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42 EXPECTATIONS AND INDEPENDENCE^ Chap. 2

Let N be a non-negative integer-valued random variable which is independent of {Xt,X 2 ,. ..}, andlet

E[N] = c, Var(AO = d^2.

Let S 0 = O, S^1! = Jfj, 5 2 = Jf, + JSf 2 ,.. ., and let Y = SN. (a) ComputeE[F|Aa E[YZ\N]. (b) Compute E[ Y], E[ Y^2 ], Var( Y). (c) Show that for any a e K+,

where

F(a) = a e e [0, 1].

CHAPTER 3

Bernoulli Processes

and Sums of Independent

Random Variables

The chaptcr just finished completes our account of the preliminaries necessary for stuâying stochastic processes. Especially in vietv ofthe monstrous looks ofthe last section, it seerns ali too advisable to inquire how the reader's patience is holâing out and to assure him that he ivill in time come to appreciate the true frienáliness ofthese concepts. For a âeeper treatment andfor the proofs which wehave omitteâ, we refer the reader to CHUNG [2].

Consider an experiment consisting of an infinite sequence of trials. Sup- pose that the trials are independent of each other and that each one hás only two possible outcomes: "success" and "failure." A possible outcome of such an experiment is (S, F, F, S, F, S,.. .), which stands for the outcome where the first trial resulted in success, the second and third in failure, fourth in success, fifth in failure, sixth in success, etc. Thus, the sample space Q of the experiment consists of ali sequences of two letters 5" and F, that is,

Q = [a>: co = (coi, co 2 ,.. .), ca, is either S or F}.

We next describe a probability measure P on ali subsets of íl. Let O < p < l, and define q = l — p. We think of p and q as the probabilities of success and failure át any one trial. The probability of the event {ca: ol — S, a> 2 = S, o> 3 = F] should then be p-p-q — p^2 q, and the probability ofthe event {co: o>i = S, o) 2 = F, co 3 = F,a> 4 = F, co 5 = S} should be p-q-q-q-p = p^2 q^3. For each n, we consider ali events which are specified by the first n trials and define their probabilities in this manner. This, in addition to the conditions in Defini- tion (1.1.6), completely specifies the probability P. For each co e Í2 and n e {1,2,...}, define Xn() = l or O according as con = S or &>„ = F. Then for each n, we havê a random variable Xn whose only values are l and 0. It follows from the description of P that Xlt X 2 ,... are independent and identically distributed with P{Xn = 1} = p, P{Xn = 0} - </. In the first three sections of this chapter we will be interested in the propertics of stochastic processes such as {Xn ;n= 1,2,...} and other processes defined in terms of {Xn}.

•l!

BERNOULLI PROCESSES Chap. 3^ Sec.^2 NUMBERS OF SUCCESSES^47

\ N,,; n e N} whose time parameter is discrete N. = (O, l ,.. .} and whose state space is discrete and again is [O, l ,.. .}. We start with simple descriptiy_e quantities. Using the definition of Nn, Corollary (2.1.25), and the result (1.5) above, we obtain

(2-2) E[Nn]^ =

E[Xn] = np

for ali n > l, and clearly, E[N 0 ] = O, só E[Na] = np for ali n > 0. Similarly, using the result in Exercise (2.3.9) and the independence of X^..., Xn,

(2.3) Var(7Vn) =^ Var(Zn) =^ npq

by (1.6). From (2.2) and (2.3) we can get, by using the formula (2.1.28) for the variance,

(2.4) E[Nl\ = Var(AO + (E[Nn])^2 = npq + n

We could have computed E[N%] directly by noting that

só that

E[N

] = i; E[xn + Ê s£[M] ;=i /=i w = np + n(n — T)p^2 = n^2 p^2 + npq.

This last computation indicates how to go about obtaining the higher mo- ments of N,. Note that we have computed the expected value and variance without knowing the distribution of Nn. However, if one needs to compute E[g(Nn)] for arbitrary g, then we need the distribution of Nn. This we do next. The use of conditional probabilities in the proof below illustrates one of the main techniques found useful in studying stochastic processes. ^ ^ £*-

(2.5) LEMMA. For any w, A: e N *

P{Nn+1 = k} = pP{Nn = k-l} + qP{Nn = k}.

