥2), P(|X −10| ≥5), P(|X −10| ≥9), and P(|X −... | prob_Page_329_Chunk5168 |
322 CHAPTER 8. LAW OF LARGE NUMBERS (d) P(|X −µ| ≥4σ). Now find the exact value using the program NormalArea or the normal table in Appendix A, and compare. 7 If X is a random variable with mean µ ̸= 0 and variance σ2, define the relative deviation D of X from its mean by D = X −µ µ . (a) Show that P(D ≥a) ≤σ2/(µ2a2). (b... | prob_Page_330_Chunk5169 |
8.2. CONTINUOUS RANDOM VARIABLES 323 (b) Suppose that in one bad week, 20 trucks break down. Find a lower bound for the probability P(|A20 −50| ≤10), where A20 is the average of the distances from Hangtown at the time of breakdown. 12 A share of common stock in the Pilsdorffbeer company has a price Yn on the nth busines... | prob_Page_331_Chunk5170 |
324 CHAPTER 8. LAW OF LARGE NUMBERS (b) We have seen that the Law of Large Numbers does not apply to the Cauchy density (see Example 8.8). Simulate a large number of experi- ments with Cauchy density and compute the average of your results. Do these averages seem to be approaching a limit? If so can you explain why thi... | prob_Page_332_Chunk5171 |
Chapter 9 Central Limit Theorem 9.1 Central Limit Theorem for Bernoulli Trials The second fundamental theorem of probability is the Central Limit Theorem. This theorem says that if Sn is the sum of n mutually independent random variables, then the distribution function of Sn is well-approximated by a certain type of co... | prob_Page_333_Chunk5172 |
326 CHAPTER 9. CENTRAL LIMIT THEOREM -4 -2 0 2 4 0 0.1 0.2 0.3 0.4 Figure 9.1: Standard normal density. we plotted these distributions for p = .3 and p = .5 for various values of n (see Figure 3.5). We note that the maximum values of the distributions appeared near the ex- pected value np, which causes their spike grap... | prob_Page_334_Chunk5173 |
9.1. BERNOULLI TRIALS 327 -4 -2 0 2 4 0 0.1 0.2 0.3 0.4 Figure 9.2: Normalized binomial distribution and standard normal density. have drawn this standard normal density. The reader will note that a horrible thing has occurred: Even though the shapes of the two graphs are the same, the heights are quite different. If we... | prob_Page_335_Chunk5174 |
328 CHAPTER 9. CENTRAL LIMIT THEOREM -4 -2 0 2 4 0 0.1 0.2 0.3 0.4 Figure 9.3: Corrected spike graph with standard normal density. formula j = ⟨np + x√npq⟩, where ⟨a⟩means the integer nearest to a. Thus the height of the spike above xj will be √npq b(n, p, j) = √npq b(n, p, ⟨np + xj √npq⟩) . For large n, we have seen t... | prob_Page_336_Chunk5175 |
9.1. BERNOULLI TRIALS 329 Using this, we have √npq b(n, p, np) ∼ √npq pnpqnq√ 2πn nne−n √2πnp√2πnq (np)np(nq)nqe−npe−nq , which simplifies to 1/ √ 2π. 2 Approximating Binomial Distributions We can use Theorem 9.1 to find approximations for the values of binomial distri- bution functions. If we wish to find an approximatio... | prob_Page_337_Chunk5176 |
330 CHAPTER 9. CENTRAL LIMIT THEOREM The individual binomial probabilities tend to 0 as n tends to infinity. In most applications we are not interested in the probability that a specific outcome occurs, but rather in the probability that the outcome lies in a given interval, say the interval [a, b]. In order to find this ... | prob_Page_338_Chunk5177 |
9.1. BERNOULLI TRIALS 331 NA (0,z) = area of shaded region 0 z z NA(z) z NA(z) z NA(z) z NA(z) .0 .0000 1.0 .3413 2.0 .4772 3.0 .4987 .1 .0398 1.1 .3643 2.1 .4821 3.1 .4990 .2 .0793 1.2 .3849 2.2 .4861 3.2 .4993 .3 .1179 1.3 .4032 2.3 .4893 3.3 .4995 .4 .1554 1.4 .4192 2.4 .4918 3.4 .4997 .5 .1915 1.5 .4332 2.5 .4938 3... | prob_Page_339_Chunk5178 |
