Two dependent random walks

Size: px

Start display at page:

Download "Two dependent random walks"

Abigail McLaughlin
5 years ago
Views:

1 Two depedet radom wals O the probability of two players hittig the same digit while walig over the same strig of radom digits. T.A. de Graaf Bachelor thesis, July, 03 Supervisor: Prof. Dr. W.Th.F. de Hollader Mathematical Istitute, Leide Uiversity

3 Cotets Itroductio. Problem defiitio Outlie Reewal theory 3. Recursio relatio Traslatio to our problem Two reewal processes 5 3. Li betwee two models Two players Structure of solutio Sigularities Asymptotic behaviour 0 4. The mai theorem Examples Results 3 5. The first o-trivial case Table of results from Maple Appedix 6 6. C++ code Maple code

4 Itroductio Suppose that two players play the followig game. From a strig of 00 digits draw radomly from to 0, each player radomly chooses oe of the first 0 digits. Afterwards they move forward by a amout equal to the value of the digit they hit, ad so o, util they are about to pass the 00th digit. The game stops there. Now, what is the probability that they ed up at the same digit? Accordig to simulatios this probability is approximately But what is the exact probability? I this bachelor thesis we search for a aswer to this questio.. Problem defiitio Let N 0 = {0} N = {0,,,...}. Let, N N be such that N. Defie (X ) N to be the i.i.d. sequece such that X {,..., } for all N, draw accordig to the uiform distributio. The startig positio is determied by the value of X, amely the startig positio is the X -th digit i the strig. Defie (S ) N as the th partial sum, S := X X, N. This sequece eeps trac of the positio i our strig. Figure gives a setch of the situatio. Every horizotal bar is a digit i the strig. The umbers uder those bars represet the positio of the digits ad the X, X,... above the bars are the digits o that positio. Note that o the bla spots there are supposed to be digits as well. But as log as you do ot hit a certai digit, it does ot matter what the value of that particular digit is. Figure : X determies the startig positio. Outlie I sectio we review some of the basics of reewal theory ad mae the traslatio of reewal theory to our problem. I sectio 3 we differetiate betwee two models. The first model describes our iitial problem, with two players walig o the same strig. I the secod model we cosider our two players to be idepedet, i.e., they each wal o their ow strig. The we will mae the followig importat observatio. The probabilities that they do ot meet util a certai time are the same i both models. So we ca use the secod model to give a aswer to the questio we raised for our iitial problem, sice i the secod model the idepedece of the strigs maes computatios less complicated.

5 3 Reewal theory. Recursio relatio We call (X ) N the iterarrival time sequece ad X the th iterarrival time. The sequece (S ) N, defied as above, is called the arrival time sequece ad S the th arrival time. Let f = (f ) N0 be the commo distributio of (X ) N : { f := P (X = ) = for =,...,, 0 otherwise, with f 0 := 0. We call f the waitig time distributio. Let (Z ) N0 deotes whether or ot there occurs a reewal at time : { if = Sm for some m N Z := 0, 0 otherwise, be the sequece that with Z 0 :=. Defie u as the probability that a reewal occurs at time : u := P (Z = ), with u 0 :=. We ca express (u ) N0 i terms of (f ) N0 via the recursio relatio u = f i u i, N. For this observatio we will use followig defiitios. Defiitio. Let g, h : N 0 R. The discrete-time covolutio product of g ad h is the fuctio j (g h) : N 0 R, j g (j i) h (i). Note that the discrete-time covolutio product is associative, commutative ad has the idetity elemet δ : N 0 R with { if j = 0, δ (j) := 0 otherwise. Defiitio. Let g : N 0 R. Defie the -fold covolutio product of g, g () : N 0 R, by g (0) (j) := δ (j), g () (j) := g (j), g () (j) := (g g... g) (j),. }{{} times Our X,..., X are i.i.d. radom variables with commo distributio f (). Now, f () is the distributio of S = X X, i.e., f () j is the probability that the ( + )th reewal taes place at time j: f () j = P (S = j). Note that the first reewal taes place at time 0 sice we have Z 0 :=. We get u = P (Z = ) = E (Z ) = Note that this sum is fiite due to the fact that P (S i = ) = f (i) = P (S i = ) = 0 for i >. f (i).