Proof. Fix n and k. Using Theorem (1.3.3) with A = {Na+l = k] and

RJ = l N,, ./], we get

/ ' l ^ (^) B , i = * } =A-} :^ - Ç^ P{Nn,, =k\Nn= j ] P{Nn^ = j }.

Since [Xlt... , Xn] is independent of Xn+1, by Definition (2.2.24), Nn =

• • • • + Xn is independent o f X (^) n + 1. Using this fact along with Nn+l = Nn Xa+1, we obtain

P{Nn+1 = k\Nn=j} = P{Xn+í =

p if y = k - l,

q if j = k,

O otherwise.

Putting this in (2.6), we obtain the desired result. (^) D

For n = O, N 0 = O and therefore P{N 0 = 0} = l, and P{N 0 = k] = O for k 3=. 0. These, along with Lemma (2.5), specify the values of P{Nt = k}. The values of P{A^i = k} found, along with Lemma (2.5), give the values of P[N 2 = k], etc. Doing this a few steps, we obtain the table shown in Figure 3.2.1. From the table we recognize the nth row as the terms in the binomial expansion of (q + p)". Then we prove this conjecture.

n

. n n n n

= 0:

= 2: = 3: = 4:

• = 0

1 q l^2 l^3 q*

k = l

0 P 2qp 3q^2 p W P

k = 2

Q 0 P^2 3qp^2 6qlp>

k = 3

0 0 0 P^3 4qp^3

k =

0 0 0 0 P

Figure 3.2.1 Using the probabilities P{Nn = k] in one row, one obtains the values P[Nn+ 1 = k] in the next row.

(2.7) THEOREM. For any n e N

P{Nn = (^) *, k = O,... , «.

Proof. For n = O, the claim is true, since NQ = 0. We now make the induction hypothesis that the formula is true for n = m and ali k. To complete the proof we need to show that, then, the claim is also true for n = m + l and ali k. For n = m + l, k = O, Lemma (2.5) and the induction hypothesis give

P{Nm+l =0) =p.Q + q.qn> = q™+i

as claimed. For n = m + l and O < k < m + l, again by Lemma (2.5) and

48 BERNOULLI PROCESSES^ Chap. 3

the induction hypthesis, we have

mi k(m- k)' ml (k- l)(m -k) (m+1) kl(m+ l - k)\l

só the claim is again true.

nkgm-k+l^ *^ i^ J_
p q \m - k + l + k.

D

We have now specified the distribution of the number of successes in the first n trials. Noting that Nm+n — Nm = Xm+l + • • • + Xm+n ajsa is the sum of n independent and identically distributed Bernoulli variables, we deduce that the distribution of Nm+n — Nm is the same as that of Nn. This we state as ~

(2.8) COROLLARY. For any m, n e

k

=

independent of m.fr

£. The distributfon figuring above is called the binomial distribution. As we ^ had noted before, P{Nn = k] is the fcth term in the binomial expansion of (q + p)". The coefficient of the &th term is usually denoted by ( j? j (read n above k), that is,

(29) ,k) k(n-k)\

(2.10) EXAMPLE. Using Corollary (2.8), we obtain

= 4} =

P{NÍ3 - N 7 = 3} = K LJ

S<> lhe corollary above enables us to compute the distribution of N,,H.a — Nni l « ' r a l i n, in. I5ul we still are not able to compute compound probabilities

Sec. (^2) NUMBERS OF SUCCESSES (^) 49

such as P{N 5 = 4, N-, = 5, N 13 = 8). The means for doing that is furnished by the following

(2.11) THEOREM. For any m, n e N

P{Nm+n ~Nm = k\N 0 ,...,Nm} = P{N,n+n ~Nm = k} = (nk )^ <T *

for k = O,... , n.

(2.12) EXAMPLE. To illustrate the use of this theorem, suppose we are interested in evaluating the joint probability

P[NS = 4,N 7 = 5,N 13 = 8}.

