Chapter_II - 5mm3 LA‘MDIK TEEZ£?2"S 56 I...

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Unformatted text preview: 5mm3 LA‘MDIK TEEZ£?2"S 56 I lntroductlon =72 :- The familiar formula ifl=1+x+f+-~=1: me<1 mu) x Chapter II for tlge sugn of a3georpgt1£ilc :eries :esults from (6.21) with n == 1. The cases Conditional Probability and =ann= reteormuas .. . " y Conditional Expectation Z(k+nf=1+2x+hL+H. kxo 1 = f < 1, 6.23 (1 _ x)! 01' lxi ( ) 2 (k + 2)(k + 1)x" = 3 for tx| < 1. (6.24) k=0 (1 '— 3‘) Sums of Numbers 1. The Discrete Case The following sums of powers of integers have simple expressions: The conditional probability Pr{AIB } 0f the 6V6nt A giVen the event 3 is defined by 1+2+ +n £112 . . . = , P A d B 2 Pr{AIB} = iiifii if Pr{B} > 0, (1.1) 2 2 _ n(n + 1)(2n + 1) 1 + 2 + ' ' ' + n — 6 ’ and is not defined, or is assigned an arbitrary value, when Pr{B} = 0. Let 2 2 X and Y be random variables that can attain only countably many differ- 1 + 23 + _ _ . + n3 : "—0‘11).._ ent values, say 0, 1, 2, . . . . The conditional probability mass function 4 mflmofixgwnr=yndfimdm i _Pr{X=xandY=y} 'fPr Y— 0 thr(x)’)— Pr{Y=y} I { “3”) a and is not defined, or is assigned an arbitrary value, whenever Pr{ Y = y] = 0. In terms of the joint and marginal probability mass func— tions px,(x, y) and p,(y) = E, ny(x, y), respectively, the definition is PxKx- Y) My) Observe that pmccly) is a probability mass function in x for each fixed y, i.e., px|y(x!y) 2 0 and E; pxtyGly) = 1, for all x, y. prytxly) = ifpycywo; Jay-=0. 1.... (1.2) 57 RESERVE - 58 ll Conditional Probability and Conditional Expectation The law of total probability takes the form PrlX = x} = Z panama). (1.3) y=0 Notice in (1.3) that the points y where paddy) is not defined are exactly those values for which 19,01) = 0, and hence, do not affect the computa- tion. The lack of a complete prescription for the conditional probability mass function, a nuisance in some instances, is always consistent with subsequent calculations. Example Let X have a binomial distribution with parameters p and N. where N has a binomial distribution with parameters q and M. What is the marginal distribution of X? We are given the conditional probability mass function pXIN(k|n) =(: )P (1 _ P)"_. it k = 0, 1, . . . , n, and the marginal distribution 1M") = (1:!)61'11 * a)“: n = O, 1, . . . , M. We apply the law of total probability in the form of (1.3) to obtain M PrlX = k} = Zopxmklwm) =:*W—”n£k—”pku ”pr—*Wfiqu —q)M-,. z grim — qwiiq 61),: m (1mm x (1 i ) =EKWA€75¥WW0 — gnu-q“ gal??? 1. The Discrete Case 59 Ml :k!(M— k)! In words, X has a binomial distribution with parameters M and pg. (pq)*(1— per)“: k = 0, 1, . . . , M. Example Suppose X has a binomial distribution with parameters p and N where N has a Poisson distribution with mean A. What is the marginal distribution for X? Proceeding as in the previous example but now using N'e" n! ’ pN(n)= n=0,1,..., we obtain Pr{X = k} = Z m~(kln)p~(n) —A n. =Z~k1m _ k),P*1( rp) _ Mean" °° [A(1—p)]””‘ _ kl ; (n — k)! : eminent) k! _... (Ap)ke—Ap _ k! In words, X has a Poisson distribution with mean )tp. fork=0,1,.... Example Suppose X has a negative binomial distribution with parame— ters p and N, where N has the geometric distribution Pivot) = (1 “" BMW“ forn = 1,2, .... What is the marginal distribution for X? We are given the conditional probability mass function +k—1 pxw(kln)=(n k )p"(1—p)*,k=o,1,.... RESERVE 60 ll Condltional Probability and Conditional Expectation Using the law of total probability, we obtain Pr{X = k} = thklnwm n=0 (n+k— 1)! 