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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., pxy(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(kn) =(: )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 — gnuq“ 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"(1p)"(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[cY = 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 sixsided 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{Az, = k} = 0 for k = 2, 3, or 12. Similarly, 20 = 7 or 11 results in an
immediate win, and thus Pr{AZo = 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{AZu = 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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