The event whose probability is desired is equal to the event {N 5 = 4, N 7 — N 5 = l, N (^) Í 3 — N 7 = 3}. By Theorem (2.11), the random variable N 13 — N^ is independent of N 0 ,... , N 7. Thus,

P{7Y 5 =4,N 7 -N 5 = l, N 13 - N 7 = 3} = P{N 13 - N 7 = 3}P{N 5 =4,N 7 ~ N 5 = 1}.

Again by Theorem (2.11), N 7 — N 5 is independent of N „,..., N 5 and there- , • fore of N, in particular. Hence, ,. 'fi' ~~~~ " '- ..... -• -, ;.. • ;.,-.. '•! '

P{N 5 = 4, N 7 - N 5 = 1} = P{N 5 = 4} P{N 7 - N 5 = 1}.

Putting ali these together,

= 8} = P[N 5 = 4}P{N 7 = 1}P{N 13 ~ (^393) = 200p»q',

where the last line followcd from Corollary (2.8), which was also put in Theorem (2. 11). Finally, suppose we want to compute E[NSN^]. Writing Ns = N 5 + (Ns — N 5 ), we have

E[N 5 NS] = E[Ns(Nf + (Ns - Ns))] = E[NI + Ns(Nt -N,)] = E[N^25 ] + E[N 5 ]E[NS - N 5 ],

where to get the last line we first used Corollary (2.1.25), and then used the independence of yV 5 and NS—NS (which is implied by Theorem (2. 1 1) above)

BERNOULLI PROCESSES Chap. 3

by Proposition (2.2.15). Putting the last two results in (2.15), we obtain

E[Y\ ATO,... , 7VJ = £ h(Nn, i,, ..., 0*0-,) • • • jr(O = f(Nm) i

independent of A^,... , Nm_í. Finally, by Corollary (2.2.22), this implies that

E[Y\N 0 , ...,Nm} = f(Nm) = E[Y\Nm],

as was to be shown. G

Theorem (2.14) above is satisfied by many processes of much less specific structure. Such processes are called Markov chains, and we will later study them at some length. The particular Markov chain we have, namely [Na; n e N], hás the further property mentioned in Corollary (2.13), and this makes our present job easier. The following is to illustrate the workings of Theorem (2.14).

(2.16) EXAMPLE. We want to compute the conditional expectation of Ni: given Ns. We have

E[NÍ , | N 5 ] = E[N 5 | Ns]

The second equality made use of Corollary (2.2.12); to pass to the third equality we used the fact that E[NS \N 5 ] = N 5 by Proposition (2.2.15), and the fact that N, (^) t — N 5 is independent of N 5 by Corollary (2.13). G

(2.17) EXAMPLE. Let us compute E[N 5 Ng] again (cf. Example(2.12)forthe same) to show that conditional expectations, used properly, can reduce the work involved. By Proposition (2.2.16), E[N 5 Ns] = E[E[N,Ng\N 5 ]]. On the otherhand,

E[N 5 NS \ Ns] = N 5 (N 5 + 3p).

Hence,

3pN 5 ] = E[Nl] + 3pE[N 5 ] = 25p^2 + 5pq + 3p-5p = 40p^2 + 5pq.

(2.18) EXAMPLE. We want to compute

z E\N 5 Nn\N 2 ,N 3 ].

G

Sec. 3 TIMES OF SUCCESSES

By Theorems (2.2.19) and (2.14),

(2.19) Z = E[E[Nstf,, | tf 0 ,... , N 5 ] | tf 2 , N 3 ]

Using the result of Example (2.16),

E[N 5 Nl 11 tf 5 ] = N 5 £[tf, j | tfs] = N 5 (N 5 + só that

Again applying Theorems (2.2.19) and (2.14) and Proposition (2.2.15) in that order, Z = E[E[N^25 1 N 0 , N,, N 2 , N 3 ] \ N 2 , N 3 ]

(that is, given N 2 and N 3 , we need only N 3 as far as predicting the future is concerned). Writing N 5 = N 3 + (tf 3 — tf 3 ), thenusing Corollary (2.13) to deduce the independence of tf, — tf 3 and tf 3 , we get