01W"— 1 W _1), p"(1-p)"(1'-B)B"“‘ = (1 — B)(1 — prp 2(“k k “ Ute)“ = (1 — 3X1 ‘PYP'U — BP)"‘" = (€:g§)(11:gp)k fork: 0,1,.... We recognize the marginal distribution of X as being of geometric form. Let g be a function for which the expectation of 300 is finite. We de- fine the conditional expected value of g(X) given Y = y by the formula ElthllY = y] = Z 8(X)Pny(XlY) . ifpxy) > 0. (1.4) and the conditional mean is not defined at values y for which py(y) = 0. The law of total probability for conditional expectation reads Eigtxn = Z E[g(X)|Y = ylpyo’). (1.5) y The conditional expected value E [3(X )IY = y] is a function of the real variable y. If we evaluate this function at the random variable Y, we ob- tain a random variable that we denote by E [g(X)|YJ. The law of total prob- ability in (1.5) now may be written in the form E[g(X)1 = EtEtgtxin}. (1.6) Since the conditional expectation of g(X) given Y = y is the expectation with respect to the conditional probability mass function Farah): condi- tional expectations behave in many ways like ordinary expectations. The following list summarizes some properties of conditional expectations. In Exercises 61 this list, with or without affixes, X and Y are jointly distributed random variables; c is a real number; g is a function for which E[lg(X )1] < or: ; h is a bounded function; and v is a function of two variables for which E[|v(X, Y)l] < 00. The properties are (1) Eiclgxxl) + czgtizflY = y] = clEtngolY = y} + czngztxnll’ = y]. (1.7) (2) ifg > 0 then E[g(X)IY = y] > 0 (1.8) (3) £1th nlY— y1= E[v(X y)|Y=y1 (1.9) (4) E[g(X)IY = y] = E[g(X)] ifX and Y are independent. (1.10) (5) Elth)h(Y)|Y = y] = hmElthflY = n (1.11) (6) Elth)h(Y)] = 2 magma = ylprty) r (1.12) = E{h(r)E1g(X)IY1}. As a consequence of (1.7), (1.11), and (1.12), with either g E 1 or I: E l, we obtain E[c|Y = y] = c, (1.13) E[h(Y)|Y = y] = My). (1.14) E[g(X)1 = Z E[g(X)|Y = ylpyo) = ElElth)|Y]}- (1.15) Y Exercises 1.1. I roll a six—sided die and observe the number N on the uppermost face. I then toss a fair coin N times and observe X, the total number of heads to appear. What is the probability that N = 3 and X = 2? What is the probability that X = 5? What is E [X ], the expected number of heads to appear? 1.2. Four nickels and six dimes are tossed, and the total number N of heads is observed. If N = 4, what is the conditional probability that ex- actly two of the nickels were heads? RESERVE 62 ll Conditional Probability and Conditional Expectation 1.3. A poker hand of five cards is dealt from a normal deck of 52 cards. Let X be the number of aces in the hand. Determine Pr{X > IIX :_> 1}. This is the probability that the hand contains more than one ace, given that it has at least one ace. Compare this with the probability that the hand con- tains more than one ace, given that it contains the ace of spades. 1.4. A six-sided die is rolled, and the number N on the uppermost face is recorded. From a jar containing 10 tags numbered 1, 2, . . . , 10 we then select N tags at random without replacement. Let X be the smallest num- ber on the drawn tags. Determine Pr{X = 2}. 1.5. Let X be a Poisson random variable with parameter A. Find the con— ditional mean of X given that X is odd. 1.6. Suppose U and V are independent and follow the geometric distri— bution p(k)=p(l~p)* fork=0,1,.... Define the random variable Z = U + V. (a) Determine the joint probability mass function pulz(u, z) : Pr[U= u,Z=z}. (b) Determine the conditional probability maSS function for U given that Z = n. ‘ Problems 1.1. Let M have a binomial distribution with parameters N and p. Con— ditioned on M, the random variable X has a binomial distribution with pa- rameters M and 71'. ' (a) Determine the marginal distribution for X. (b) Determine the covariance between X and Y = M — X. 1.2. A card is picked at random from N cards labeled 1, 2, . . . , N, and the number that appears is X. A second card is picked at random from cards numbered 1, 2, . . . , X and its number is Y. Determine the condi— tional distribution of X given Y = y, for y = 1, 2, . . . . Problems 63 1.3. Let X and Y denote the respective outcomes when two fair dice are thrown. Let U: mian, Y}, V: max{X, Y},andS= U+ V, T: V—~ U. (a) Determine the conditional probability mass function for U given V = v. (b) Determine the joint mass function for S and T. 1.4. Suppose that X has a binomial distribution with parameters p = i and N, where N is also random and follows a binomial distribution with parameters q = 41 and M = 20. What is the mean of X? 1.5. A nickel is tossed 20 times in succession. Every time that the nickel comes up heads, a dime is tossed. Let X count the number of heads ap- pearing on tosses of the dime. Determine Pr{X = 0}. 