6 4 Now we are able to obtai the recursio relatio coectig (f ) N0 ad (u ) N0. For N, u = P (Z = ) = P (S = ) + P (S = i) P (Z = S = i) = P (X = ) + P (X = i) P (Z i = ) = f + f i u i = f i u i.. Traslatio to our problem I this sectio we mae the traslatio of reewal theory to our problem. For ow, we cosider oly oe player i our game. I reewal theory X deotes the legth of the first iterarrival time. I our case it determies the startig positio of our player. Not by coicidece we choose, for all i N, X i radomly betwee ad ad also have exactly startig positios. Now X has the same distributio as X, X 3,..., which maes computatios easier. I reewal theory we say that a reewal occurs at whe S = m for some m N 0. This correspods i our problem to the evet that the player hits digit. Now, our iterarrival time sequece (X ) N deotes the successive legths of our jumps ad therefore determies the digits that are beig hit i our strig. I this way we ca cosider our problem as a reewal process. Note that S, N, deotes the positio after the th jump. Suppose that S + X N, but S + X + > N. The S is the ed positio i the strig. Defiitio.3 Defie the geeratig fuctios of (f ) N0 ad (u ) N0, respectively, by F (z) := f z, U (z) := u z, z R. Lemma.4 The geeratig fuctios of (f ) N0 ad (u ) N0 are related by F (z) = U (z), U (z) = U (z) F (z). Proof. Note that sice f 0 = 0 ad u = f iu i, we have U (z) = u z = u 0 + [ ] f i u i = = + U (z) F (z). z

7 5 Lemma.5 For N, lim u = +. Proof. Let N. By the Reewal Theorem we ow that lim u = E(X ) = = ( + ) +. 3 Two reewal processes 3. Li betwee two models I this sectio we loo at the case of two players. We differetiate betwee two models. Defiitio 3. We call the case of two players walig o the same strig model : M. Defiitio 3. We call the case of two players walig o two differet strigs model : M. Now we are at a poit of formulatig a theorem that is essetial to us i solvig our problem. This theorem maes lemma 3.4 below possible ad therefore we are able to solve our problem via the path setched i sectio 3. below. Theorem 3.3 [The ey theorem] For N, Proof. Defie, for player p =,, (S (p) P (player ad do ot meet up to time M ) = P (player ad do ot meet up to time M ). digit that is beig hit by oe of the players, i.e., l := max{s (p) i Let r p N be such that l = S r (p) p l. Let r := max{r, r }. ) N as before. Let N. Let l be the largest : S (p) i 0, i N, p =, }., p {, }, i.e., the umber of jumps of player p util it hits Note that oly the digits that are beig hit by at least oe of the players are relevat. For example, if player does ot hit digit m N, i.e., m N : S () m = m, the we are able to cosider the value of digit m udetermied for player. This meas that the value of digit m for player does ot ecessarily have to be equal to the value of digit m for player. Hece P(player ad do ot meet up to time M ) = P(player ad do ot meet o i =,..., M ) = P( i {,..., r} : S () i S () i M ) = P( i {,..., r} : S () i = S () i M ) = P( i {,..., r} : S () i = S () i M ) = P(player ad do ot meet up to time M ). See Semi-Marov Chais ad Hidde Semi-Marov Models Toward Applicatios, Vlad Stefa Barbu ad Niolaos Limios.

8 6 3. Two players Sice we are ivestigatig the probability that the two players do ot meet up to time, we are able to use model. Defie, for player p =,, (Z (p) ) N0 as before. We defie a joit reewal as the evet that the two players hit the same digit: u := P (both players hit digit ) = P(Z () =, Z () = ). Figure : Two players hit the same digit For this joit reewal process we also have u = f i u i, N, with u 0 := 0, for a uow distributio f : N 0 R. Here f i, i N, gives the probability that the two players meet for the first time o digit i. So the probability that two players do ot meet up to time N is give by f i. Ad therefore is the probability that two players ed up at the same digit give by Lemma 3.4 For N, u = u. N i>n f i = i>n Proof. The probability to hit digit is for both players u. So, uder the assumptio of idepedet strigs, the probability that both players hit digit is u. Defie the geeratig fuctios of (f ) N0 ad (u ) N0 respectively, by f i. F (z) := f z, U (z) := u z, z R. The, aalogously to the relatio betwee F (z) ad U(z) i lemma.4, we get F (z) = U (z), U (z) = U (z) F (z).

