The Arcsine Distribution

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1 The Arcsine Disribuion Chris H. Rycrof Ocober 6, 006 A common heme of he class has been ha he saisics of single walker are ofen very differen from hose of an ensemble of walkers. On he firs homework, we considered a good eample of his, whereby we performed simulaions of symmeric Bernoulli walkers on he inegers, saring a 0. We defined α N o he proporion of seps from o N for an individual walker which saisfy > 0. We showed ha in he limi as N, α N has PDF p(α π α( α for 0 < α <. ( This is referred o as he arcsine disribuion, since he corresponding CDF is C(α α +sin π. Inuiively, we would epec ha α N would be mos likely o o be /, and we cerainly know ha for an ensemble of walkers, he concenraion of hem on he lef will mach he concenraion on he righ. However, as shown in figure, we see ha ha opposie is rue: / is he leas likely fracion, and i is mos probable ha he walker spends all he ime on one side of he origin. I is possible o make use of Laplace ransforms and coninuum mehods o show his, bu in his lecure we provide a rigorous proof of he resul using discree mehods. Much of he proof ress on couning possible pahs of Bernoulli random walkers, and we begin by considering several differen pah couning argumens ha will be useful in he proof. Also considered on he firs homework was he disribuion of α N for he case of he Cauchy walk, which we found o eacly follow he above form as well. While we will no consider i here, i is worh noing ha he arcsine disribuion holds for any symmeric PDF for he seps, and i is possible o prove his by argumens involving he rescaling of he ime variable. The Reflecion Principle and pah couning Consider a symmeric Bernoulli walk on he inegers. We hink of a walker as racing ou a pah on he inegers in he (, plane. Le N(, be he number of pahs o he poin ha ake seps. We know ha { N(, ( ( + ( + for + even 0 for + odd. Now inroduce he quaniy X y (, o be he number of pahs from (0, 0 o (, which cross a poin y >. A ypical pah saisfying his propery is shown in figure. From he diagram, we see ha i is possible o consruc a corresponding pah from (0, 0 o (y, simply by reflecing up he secion of he pah from he las crossing of y. Conversely, given any pah from (0, 0 o

2 M. Z. Bazan Random Walks and Diffusion Lecure p(α α Figure : The arcsine disribuion, as given by equaion. (y, we can find consruc a pah from (0, 0 o (, ha crosses y, by reflecing down he secion of pah from he las crossing of y. This esablishes a bijecion, and hence we see ha X y (, N(y,. Now le F (, be he number of pahs from (0, 0 o (, ha make heir firs passage o a ime. Any pah which saisfies his propery mus have gone hrough he poin (,, and make a final posiive sep o (,. To find F (,, we are herefore ineresed in enumeraing all possible pahs from (0, 0 o (, ha do no cross. We see ha his is precisely which can be simplified o F (, N(, X (, F (, N(, N( +, ( ( ( + ( + ( + ( N(,. ( From his, i is possible o calculae he number of pahs f( ha firs reurn o he origin a ime. Since a pah can only reurn in an even number of seps, we assume is even. Consider hose pahs which make heir firs sep o (,. Any of hese pahs ha reurn o he origin a ime can be hough of as aking a firs passage pah of lengh o afer his iniial sep. By symmery, we know ha here are an equal number of pahs which iniially go o (, and hen

3 M. Z. Bazan Random Walks and Diffusion Lecure 4 3 y y k k Figure : The reflecion principle: he blue pah o (, which crosses y > can be made ino a pah o (y, by reflecing upwards he secion from he las crossing of y, shown in purple. Figure 3: Any reurning walk of lengh has a firs reurn a a ime k. I can be viewed a firs reurn pah of lengh k (shown in blue followed by reurning pah of lengh n k (shown in purple. reurn a ime. We herefore have f( F (, ( ( + ( ( ( ( [( ] N(0, (. Thus he number of pahs ha firs reurn a ime is /( of hose which reurn a ime. Le he number ha reurn a ime be R(. Any such pah mus have a unique firs reurn o 0, a ime k say, and as such, i can be viewed as a firs reurn segmen of lengh k coupled wih a reurn pah of lengh k, as shown in figure 3. The number of such pahs is jus he produc f(kr( k. Since every pah reurning o he origin a mus have a firs reurn, we know ha R( k,4,... f(kr( k. (3 Now consider W ( o be he number of pahs which never reurn o he origin by ime. By symmery, we know ha his will be eacly wice hose pahs which end up in he region > 0 and never reurn o he origin. Any such pah will end up a some poin. Again, we proceed by making a correspondence. Consider such a pah which goes o (,. I is possible o consruc