1.6. A dime is tossed repeatedly until a head appears. Let N be the trial number on which this first head occurs. Then a nickel is tossed N times. Let X count the number of times that the nickel comes up tails. Determine Pr{X = 0}. Pr[X =1], and E[X]. 1.7. The probability that an airplane accident that is due to structural failure is correctly diagnosed is 0.85 , and the probability that an airplane accident that is not due to structural failure is incorrectly diagnosed as being due to structural failure is 0.35. If 30 percent of all airplane acci- dents are due to structural failure, then find the probability that an airplane accident is due to structural failure given that it has been diagnosed as due to structural failure. 1.8. Initially an urn contains one red and one green ball. A ball is draWn at random from the urn, observed, and then replaced. If this ball is red, then an additional red ball is placed in the urn. If the ball is green, then a green ball is added. A second ball is drawn. Find the conditional proba- bility that the first ball was red given that the second ball drawn was red. 1.9. Let N have a Poisson distribution with parameter A = 1. Con- ditioned on N = 14, let X have a uniform distribution over the integers O, 1, . . . , n+1. What is the marginal distribution for X? RESERVE 64 ll Conditional Probability and Conditional Expectation 1.10. Do men have more sisters than women have? In a certain soci— ety, all married couples use the following strategy to determine the num- ber of children that they will have: If the first child is a girl, they have no more children. If the first child is a boy, they have a second child. If the second child is a girl, they have no more children. If the second child is a boy, they have exactly one additional child. (We ignore twins, assume sexes are equally likely, and the sex of distinct children are independent random variables, etc.) (a) What is the probability distribution for the number of children in a family? (b) What is the probability distribution for the number of girl children in a family? (c) A male child is chosen at ran— dom from all of the male children in the population. What is the proba- bility distribution for the number of sisters of this child? What is the prob- ability distribution for the number of his brothers? 2. The Dice Game Craps An analysis of the dice game known as craps provides an educational ex_ ample of the use of conditional probability in stochastic modeling. In craps, two dice are rolled and the sum of their uppermost faces is ob— served. If the sum has value 2, 3, or 12, the player loses immediately. If the sum is 7 or 11, the player wins. If the sum is 4, 5, 6, 8, 9, or 10, then further rolls are required to resolve the game. In the case where the sum is 4, for example, the dice are rolled repeatedly until either a sum of 4 reappears or a sum of 7 is observed. If the 4 appears first, the roller wins; if the seven appears first, he or she loses. Consider repeated rolls of the pair of dice and let 2,, for n = 0, 1, . . . be the sum observed on the nth roll. Then Zn, 2,, . . . are independent identi— cally distributed random variables. If the dice are fair, the probability mass function is (2.1) 2. The Dice Game Craps 65 LetA denote the event that the player wins the game. By the law of total probability, 12 Pr{A} = ZPrlAlzi = klpztk). (2.2) k=2 Because Zo = 2, 3, or 12 calls for an immediate loss, then Pr{A|z, = k} = 0 for k = 2, 3, or 12. Similarly, 20 = 7 or 11 results in an immediate win, and thus Pr{A|Zo = 7} = Prpilz, = 11} = 1. It remains to consider the values Zn 3 4, 5, 6, 8, 9, and 10, which call for additional rolls. Since the logic remains