9 7 3.3 Structure of solutio Recall that our probability of iterest is N f i. We would lie to ow the uow distributio f. I sectio 5. below we determie the distributio of f i the case of = ad N = 4, 5, 6. Ufortuately we have ot maaged to do this aalytically for geeral ad N. However we have maaged to do this umerically with the help of Maple. We did this accordig to the followig path: f () F () U (3) u (4) u (5) U (6) F (7) f. The first two steps (), () are easy: the choice of f = (f ) N0 determies F (z), ad the relatio U(z) = F (z) determies U(z). Sice ow that x =, x <, x we have for F (z) < that U(z) = F (z). As the coefficiets of the geeratig fuctio U(z) determie u = (u ) N0, step (3) is ow possible. But ote that it might be rather difficult to compute these coefficiets i closed form. How difficult this is depeds o F (z). Step (4) is doe by lemma 3.4. Steps (5) through (7) are possible by the geeratig fuctios U(z), F (z) ad their relatio. Steps (5) ad (6) are easy. Step (7) is hard, agai due to the fact that from F (z) it is i geeral ot easy to deduce its coefficiets f i, i N 0, i closed form. 3.4 Sigularities Recall our distributio We get Write z = ε. The F (z) = f : N 0 [0, ], F (z) = ( ε)i { for =,...,, 0 otherwise. zi = z( z ). z = ( ε) + ( ε) + ( ε) ( ε) = ( ε ε 3ε... ε) + O ( ε ) = ( ) ( + ) ε + O ( ε ) = + ε + O ( ε ),

10 8 ad hece U(z) = F (z) = + ε + O ( ) ε. Let h : N 0 R be such that u = + + h for N 0. The H(z) := z i h i = U(z) + ε. We expect h to coverge expoetially fast to zero, sice for R the radius of covergece of H(z) we have z R with z < R : H(z) = z i h i <. So we expect h = ( R ) +o(). Let us first ivestigate the sigularities of H(z). Lemma 3.5 H(z) has o sigularity at z =. Proof. Let us express the Taylor polyomial of F (z) aroud z = as with The F (z) = F (i) () (z ) i i! = + F ()(z ) + Q(z), Q(z) = F () (z ) + So H(z) ideed has o sigularity at z =. i=3 F (i) () (z ) i. i! H(z) = U(z) + z = F (z) + z = F ()( z) + F (z) ( F (z))f ()( z) Q(z) z = ( F (z))f F () ()( z) F (). Lemma 3.6 For odd, there is o z R for which H(z) has a sigularity. Proof. A sigularity of H(z) differet from z = must be a solutio of F (z) = 0, z, i.e., a solutio of ( z) = z( z ), for z R, sice we are oly iterested i real solutios at the momet. Put g(z) := z( z ). The g() = 0, g(0) = 0, g (z) = ( + )z. Note that [( z)] = ad g () =. There is o itersectio o [0, ), sice g(z) < ( z) for 0 z <. For z > we have ( + )z < ( + )z > +,

11 9 ad so o itersectio occurs o (, ) either. For odd ad z < 0, we have z( z ) < 0, ad so also o itersectio occurs o (, 0). Lemma 3.7 For eve, H(z) has a sigularity at z R, with z <. Proof. From the same reasoig as i the case of odd, we ca coclude that there is o itersectio o [0, ) or o (, ). But g( ) = 0 ad ( ) = 3, ( ( ) ) = + +. We ca prove by iductio that, for eve, Ad sice g (z) < for z <, we do have a itersectio i [, ). Figure 3: O the left: is odd, o the right: is eve It is remarable that for odd H(z) has o sigularity at ay z R. At first glace this could mea that for odd h coverges faster tha ay expoet to 0, while for eve h coverges as ( z ) +o(). But recall that we are looig for the radius of covergece of H(z). Therefore, we must also ivestigate the complex solutios of F (z) =. Let N. Defie V := {z C : F (z) = } ad R := mi z, z V \{} as the radius of covergece of H(z). Lemma 3.8 For N, R >. Proof. Let N. We have to prove that for z V, z : z >. We ow that if z V, the F (z) =. By the triagle iequality we get (z z ) ( z z ) = ( z z ). Put z = re iφ. The (r r ) ad hece r. We ow that the triagle iequality becomes a equality if ad oly if the agles of all z, z,..., z are equal. We ow that z = e 0 is a solutio. Put z j = r j e j(iφ). Now (z z ) = ( z z ) if ad oly if j(iφ) = 0 for all j {,..., }, i.e., if ad oly if φ = 0. This brigs us bac to the solutio z =. So for all z V with z =, we have z =.