4 M. Z. Bazan Random Walks and Diffusion Lecure 4 4 anoher pah o (, by roaing his pah by 80 around he poin (/, /, as shown in figure 4. If he original pah was non-reurning, we know ha he new pah mus be a firs passage pah o (,. By symmery, here are an equal number of non-reurning pahs in he region < 0, and hence, by enumeraing over all possible final posiions, we see ha W (,4,...,4,... F (, N(, 0 ( ( ( ( N(0, [N(, N( +, ] where we have used he fac ha he erms in he summaion cancel in pairs, leaving only a conribuion from he iniial and final erms. Ineresingly, we see ha he number of pahs which reurn a ime is eacly he same as hose which never reurn by ime. A relaed resul ha will be useful in he ne secion is o find he number of pahs of even lengh which lie wholly in he domain 0. This can be wrien as hose pahs which saisfy 0, 0, 3 0,..., 0, 0 where i denoes he posiion of he ih sep. Since is even, we know ha if 0, hen 0 auomaically. Thus our condiion is equivalen o requiring 0, 0, 3 0,..., 0. We see ha for any pah of his form, here is a corresponding non-reurning pah in he domain > 0, formed by firs aking a sep o (, and following i wih a pah of lengh saisfying he above consrains, as shown in figure 5. Thus we have shown ha (Number of pahs saisfying 0, 0,..., 0 W (. For each pah conribuing o he lef hand side of his epression, we can consruc wo pahs of lengh which remain in he region 0, since he final sep can be eiher posiive or negaive. Hence (Number of pahs saisfying 0, 0,..., 0 W (. (4 Deriving he arcsine disribuion We are now in a posiion where we can derive he arcsine disribuion for he discree case. We wan o consider he number of pahs P (k, n which spend k unis in he posiive domain, and n k unis in he negaive domain, where n and k are once again even.

5 M. Z. Bazan Random Walks and Diffusion Lecure 4 5 Figure 4: The blue line is non-reurning walk o (,. By roaing he walk by 80 around he poin (/, / we can creae a firs passage walk o (,, shown in purple. Figure 5: A posiive non-reurning walk of lengh can be hough of as a single sep o (, shown in blue followed by a posiive walk of lengh (shown in purple. Before proceeding, we mus however be careful o correcly define how we deal wih poins of he walk which cross 0. For poins a 0, i is unclear wheher hese should coun for he posiive domain or he negaive domain, and his has he poenial o cause problems for our couning process. Raher han couning random walk posiions, i is herefore mahemaically convenien o coun wheher he random walk line segmens in he (, plane lie in he domain > 0 or < 0, he posiion of a line segmen is always well-defined. We can also inerpre his as a ie-breaker condiion: a walker a 0 couns as posiive if is previous sep was posiive, and negaive if is previous sep was negaive. Using his rigorous definiion of P (k, n, we can carry ou an inducive proof o show ha P (k, n R(kR(n k. We know ha 0 k n, and we wish o carry ou inducion on n. Suppose ha n k. Then we see immediaely by making use of equaion 4 ha P (k, k R(k R(kR(0. (By symmery, we also know ha P (0, k R(0R(k. Now consider a general pair (k, n, and assume ha he resul is rue for all cases where he oal pah lengh is less han n. Consider any pah. There mus be a firs reurn a some even r wih r < n. There are wo cases: These firs r verices are spen on he posiive side, and he remaining secion is n r long and has eacly k r sides above he ais. The number of such pahs will be f(rp (k r, n r. These firs r verices are spen on he negaive side, and he remaining secion is n r long and has eacly k sides above he ais. The number of such pahs will be f(rp (k, n r.