the same in each of these cases, we will argue only the case in which Z0 = 4. Abbreviate with at = Pr{A|Zu = 4}. Then a is the probability that in successive rolls 2,, 2,, . . . of a pair of dice, a sum of 4 appears before a sum of 7. Denote this event by B, and again bring in the law of total probability. Then 12 e = Pr{B} =;Pr{aiz, = k}pz(k). (2.3) Now Pr{BIZ, = 4} = 1, while Pr{BIZ, = 7} z o. If the first roll results in anything other than a 4 or a 7, the problem is repeated in a statistically identical setting. That is, Pr{BIZl 7— k}= o: for k is 4 or 7. Substitution into (2.3) results in a=pzr4l><1+pzm><0+ Z pz(k)>< a #41 = 132(4) + [1 * 122(4) - peanut, 01' 192(4) 0! = 172(4) + 132(7). (24) The same result may be secured by means of a longer, more computa- tional, method. One may partition the event B into disjoint elemental events by writing Be {Z.=4}U{Z,#=4er7,z,:4} U{Zt¢40r7,Zz=#4or7,Z,=4}Um. RESERVE 66 ll Conditional Probability and Conditional Expectation andthen PHB} = Pr{zl = 4} + Pf{Zl =# 4 or 7, Z2 = 4} +Prizl#4or7.zz#=4or7,23=4}+..._ Now use the independence of 21, Z), . . . and sum a geometn'c series to se— cure PIlB} = 192(4) + [1 — 132(4) — Pz(7)]Pz(4) + [1 — FAQ—12(7)}: 2(4) + ‘ ‘- _ M4) [92(4) + 122(7) in agreement with (2.4). Extending the result just obtained to the other cases having more than one roll, we have p200 PzUC) + 132(7) Finally, substitution into (2.2) yields the total win probability Pr{AIZn : k} = for k = 4, 5, 6, 8, 9, 10. P2002 PF{A} = P (7) + P (11) + ——-. (2.5) 2 Z k=4.s:s.:s,9.m P7. (‘0 + 192(7) The numerical values for pz(k) given in (2.1), together with (2.5), deter— mine the win probability Pr{A} = 049292929 - - - . Having explained the computations, let us go on to a more interesting question. Suppose that the dice are not perfect cubes but are shaved so as to be slightly thinner in one dimension than in the other two. The numbers that appear an opposite faces on a single die always sum to 7. That is, l is opposite 6, 2 is opposite 5, and 3 is opposite 4. Suppose it is the 3—4 di- mension that is smaller than the other two. See Figure 2.1. This will cause 3 and 4 to appear more frequently than the other faces, 1, 2, 5, and 6. To see this, think of the extreme case in which the 34 dimension is very thin, leading to a 3 or 4 on almost all tosses. Letting 1’ denote the result of toss— ing a single shaved die, we postulate that the probability mass function is given by 2. The Dice Game Craps 67 Pro) = Pr(4) = E + 25 E19“ Pr“) = Pra) = 171(5): 155(6): 3 *3 =13— where e > 0 is a small quantity depending on the amount by which the die has been biased. A Cubic Die A Shaved Die Figure 2.1 A cubic die versus a die that has been shaVed down in One dimension. If both dice are shaved in the same manner, the mass function for their sum can be determined in a straightforward manner from the following joint table: ' RESERVE 68 II Condltlonal Probability and Conditional Expectation It is easily seen that the probability mass function for the sum of the dice is ' 13(2) = pi = 1702). m3) = 2p3 = p(11). 2700 = p-(p— + 212+) = 1200). 10(5) = 41241. = 19(9). 1203') = p3 + (p. + it)2 = 226). pm = 4123+ 2191- To obtain a numerical value to compare to the win probability 0.492929 - - - associated with fair dice, let us arbitrarily set 3 = 0.02, so that p_ = 0.146666 - - - and 1,0,, = 0.206666 - - - . Then routine substitu— tions according to the table lead to p(2) = p(12) = 002151111, p(5) = p(9) = 012124445, p(3) = p(11) = 004302222, p(6) = p(8) = 0.14635556, (2.6) 13(4) : p(10) = 008213333, p(7) = 017146667, and the win probability becomes Pr{A} = 0.5029237. The win probability of 0.4929293 with fair dice is unfavorable, that is, is less than %. With shaved dice, the win probability is favorable, now being 0.5029237. What appears to be a slight change becomes, in fact, quite sig— nificant when a large number of games are played. Sec III, Section 5. Exercises 2.1. A red die is rolled a single time. A green die is rolled repeatedly. The game stops the first time that the sum of the two dice is either 4 or 7. What is the probability that the game stops with a sum of 4? 