Z = E[Nl + 2N,(tf 5 - tf,) + (tfs - tf 3 )^2 + 6pN 5 [ tf 3 ] = tf l + 2N 3 E[N 5 - N 3 1 tf 3 ] + E[(NS - tf 3 )^21 tf 3 ] + 6pE[N 5 \ N 3 ] = tf^2 -f 2N 3 E[N 5 - tf 3 ] + E[(NS - tf 3 )^2 ] + 6p(N 3 + 2p) == tf^2 + 2N 3 -2p + 4p^2 + 2^ + 6/?tf 3 + Up^2

• I6p^2 + 2pq. Q

Wc now have a number of theorems at our disposal to compute the prob- abililics or expectations rclatcd to the sequence {N,,}. In the next section we will obtain results o f a similar nalurc about the sequence of times (random) M! which the successes occui'.

'. '1'inies of Successes

l cl | A ' , , ; / / ' l \ l H" a Hcrninilli procirss w i t h probability/; of success. For l i x c i l <o • í i, considcr l h o i v a l í / ; i l i < m A' (^) t(m). \ ..(<»), A ',(<•;>),... ofthis process. I l u. i', a se<|iicnu' nl' oiu-s a i u l /rio:,. |)ciiolc hy '/',(o>), T 2 (co),... the n n l u i ". corresponding l « > l h e :,IHIT:,NÍVI' DiK's. l''or example, if X^at) = O, \ .(.->) l , A , ( o > ) d, A ',((„) I , A '. , ( (^) ( , < ) l Ihcn r,(co) = 2, T 2 (co)

54 BERNOULLI PROCESSES Chap. 3

•• = 4, T 3 (a)) = 5,.... Then Tk is the trial number at which the kíh success

occurs. In this section we are interested in the sequence of random variables

(3.1) EXAMPLE. We have the following model as an approximation of the

traffic flow through a fixed point on an uncongested highway. The time axis

is divided into intervals of length l second each. If there is a vehicle crossing

during the «th second, we put Xn = l ; otherwise put Xn = 0. The traffic flow

is such that the numbers of vehicles passing the fixed point in disjoint time

intervals are independent. If, furthermore, the traffic density remains constant

over time, then the sequence [Xn; n > 1} is a Bernoulli process. Now, Nn is

the number of vehicles passing the fixed point in the interval (O, n], and Tk is

the second during which the kih vehicle crosses the given point. Q

There is a fundamental relation between the times of successes Tk and the

numbers of successes Nn. Fix <a e Q; suppose that for the realization ca, the

kth success hás occurred at or before the «th trial; that is, suppose Tk(co) < n.

Then the number of successes in the first n trials must be at least k ; that is, we

must have Nn(co) > k. Thus, if Tk(a>) < n, then N^co) > k. Conversely, if

N*(co) > k, then Tk((o) < n.

Another relation between the Tk and the Nn is obtained as follows. Fix

ca e Í2 again, and suppose Tk((o) = n. This implies that for that realization

ca, there were exactly k — l successes in the first n — l trials and a success

occurred at the nih trial, that is, Nn_l(cai) = k — l and Xn((o) = 1. Conver-

sely, if Nn_l((ai) — k — l and Xn(ca) = l, then Tk(co) = n. We state these

two relationships next as a lemma, and then use them to obtain the distri-

bution of the Tk from the known distribution of the Na.

(3.2) LEMMA. For any co c Í2, k —•• l , 2,... , and n > k, we have

(a) Tk(a>) < n if and only if Ntt(co) > k,

(b) Tk(co) = n if and only if Nn_i((o) = & - l, Xn((o) = 1.

(3.3) THEOREM. For any A: e {1,2,...},

P{Tk <n} = ±(n. \p'q'"i, n = k, k + l ,... ;

j = k \ J /

P{Tk = „} = J

V v^ * /

Proof. Fix k e {1,2,...} and n > k. By Lemma (3.2a) the event [Tk < n]

is equal to the event {Nn > k}. Hence,

P\Tk - ' n] :•-• P{Nn ^ k} = ± P{Nn = j ] ,

j k

Sec.3 TIMES OF SUCCESSES 55

which implies (3.4) when P{Nn = j ] is replaced by its value, using Theorem

By Lemma (3.2b) the events {Tk = n] and {7Vn_i = k — l, Xn = 1} are

the same, and by Definition (2.2.24) of independence, Xn and Na-l are in-

dependent. Hence,

P{Tk = n} = P{AT,_, = k - 1} P{Xn = 1},

which yields the formula (3.5) when the two probabilities on the right are

evaluated using (2.7) and the definition of the Xn. Q

The preceding theorem, in principie, enables us to compute probabilities

and expectations concerning only one of the Tk. In order to consider more

complex events involving several of the Tk, we need to know more about the

structure of the process \Tk\ k = 1,2,...}. In fact this will further yield

theorems which considerably simplify the computations of quantities such as

E(Tk\ Var(7U etc.