332 CHAPTER 9. CENTRAL LIMIT THEOREM normal density between the standardized values corresponding to (i −1/2) and (j + 1/2); these values are i∗= i −1/2 −np √npq and j∗= j + 1/2 −np √npq . Thus, P(i ≤Sn ≤j) ≈NA i −1 2 −np √npq , j + 1 2 −np √npq ! . It should be stressed that the approximations obtained by using the Ce... | prob_Page_340_Chunk5179 |
9.1. BERNOULLI TRIALS 333 probability .6 and that the acceptances can be modeled by Bernoulli trials. If the college accepts 1700, what is the probability that it will have too many acceptances? If it accepts 1700 students, the expected number of students who matricu- late is .6 · 1700 = 1020. The standard deviation fo... | prob_Page_341_Chunk5180 |
334 CHAPTER 9. CENTRAL LIMIT THEOREM i.e., ¯p is just a linear function of S∗ n. Since the distribution of S∗ n is approximated by the standard normal density, the distribution of the random variable ¯p must also be bell-shaped. We also know how to write the mean and standard deviation of ¯p in terms of p and n. The me... | prob_Page_342_Chunk5181 |
9.1. BERNOULLI TRIALS 335 0.48 0.5 0.52 0.54 0.56 0.58 0.6 0 5 10 15 20 25 Figure 9.5: Polling simulation. So if the pollster chooses n to be 1200, say, and calculates ¯p using his sample of size 1200, then 19 times out of 20 (i.e., 95% of the time), his confidence interval, which is of length 6%, will contain the true ... | prob_Page_343_Chunk5182 |
336 CHAPTER 9. CENTRAL LIMIT THEOREM 0.48 0.5 0.52 0.54 0.56 0.58 0.6 Figure 9.6: Confidence interval simulation. of their efforts. The reader will note that most of the approximations to p are within 3% of the actual value of p. The sample sizes for these polls were typically around 1500. (In the table, both the predict... | prob_Page_344_Chunk5183 |
9.1. BERNOULLI TRIALS 337 Year Winning Gallup Final Election Deviation Candidate Survey Result 1936 Roosevelt 55.7% 62.5% 6.8% 1940 Roosevelt 52.0% 55.0% 3.0% 1944 Roosevelt 51.5% 53.3% 1.8% 1948 Truman 44.5% 49.9% 5.4% 1952 Eisenhower 51.0% 55.4% 4.4% 1956 Eisenhower 59.5% 57.8% 1.7% 1960 Kennedy 51.0% 50.1% 0.9% 1964... | prob_Page_345_Chunk5184 |
338 CHAPTER 9. CENTRAL LIMIT THEOREM Exercises 1 Let S100 be the number of heads that turn up in 100 tosses of a fair coin. Use the Central Limit Theorem to estimate (a) P(S100 ≤45). (b) P(45 < S100 < 55). (c) P(S100 > 63). (d) P(S100 < 57). 2 Let S200 be the number of heads that turn up in 200 tosses of a fair coin. E... | prob_Page_346_Chunk5185 |
9.1. BERNOULLI TRIALS 339 time the numbers are equally likely to fall above or below this range. Assume that each member decides to come with a given probability p, and that the decisions are independent. How many members are there? What is p? 9 Let Sn be the number of successes in n Bernoulli trials with probability .... | prob_Page_347_Chunk5186 |
340 CHAPTER 9. CENTRAL LIMIT THEOREM (b) Modify your program in (a) to plot the cumulative distribution F ∗ n(x) of the standardized random variable S∗ n = Sn −np √npq . (c) Define the normal distribution N(x) to be the area under the normal curve up to the value x. Modify your program in (b) to plot the normal distribu... | prob_Page_348_Chunk5187 |
9.2. DISCRETE INDEPENDENT TRIALS 341 Let Sn = X1 + X2 + · · · + Xn be the sum of n independent discrete random variables of an independent trials process with common distribution function m(x) defined on the integers, with mean µ and variance σ2. We have seen in Section 7.2 that the distributions for such independent su... | prob_Page_349_Chunk5188 |
342 CHAPTER 9. CENTRAL LIMIT THEOREM -4 -2 0 2 4 0 0.1 0.2 0.3 0.4 -4 -2 0 2 4 0 0.1 0.2 0.3 0.4 n = 2 n = 10 Figure 9.7: Distribution of standardized sums. -4 -2 0 2 4 0 0.1 0.2 0.3 0.4 -4 -2 0 2 4 0 0.1 0.2 0.3 0.4 n = 3 n = 10 Figure 9.8: Distribution of standardized sums. Approximation Theorem As in the case of Ber... | prob_Page_350_Chunk5189 |