6 M. Z. Bazan Random Walks and Diffusion Lecure 4 6 Summing up all hese possibiliies, we see ha P (k, n k f(rp (k r, n r + r,4,... n k r,4,... f(rp (k, n r. For every case on he righ hand side, he oal pah lengh is less han n, and we can herefore assume he resul is rue for hese cases by our inducive hypohesis. This gives P (k, n k r,4,... R(n k k f(rr(k rr(n k + r,4,... n k r,4,... f(rr(k r + R(k R(n kr(k + R(kR(n k R(kR(n k f(rr(kr(n r k n k r,4,... f(rr(n k r where we have made use of equaion 3 o evaluae he sum. We have herefore proved he inducive sep and hence P (k, n R(kR(n k for all permissible values of k and n. 3 Obaining a coninuum resul as n ges large To find a PDF of our original α N, we consider he limi as n and k become large. By making use of Sirling s formula, which saes ha k πkk k e k, we have k πkk k e k R(k [( k ] [ πk ( k k/ e k/] πkk k e k kπk k k e k k+ πk. Using his resul, we know ha he probabiliy of a pah of lengh n having k line segmens posiive is given asympoically by Thus, in erms of α N for a long pah, we see R(kR(n k n k+ n k+ π k(n k n π k(n k. P(α < α N < α + dα P(Nα < k < N(α + dα which is he arcsine disribuion. N(α+dα Nα dk π k(n k [N(α + dα Nα] π Nα(N Nα π α( α

7 M. Z. Bazan Random Walks and Diffusion Lecure Relaionship o coninuum approach In he following lecures, we will be considering coninuum firs passage mehods, and i is ineresing o see how he above resuls relae o his. Suppose ha our Bernoulli walker is sepping on a laice of size δ and making a sep a ime inervals τ. We can inroduce a probabiliy densiy ρ(,, and make he correspondence ρ(xδ, T τ o T N(X, T. If he laice spacings saisfy a relaion Dτ δ for some consan D, hen in he limi δ 0 we find ha ρ saisfies a diffusion equaion ρ Dρ and for a walker iniially locaed a 0, such ha ρ(, 0 δ(, we have ρ(, e 4D. 4πD Recall from equaion ha he number of firs passage pahs o some locaion is given by F (, N(, /. In he coninuum case, we have an analogous resul ha firs passage ime o a locaion is given by F ( e 4D. 4πD 3 This is referred o as he Smirnov densiy. We see ha for large, F ( 3/, and hence our epeced waiing ime is infinie. 5 The image mehod The Smirnov densiy can also be derived using a coninuum approach. Consider a walker a 0 a 0, and as above, le is PDF be given by ρ(,. If he walker says in he region > 0, hen i jus diffuses normally, according o ρ Dρ. However, if i reaches he poin 0, i achieves firs passage and is removed his can be modeled by a boundary condiion ρ(0, 0 for all. Borrowing ideas from elecrosaics, we propose ha his problem has soluion ρ(, ( e ( 0 4D e (+ 0 4D 4πD The firs erm would be our soluion for he case of an infinie domain wih no boundaries. However, o correcly handle he boundary, an equal and opposie image has been inroduced a 0. This erm immediaely saisfies he diffusion equaion. We see ha a 0, ρ is zero, and our boundary condiion is also saisfied. We can now use his soluion o find ou how fas he walkers achieve firs passage, by looking a he probabiliy curren a 0: F 0 ( D ρ πD 3 e 0 4D. This eacly maches he answer we obained from he discree argumen.

Math 2142 Exam 1 Review Problems. x 2 + f (0) 3! for the 3rd Taylor polynomial at x = 0. To calculate the various quantities:

Math 2142 Exam 1 Review Problems. x 2 + f (0) 3! for the 3rd Taylor polynomial at x = 0. To calculate the various quantities: Mah 4 Eam Review Problems Problem. Calculae he 3rd Taylor polynomial for arcsin a =. Soluion. Le f() = arcsin. For his problem, we use he formula f() + f () + f ()! + f () 3! for he 3rd Taylor polynomial