12 0 Table: A list of radii of covergece ad real solutios of F (z) = for differet, rouded off to 6 decimal places. R z Lemma 3.9 For N, > eve, R <. Proof. Let N >, eve. By lemma 3.7 we ow that z defiitio R z, we have R (, ). (, ). Ad sice by We would have wated lemma 3.9 for the case of odd as well, but ufortuately we have ot maaged to proof this yet. 4 Asymptotic behaviour 4. The mai theorem Lemma 4. For N, lim sup h =. R Proof. This is the well ow Cauchy-Hadamard theorem for power series. The mai result for our iitial problem is the followig. Theorem 4. For N, lim Proof. Let N. Recall that for N, ( u = Substitute u = u = ( + + h ). The ( ) + + h = i> f i ) = R. f i u i. ( ) f i + + h i, See Real Aalysis, Dipa Chatterjee.

13 or 4 ( + ) h + h = ( ) 4 f i ( + ) h i + h i. Use f i = to write the latter equatio as ( ) 4 ( + ) h + h f i = 4 + i> f i (h i h ) + ( f i h i h ). Sice we are ivestigatig the asymptotic behaviour as, we may eglect the o-leadig 4 terms. For istace, i the left-had side + h + h 4 is egligible with respect to (+) whe. Similarly, i the right-had side h i h = (h i h )(h i + h ) is egligible with respect to 4 + (h i + h ) whe. Thus, we get + f i i> f i (h i h ), with beig defied as: a b whe lim a b =. Lemma 4. says that h = ( R ) +ε for some ε = o (), i.e., lim ε = 0. Hece f i (h i h ) = R Claim: For some δ = o (), i.e., lim δ = 0, Sice lim ε ( ( ) i+ε i ( f i R R ) +ε ) ( ) +ε ( ) = f i R i ε i+ε. f i R i ε i+ε f i R i+δ. = 0 ad sup i ε i = sup i ε i <, we have lim sup ε i i = 0. Put δ = sup i ε i. The i ε i + ε i + δ = i + sup i ε i for i, N. Hece or lim [ + lim i> [ i> ] [ ( ) +ε ) f i = lim f R i (R ] i+δ ] ( ) [ +ε ) f i = lim f R i (R ] i+δ = R.

14 I the last equality we use that [ lim ) f i (R ] i+δ = lim [ ( = lim = lim [ R δ R δ ( f i R i ) lim F (R ) R δ R δ [ ( f i F () ] )] )] R i R δ Now F () = ad, to obtai the value of F (R ), we ote that by lemma.5, lemma 3.4 ad lemma 3.8 we get lim u R =. So F (R ) = U(R ) = =. u R Hece lim [ ) f i (R ] [ i+δ = lim R δ. ] =. 4. Examples Let us wor out the case we met i our itroductio: = 0, N = 00. The Maple code i the appedix gives us the probability f i i>00 To determie whether our asymptotic result is gettig close, we eed the radius of covergece of H(z). From the table of sectio 3.4 we ow that R 0.8. Now we compute ( ) , ad so we get R 0 i>00 f i This is very far off. We would have wated ( R 0 ) , i.e., R So our asymptotic result is far from accurate. Let us examie two other examples: = for N = 4 ad N = 6. We ow from sectio 3.4 that R =. From sectio 5. below we ow that i>4 f i = ad i>6 f i = We get ( ) 4 ( ) 6 = 0.065, = So that is exactly what we wated. Let us see whether this is still the case for N = 0. Our Maple code gives: We get ( ) 0 = This is still the exact result that we were looig for. i>0 f i =

15 3 5 Results 5. The first o-trivial case Let us ivestigate the first o-trivial case, =, for N = 4, 5, 6. First, it is easy to calculate the exact probability of two players meetig for the first time. Defie, for player i =,, ) N as before. We have (X (i) f = P(players meet at first digit) = ( ) = 4 f = P(players meet for the first time o digit ) = P(oe player starts at digit, the other at digit, the first digit has value ) + P(both players start at digit ) = P(X () =, X () =, X () = ) + P(X () =, X () =, X () = ) + P(X () =, X () = ) ( ) 3 ( ) = + = f 3 = P(players meet for the first time o digit 3) = P(X () =, X () =, X () =, X () = ) + P(X () =, X () =, X () =, X () = ) ( ) 4 = = 8 ( ) 5 f 4 = P(players meet for the first time o digit 4) =... = = f 5 = P(players meet for the first time o digit 5) =... = ( 6 ) 6 = 3. So i this particular case the probability that both players ed up at the same digit, i.e., i>n f i = N f i is for each N easy to calculate. Case N f i =, N = =, N = =, N = Next, let us also aalyze this case accordig to the path we setched i sectio 3.. We get F (z) = (z + z ), U(z) = (z + z ). We ow that, for x <, xi = x. So, for (z + z ) <, we have U(z) = ( ) i (z + z ). This gives us U(z) = + z z +... Defie r := ad q :=. With the biomium of Newto we ca see that, for N, u = r ( ) r + i q i r+i.