Fix w e Q and suppose r, (o) = 3, T 2 (co) = 4, T 3 (a)) = 5, T 4 (ca) = 12.

Then, in order that T 5 (co) = 17, we must have X^ 3 ((o) = O, X 14 (co) = O, Xl 5 (co)

= O, Xl6((a) = O, Xn(co) = 1. More generally, in order that T 5 (co) = 12 + m,

we must have Xl2+i(co) = O,... ,X12+m^ 1 (co) = O, X12+m(co) = 1. We note

that, given T 4 , the information concerning Tlt T 2 , and T 3 had no value in

predicting T 5. In fact the same is true in predicting T 5 — T 4. These observa-

tions lead to the following two theorems. (Throughout the following we have

7' 0 = O, and TI, T 2 ,... as the times of the first, second,... successes in a

Bernoulli process {Xn', n > 1} with probability p of success in any one trial.)

(3.6) THEOREM. For any Â; e N and n>k,

P[Tk+í =n\T 0 ,...,Tk} = P{Tk+1 = n \Tk}

independent of 7',,,... , Tk..í.

Proof. Fix A' and n and consider lhe condilional probability f ( T a ,... , Tk)

on lhe left, tliat is, for iiUegers O «„ <i, • • • • '. ak •• a We have

/(«,„... , ak) •- P'J\ n //| 7',, ti ....... Tk ak}.

II /; í/j í/, then this conditional probahilily is xero. Otherwise, if n > ak

ii, in order that 7'A ••—• a, Tk.tl :^ //, we must have Xa+í = O,... , Xn_l = O,

V,, l. llcncc,

/ < " ....... "J O,^... ,^ X,,^^ =^ O,^ AT,^ =^ 1}^ =

58 BERNOULLI PROCESSES Cfldp. 3

by Corollary (3.10), 7, T 5 — TI, T 7 — T 5 are independent, we can write

P{T, = 3, r, - r, = 6, r 7 - r 5 = 8} = P{Tl = 3}P[T 5 -Tl= 6}P{T 7 - T 5 = 8} = P{T 1 = 3}P{T, = 6}P{T 2 = 8],

where the last equality is again obtained from (3.10). Now using Theorem (3.3) to evaluate these, we obtain

P{Tl = 3, T 5 = 9, T 7 = 17} = P{r, = 3}P{T 4 = 6}P{T 2 = 8}

= G»?^2 )(iQpVX7/»V) =^ D

(3.17) EXAMPLE. A certain component in a large system hás a lifetime whose distribution can be approximated by n(ni) = pqm~l, m > 1. When the component fails, it is replaced by an identical one. Let Tí, T 2 , • • • denote the times of failure; then Uk = Tk — TV ^ is the lifetime of the A:th item replaced. Since the components are identical,

P[Uk = m] = pq" m>,

and the items have independent lifetimes. Thus [Tk] can be looked upon as the times of "successes" in a Bernoulli process. Past records indicate that the first three failures occurred at times 3, 12, and 14. We would like to estimate the time of the fifth failure. By Theorem (3.6),

Writing T 5 = T 3 + (T, — T 3 ), using Corollary (2.2.12) to write the condi- tional expectation as the sum of the conditional expectations, noting that E[T 3 \ T 3 ] = T 3 by Proposition (2.2.15) and that E[T 5 - T 3 \ T 3 ] = E[T 5 — J 3 ] by the independence of T 5 — T 3 from T 3 (this is by Corollary (3.10)), we have

E[TS \ T 3 ] = E[T 3 \ T 3 ] + E[T 5 - T 3 \ T 3 ] = T 3 + J-.