9.2. DISCRETE INDEPENDENT TRIALS 343 Central Limit Theorem for a Discrete Independent Trials Pro- cess The Central Limit Theorem for a discrete independent trials process is as follows. Theorem 9.4 (Central Limit Theorem) Let Sn = X1 + X2 + · · · + Xn be the sum of n discrete independent random variables with common di... | prob_Page_351_Chunk5190 |
344 CHAPTER 9. CENTRAL LIMIT THEOREM “correct” grade and the recorded grade. So there are two average grades for the student, namely the average of the correct grades and the average of the recorded grades. We wish to estimate the probability that these two average grades differ by less than .05 for a given student. We ... | prob_Page_352_Chunk5191 |
9.2. DISCRETE INDEPENDENT TRIALS 345 Theorem 9.5 (Central Limit Theorem) Let X1, X2, . . . , Xn , . . . be a se- quence of independent discrete random variables, and let Sn = X1 + X2 + · · · + Xn. For each n, denote the mean and variance of Xn by µn and σ2 n, respectively. De- fine the mean and variance of Sn to be mn a... | prob_Page_353_Chunk5192 |
346 CHAPTER 9. CENTRAL LIMIT THEOREM -4 -2 0 2 4 0 0.1 0.2 0.3 0.4 0.5 0.6 -4 -2 0 2 4 0 0.1 0.2 0.3 0.4 -4 -2 0 2 4 0 0.1 0.2 0.3 0.4 Figure 9.9: Sums of randomly chosen random variables. | prob_Page_354_Chunk5193 |
9.2. DISCRETE INDEPENDENT TRIALS 347 50 55 60 65 70 75 80 0 0.025 0.05 0.075 0.1 0.125 0.15 Figure 9.10: Distribution of heights of adult women. attributes, such as height, are strongly influenced by genetic factors: tall parents tend to have tall offspring. Thus in this case, there seem to be two large effects, namely th... | prob_Page_355_Chunk5194 |
348 CHAPTER 9. CENTRAL LIMIT THEOREM a given species have their genetic information encoded in sets of physical entities, called chromosomes. The chromosomes are paired in each somatic cell. For example, human beings have 23 pairs of chromosomes in each somatic cell. The sex cells contain one chromosome from each pair.... | prob_Page_356_Chunk5195 |
9.2. DISCRETE INDEPENDENT TRIALS 349 a population is that the variance does not seem to increase or decrease from one generation to the next. This was known at the time of Galton, and his attempts to explain this led him to the idea of regression to the mean. This idea will be discussed further in the historical remark... | prob_Page_357_Chunk5196 |
350 CHAPTER 9. CENTRAL LIMIT THEOREM W denoting non-genetic effects. The Hardy-Weinberg Law states that for each Xi, the distribution in the offspring generation is the same as the distribution in the parent generation. Thus, if we assume that the distribution of W is roughly the same from generation to generation (or if... | prob_Page_358_Chunk5197 |
9.2. DISCRETE INDEPENDENT TRIALS 351 Figure 9.11: Two-stage version of the quincunx. is...that the processes of heredity must work harmoniously with the law of deviation, and be themselves in some sense conformable to it. Galton invented a device known as a quincunx (now commonly called a Galton board), which we used i... | prob_Page_359_Chunk5198 |
352 CHAPTER 9. CENTRAL LIMIT THEOREM opened, all of the shot will fall, producing the same distribution as would occur if the shot were not temporarily stopped at the line AB. But the action of stopping the shot at the line AB, and then releasing the compartments one at a time, is just the same as convoluting two norma... | prob_Page_360_Chunk5199 |
9.2. DISCRETE INDEPENDENT TRIALS 353 Figure 9.12: Galton’s explanation of reversion. | prob_Page_361_Chunk5200 |
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