16 4 So U(z) = u iz i = u i zi. Thus F (z) = U(z) = [ r ( ) r + i q i for (z + z ) <, i.e., < z <. r+i ] z We would lie to have F (z) i closed form, so that we ca determie the coefficiets. We have ot maaged to do so, but what we ca do is determie f i, for all i N 0, by computig the Taylor series of F (z) aroud z = 0: d F dz (0) = = F (0) = df dz (0) = [ r ( r + i q i ) r+i ] z [ [ r ( ) ] ] r + i q i r+i z [ [ r ( ) ] ] r + i q i r+i z [ [ r ( ) ] ] [ [ r + i r q i r+i z [ [ r z=0 z=0 ( ) ] ] 4 r + i q i r+i z [ [ r ( ) ] ] [ [ r + i r q i r+i z [ [ r ( ) ] ] 4 r + i q i r+i z = 0 = 4 ( ) r + i q i r+i ( ) r + i q i r+i ] ] z ] ] z z=0 z=0 So we ca write F (z) aroud z = 0 as F (z) = d 3 F dz 3 (0) =... = 3 4 d 4 F dz 4 (0) =... = 3 d 5 F dz 5 (0) =... = 5 4. F (i) (0) z i = i! 4 z + z + 8 z3 + 6 z4 + 3 z5 + O(z 6 ). Thus we recovered the distributio of f i, i N 0, foud above.

17 5 5. Table of results from Maple With the help of the Maple code i the appedix, we ca determie for every ad N what the probability is that both players ed up at the same digit. These results are highly accurate. Ufortuately they are ot exact, due to the fact that Maple ca ot wor with the ifiite summatio of the geeratig fuctio U(z). However, if we add up a large umber of summads, say 300, the we get a very accurate result for small N. As we see i our table, the results become less accurate whe N becomes larger. We eed to add up a larger umber of summads to get a good result. This requires a powerful computer. Iput Result Maple Result C++ Absolute differece =, N = =, N = =, N = = 0, N = = 5, N = = 0, N = = 40, N = = 00, N = = 00, N = I the case of = 00, N = 000, we compare the results whe we add up to eve higher umbers. Let M N be the umber of summads, i.e., U(z) = M u z. M Result Maple Absolute differece with result C

18 6 6 Appedix 6. C++ code #iclude <iostream> usig amespace std; //create a radom umber betwee..s usiged log it lehmer(log it s){ static usiged log log a = 007, b = 49430, c = , z = b; if (s < 0){ s = -s; a = s; } z = (a + b * z) % c; retur (z % s) + ; } it mai(it argc, char * argv[ ]){ it N = 00; it = 0; usiged log it cycles = ; usiged log it i, j, l, equal = 0; usiged log it list[n], umber; for (i = ; i <= cycles; i++){ //create clea radom-iteger-filled list for (it i = ; i <= N; i++){ list[i] = lehmer(); } //pic startig places ad ru util we would pass the 00th digit j = lehmer(); while (j < N){ umber = list[j]; if (j + umber <= N){ j += umber; } else { brea; } } l = lehmer(); while (l < N){ umber = list[l]; if (l + umber <= N){ l += umber; } else { brea; } } } //if both edpoits are the same, icremet umber of same edpoits if (l == j){ equal++; } } pritf("%f\", (double)equal/(double)cycles);

19 7 6. Maple code > with(statistics): := 0: N:= 00: X:= RadomVariable(DiscreteUiform(,)): f:= x->probabilityfuctio(x,x): u:= proc() optio remember; if the else add(f(i)*u(-i),..) fi; ed: ubar:= ->u()ˆ: ubartimesz:= (,z)->ubar()*zˆ: Ubar:= z->add(ubartimesz(,z),..300): Fbar:= z->-/ubar(z): evalf(series(fbar(z),z=0,n+),7): sum(%,z=);

62. Power series Definition 16. (Power series) Given a sequence {c n }, the series. c n x n = c 0 + c 1 x + c 2 x 2 + c 3 x 3 +

62. Power series Definition 16. (Power series) Given a sequence {c n }, the series. c n x n = c 0 + c 1 x + c 2 x 2 + c 3 x 3 + 62. Power series Defiitio 16. (Power series) Give a sequece {c }, the series c x = c 0 + c 1 x + c 2 x 2 + c 3 x 3 + is called a power series i the variable x. The umbers c are called the coefficiets of