Hence, the quantity of interest is

E[T 5 \T 1 =3,T 2 = 12, T 3 = 14] = 14 + J-.

If/> = 0.60, this is 17.33. D

Sec. 3 TIMES OF SUCCESSES 59

(3.18) EXAMPLE. In the preceding example suppose that each item costs c dollars when new. Assuming that a dollars today is worth l dollar during the next period, we now compute the expected total discounted value of ali future expenditures on this component. It is assumed that the discount rate remains the same throughout. There- fore, c dollars spent at time n is worth ca" dollars at present. Note that a is less than l ; if the "inflation" rate is 0.06, for example, then a = 1/(1 + 0.06). Fixco e Cl; the Mi replacement occurs at time Tk(ca). That kth item costs c dollars, and therefore its discounted value is

Thus, the total discounted cost is

co e £2. k=\

The expected value of a sum being the sum of the expectations (cf. Theo- rem (2.1.34)), we have

E[C\ = f] cE[uT*]. k=l

Writing Tk = T, + (T 2 - T,) + ---- h (Tk - TV,), noting that TlfT 2 - TI,... ,T.k — Tk^i are independent by Proposition (3.9), and applying Pro- position (2. l .26) on the expected value of such products, we obtain

E[y, -]

T"] =

Since T,,T 2 — TI,... have the same geometric distribution,

Ik-ncc,

c (^) \ «e.ou/ , a/; c n

We close this section with a gcneralization of the computations appearing in l'xaniplc (3.18). Suppose that the value of the amount spent at time n is «•t|ii;il Io /'(") in tcrms of present worth. In the preceding example we had / ( / ; ) ca". Now we only assume that f (ri) > O for ali n. In fact, the result ( l l')) helow hokls for any / for which J] /(«) K^ absolutely convergent.

60 BERNOULLI^ PROCESSES^ Chap. 3

We would like to compute

Without further information onf, it looks as if we have to use Theorem (3.3). But we can do better if we examine the sum S = 2*=i f(Tk). If, for co e Q, r, (o) = 2, r 2 (eo) = 5, r 3 (<») = 10, r 4 (co) = 11, etc., then S(<a) = /(2) + /(5) + /(IO) + /(!!)+ • • •. Note that 2, 5, 10, 11,... are precisely those índices n for which Zn(co) = l, and that 5(to) = /(l) Z,(o) + /(2) Z 2 (co) + /(3) X 3 (co) + • • •. This is true in general, and we can write

* = 1^ S^ f(Tk)^ = S

The rest is easy :

= S f(n~)E[Xn] = p 2 /(«). n = l n = l

4. Sums of Independent Random Variables

In comparing the stochastic structures of the process [Nn; n e N] of Sec- tion 2 and the process {T,,; n e N] of Section 3, as revealed for example by Corollaries (2.13) and (3.10) respectively, we see that they are very much alike. As we mentioned before, this is due to the facts that N! — N 0 , N 2 — N I ,... are independent and identically distributed and that TI — T 0 ,T 2 — TI,... are also independent and identically distributed. It is this property that we now will put in a general setting. Let Ylt Y 2 ,. • • be a sequence of independent and identically distributed random variables, and define for each n e N

O

Yl

if n = O, i f « > 1.

We are interested in the stochastic process [Zn; n e N] and, especially, its limiting behavior as n approaches infinity. These limit theorems have applica- tions in data collection and statistical estimation, and we shall point out the way they are used. The following property is immediate from the definition of the Zn and Ddiiiilion (2.2.24) of indcpcndence.

SUMS OF INDEPENDENT RANDOM VARIABLES 61

(4.2) PROPOSITION. The stochastic process {Zn; H e N) hás stationary and independent increments; that is, (a) ZBI — ZBO, ZB2 — Zn i ,... , Znt — Znt^ are independent for any in- tegers O < HO < nl < • • • < nk\ and (b) the distribution of ZB+m — Zn is independent of n.