2.2. Verify the win probability of 0.5029237 by substituting from (2.6) into (2.5). 2.3. Determine the win probability when the dice are shaved on the 1—6 faces and 13., = 0.206666 - - . and p- = 0.146666 - - - . Problems 69 Problems 2.1. Let X1, X2, . . . be independent identically distributed positive ran- dom variables whose common distribution function is F. We interpret X1, X2, . . . as successive bids on an asset offered for sale. Suppose that the policy is fOIIOWed of accepting the first bid that exceeds some prescribed number A. Formally, the accepted bid is X”, where N= min{kE: 1;X,>A}. Set a = Pr[X1> A} andM = E[XN]. (a) Argue the equation M=ixdF(x)+(1— 00M A by considering the possibilities, either the first bid is accepted, or it is not. (b) Solve for M, thereby obtaining no M = a" I xdF(x). A (c) When Xl has an exponential distribution with parameter A, use the memoryless property to deduce M = A + A". ((1) Verify this result by calculation in (b). 2.2. Consider a pair of dice that are unbalanced by the addition of weights in the following manner: Die #1 has a small piece of lead placed near the feur side, causing the appearance of the outcome 3 more often than usual, while die #2 is weighted near the three side, causing the out— come 4 to appear more often than usual. We assign the probabilities Die #1 12(1) = 17(2) = 11(5) = p(6) 2 0.166667, p(3) = 0.186666, p(4) = 0.146666; RESERVE 70 ll Conditional Probability and Conditional Expectation Die #2 10(1) = 19(2) = p(5) = p(6) = 0.166667. p(3) = 0.146666, p(4) = 0.186666. Determine the win probability if the game of craps is played with these loaded dice. 3. Random Sums Sums of the form X = g, + . ~ - + EN, where N is random, arise frequently and in varied contexts. Our study of random sums begins with a crisp definition and a precise statement of the assumptions effective in this sec- tion, followed by some quick examples. We postulate a sequence g“ (3,, . . . of independent and identically dis— tributed random variables. Let N be a discrete random variable, inde- pendent of .51, .52, . . . and having the probability mass function p~(n) = Pr{N: n} forn= 0,1,....Definetherandomsumbe O ifN=0, 31 X_[Ei+---+§N ifN>0. (') We save space by abbreviating (3.1) to simply X = .51 + - - - + E”, understanding that X = 0 whenever N = 0. Examples (:1) Queueing Let N be the number of customers arriving at a service facility in a specified period of time, and let g, be the service time required by the ith customer. Then X = g, + - - - + if” is the total demand for service time. (b) Risk Theory Suppose that a total of N claims arrives at an in- surance company in a given week. Let 6,. be the amount of the ith claim. Then the total liability of the insurance company is X = ‘51 + ' ' ' + giv- (c) Population Models Let N be the number of plants of a given species in a specified area, and let g, be the number of seeds pro— duced by the ith plant. Then X = E, + - -- + if” gives the total number of seeds produced in the area. (d) Biometrics A wildlife sampling scheme traps a random number N of a given species. Let {3, be the weight of the ith specimen. Then X = g, + - - - + 51,5 the total weight captured. 3. Random Sums 71 When 15,, £2, . . . are discrete random variables, the necessary background in conditional probability was covered in Section 1. In order to study the random sum X = g, + . - - + if” when 5,, 5,, . . . are continuous random variables, We need to extend our knowledge of conditional distributions. 3.1. Conditional Distributions: The Mixed Case Let X and N be jointly distributed random variables and suppose that the possible values for N are the discrete set n = 0, 1, 2, . . . . Then the ele— mentary definition of conditional probability (1.1) applies to define the condition...
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