Conversely, if a process {Zn} hás stationary independent increments, then taking H 0 = O, H, = l,.. ., nk = k shows that Zt — Z 0 , Z 2 — Zj,... are independent and identically distributed. Hence, ali discrete parameter pro- cesses with stationary independent increments are obtained in this manner. Proposition (4.2) can be used to prove, just as in Theorem (2.14), the fol- lowing result: in words, as far as predicting the value of a random variable W depending on the future Zn, Zn + 1 ,... of the process {ZB} is concerned, allpast information concerning Z 0 ,... , ZB_t becomes worthless once the present value Za is given.

(4.3) PROPOSITION. Suppose W = g(Zn, Zn+1,.. .) for some function g > O and some integer n e N. Then

_E\Yr_^ CTM/I 7 j ^Q, • • • ,^ 7 1 _-in_^ — I-j\jV P\W\ 7 j ^rt^ l j'

The following computations are easily made by using the definition of Zn given by (4.1). Suppose

(4.4) E[Yn] = a, Var(Fn) = b\

Then (4.1) and Corollary (2.1.25) give

(4.5) E[Zn] = na,

and (4.1), the independence of Y1} Y 2 ,.. ., and (2.3.9) imply

(4.6) Var(Z,,) = nb^2.

(4.7) EXAMPLC. Successivc customcrs arriving at a gás station for fuel spcnd independent amounts. Lct Y{, K 2 ,... be the amounts (measured in ilollars) demanded by the lirst, second,... customers. Then Z,, is theamount ilcmandcd by the first n customers. If an avcrage demand is 4.87 dollars, Ilicn the amount demanded by the first 9 customers hás the expected value ') 4.87 = 43.83 dollars. G

In lhe preceding example it is easy to convince ourselves (by considering 11 u- iiulepcndcncc of the actions of different customers) that YÍ,Y 2 ,... are indi-pcndent and identically distributed. But how do we find out the expected

64 BERNOULLI PROCESSES Chap. 3

probability of success is

p = P{Xk = 1} = P{Yk <t} = (p(t).

Now, note that Fn(t, co) = Nn(co)/n is the average number of successes in the first n trials. Applying Theorem (4.10) to the process [Na], we get (as in Exam-

lim Fn(t, 03) = Hm :^ == p =

For fixed n and w, t — > Fn(í, co) is a step function increasing at each num- ber Yk(aj) by a jump of magnitude l /n. It is called the empirical distribution based on n observations. Proposition (4.13) shows that for large n it approxi- mates the distribution function ç?. We have seen in Proposition (4.9) that for large n the average ZJn is close to the expected value a of the increments Yk. Somewhat better information is contained in the following famous result. It is called the central limit theorem.

(4.14) THEOREM. For any t e K,

lim

We omit the proof. In words, for large n, the distribution of the sum Zn is approximately the normal distribution with mean na and variance nb^2. As with the weak law and strong law of large numbers, this theorem also holds for any distribution for the Yn provided that the variance b^2 is finite. The Standard normal distribution ç is defined by

The corresponding density function / (defined by f (y) = drawn in Figure 3.4.2. Since f (y) = f (—y) for ali y, we have

~y'n^ ) is

(4.16) (— X) = l — <p(x), — <^ X^ <oo.

Figure 3.4.2 The standard normal density function /. In the table of Figure 3.4.3, the value tp(x) is the área under/to the left of x.

Sec. 4

In view of this only.

x

SUMS OF INDEPENDENT RANDOM

relation, it is sufficient to This is done in the table in Figure

<p(x)

x

<p(x)

list the 3.4.3.

x

VARIABLES

values

Ç>«

of <p(x) for x

x

positive

Figure 3.4.3 Values of the standard normal distribution ç for positive x. For negative x use this table along with (4.16).

(4.17) EXAMPLE. Consider a Bernoulli process with success probability /; = 0.5. Then applying (4.14) to the process [Nn\ K e N) of success counts, we get

Next, applying (4.14) to the process {Tn; n & N} of the times of successes, noting'that now a = E(Tn+í — Tn) = l/p = 2 and b^2 = Var(Tn+1 — Tn) = <t/p^2 = 2 by (3.11) and (3. 13),

l • or n = 50, for example, rr ^

is approximately equal to (using the table in Figure 3.4.3)

f^1 -^2 l -/^e~x*;í dx^ = 0.8849.

Similarly,

< 77} = 2.

66 BERNOULLI PROCESSES Chap. 3

is approximately equal to

f"" l -x' f ~ l J _«, A/271 J 2.3 -N/27Z

= 1- f" L..;—

= l - 0.9893 = 0.0107. (^) n

5. Exercises

(5.1) Consider a possible realization a> = (S, F, F, F, S, S, F, S,.. .) of a sequence of independent trials with two possible outcomes, S and F. For this particular ro, what are the values of the random variables (a) XltXt, ...,XS; (b) JVo, #„...,#,; (c) Tt.Ti.Ti.T^TJ (5.2) In Example (1.2) let the defective rate be p = 0.05. What is the probability that the first, second, and third items inspected are ali defective? What is the probability that exactly one of the first, second, or third items is defective? (5.3) For the process described in Example (1.3), compute and interpret the following quantities. (a) P{Ni =0,N 2 = O, N 3 = 1,N 4 = 1}

(b) P[N 1 = l, # 3 =2, JV* = 2}

(c) P{NS=6,N 15 = 12}. (5.4) For the problem introduced in Example (1.4), what is the expected number of bearings, among the first 400 produced, which do not meet the specifications?

(5.5) For p = 0.8, compute (a) EÍN 3 ], £[JV 7 ], E[N 3 + 47V 7 ] (b) Var(AT 3 ), Var(7V 7 - N 3 ) (c) E[6N 4 + N 1 \N 2 ].

(5.6) Consider Bernoulli trials with probability p = 0.8 for success. Suppose that the first five trials resulted in, respectively, S, F, F, S, S. What is the expected value of N 3 + 2N-, given this past history?

(5.7) Generalize the result of Example (2.16) by showing that for any n, m e N, E[N,+m\N»\ = N» + mp.

(5.8) Repeat the steps of the proof of Theorem (2.14) to show the truth of the particular result that

E[3N 5 + NI | N 0 , Nlt N 2 ] = E[1NÍ + NI \ N 2 ]. (5.9) /lypcrtfconwtric distribution. Let k, m, n be integers with k < m -- n. Show l l i. i l l'or ;iny inlcgcr j with O <i j < m, O < k — j < n,

Sec. 5 EXERCISES 67

' y) (m + n
( k ) This defines a probability distribution in j; it is called the hypergeometric distribu- tion. Note that Nm can be replaced by the sum -Xni + • • • + Xnm of any m variables selected from Xlt... , Xm+n. (5.10) In Example (1.2), suppose the items are packaged in boxes of 100 each. A sampling inspection plan calls for rejecting a box if a sample of 5 drawn from that box contained one or more defectives. What is the probability that a box con- taining exactly 4 defectives is rejected? (5.11) In Example (3.1) suppose the rate of crossings is 4 vehicles per minute. Compute and interpret the following quantities. (a) p = P{Xn = 1}, (b) P{T 4 (c)

T 3 = 12}

T 3 ], E[T 13

(d) Var(r 2 + 5T 3 ).

T 3 ]

(5.12) Repeat the proof of Theorem (3.6) to show that

if the event [T 1 < 16} occurs. (5.13) Show, by following the steps of Example (3.17), that

(5.14) Compute the following (and compare with (2.2.14)): (a) P{7\ =k,T 2 = m, T 3 = n] (b) P{T 3 =n\Tí=k,Tz = m} (c) E[T 3 \T 1 =k,T 2 =m] (d) £fe(r 3 )| r,, ra] for g(b) = a», a e [0, 1]. (5.15) If a random variable T hás the geometric distribution, then

P[T> n + m\T> n] = q™ - P{T > m]

for ali n and m. Show that the converse is also truc: if a discreto random variable T is such that P[T>n+m\T> n] = P{T > m} for ali m, n e N, then T hás a geometric distribution. (5.16) The probability that a given driver stops to pick up a hitchhiker is p = 0.04; diíferent drivers, of course, make their decisions to stop or not independently ol' each other. Given that our hitchhiker counted 30 cars passing her without stopping, what is the probability that she will be picked up by the 37th car or before? (5.17) Continuation. Suppose the arrivals of cars themselves are as described in l-xample (3.1) with 4 vehicles per minute. Then "success" for the hitchhiker occurs