MATH 540 Radom Walks Part 1 A radom walk X is special stochastic process that measures the height (or value) of a particle that radomly moves upward or dowward certai fixed amouts o each uit icremet of time. Specifically, we let X 0 be the iitial height, which may be radom or a fixed costat J, but which is idepedet of aythig that follows. O each step, the height either icreases a uits with probability p or decreases b uits with probability q = 1 p, with the outcomes o each step beig idepedet of each other. The X deotes the height after steps of the process. The uderlyig sample space is the set of all possible outcomes that deote the sequece of chages i height: Ω = { ( 1, 2,...): i {a, b} for i 1 }. These outcomes are ofte called paths. The for every step of every path have Ω, we P ( ) = p ad P( i = b) = q. i = a For example, = (a, a, b,...) are the paths that have the first three steps goig up, the up agai, the dow (with aythig possibly happeig after that). The, P({(a,a, b,...)}) = p 2 q. If we oly allow a fiite umber of steps, the the sample space is Ω = { ( 1, 2,..., ) : i {a, b} for 1 i }. I this case, there are oly 2 possible paths i Ω, while Ω itself is ucoutable. For a specific Ω, we let X 0 ( ) deote the iitial height of path. The the height after steps is give by X ( ) = X 0 ( ) + i. For graphical purposes, we assume movemets are uiform throughout time t 0. For a time t, we choose the uique iteger k such that k t < (k +1). The X t ( ) = X k ( ) + (t k )( X k +1 ( ) X k ( )). The extesio of the process for cotiuous time t 0 simply coects the discrete time poits (k, X k ( )) ad (k + 1, X k +1 ( )) i a liear maer. For ay sigle path, the graph of the process X t ( ) forms a trajectory that shows the height of the particle as a fuctio of time t as it travels accordig to path.
A trajectory with X 0 = 3, a = 3, b = 2, that follows path (a, a, b, a, b,... ). We also ca modify the process to allow the height to remai costat with probability r = 1 p q as well as to go up with probability p or dow with probability q o each step, ad to allow the icremetal steps to have legth c. I this case, the sample space is 7 6 5 4 3 2 1 { } Ψ = ( 1, 2,...): i {a, 0, b} for i 1 2 4 6 8 10 12 14 A trajectory with X 0 = 3, a = 4, b = 3, c = 2 that follows path (a, b, b, 0, a, b, 0 ). But ow the icremetal poits are (c, X c ( )), (2c, X 2c ( )), (3c,X 3c ( )), etc. So for all itegers 1, we ow have X c ( ) = X 0 ( ) + i, where = ( 1, 2, 3,...) Ψ. For a time t 0 such that kc t < (k +1)c for the iteger k 0, we have X t ( ) = X kc ( )+ (t kc) c ( X (k+1)c ( ) X kc ( )).
Iitially, we shall study some properties of a radom walk process that is stopped after a fixed umber of steps. I particular, we shall determie the mea ad variace of the fial height, the mea ad variace of the trajectory arclegths, ad the mea siged area betwee the trajectories ad the x -axis. Mea ad Variace of the Height after Steps We ow shall derive the mea ad variace of the height X c of the geeral radom walk after idepedet steps of horizotal legth c. Our sample space ow is Ψ = ( 1,..., ): i {a, 0, b} Ψ, X 0 ( ) is the specific startig height that the proceeds alog path = ( 1,..., ). Thus the fial height after steps of legth c is give by X c ( ) = X 0 ( ) + i. To derive E[ X c ] ad Var( X c ), we merely eed to cosider the chage i height o each step. That is, we eed E[ i ] ad Var( i ). But each i is a radom variable with rage {a, 0, b} havig expected value E[ i ] = a p + 0 r + ( b) q = a p bq. Because E[ i 2 ] = a 2 p + 0 2 r + ( b) 2 q = a 2 p + b 2 q, the variace of i is give by Var ( ) = E[ i 2 ] E[ i ] i ( )2 = a 2 p + b 2 q ap bq ( ) 2. Usig the liearity of expectatio ad the liearity of variace whe applied to a sum of idepedet radom variables, we have Theorem. For a geeral radom walk X that begis at height X 0, the mea ad variace of the height after steps of legth c are ad E[ X c ] = E[X 0 ] + E[ i ] = E X 0 [ ] + (a p bq) Var( X c ) = Var( X 0 ) + ( a 2 p + b 2 q ( a p bq) 2 ) (= Var( X 0 ) + (a + b) 2. pq if r = 0, p + q = 1) We see that the steps that remai costat with probability r have o effect o the average or variace of the height other tha decreasig the values of p ad q whe r 0.
Applicatio to Gamblig 1: A Fiite Number of Bets A group of bettors all bet o the same game with the probability of wiig beig p, ad the probability of losig beig q = 1 p. (There are o ties, so r = 0.) Each perso makes a sequece of bets of $b. The $b is lost if ad oly if the bet is lost. If the bet is wo, the the player gets $a. Each bet represets a step of legth 1, so that c = 1 i this case. Here are some fudametal questios: (a) How should the casio choose the payoff of $a so that the casio makes a profit i the log ru? (b) What is the distributio of the player s fial fortue? (c) How may wis does the player eed to at least break eve? (d) What is the probability that the player breaks eve? Aswers. Let deote a idividual player who starts with iitial fortue $ X 0 ( ). The average fial fortue of all players after bets is E[ X ] = E[ X 0 ] + (a p bq). (a) I order for the casio to make a profit off of oe player, it eeds X ( ) < X 0 ( ). But i order for the casio to make a profit i the log ru, the casio eeds for the players to lose moey o average. So the casio eeds E[ X ] < E[ X 0 ]. That is, it eeds E[ X 0 ] + (a p bq) < E[ X 0 ] or (a p bq) < 0 or a p < bq or a < q p b The ratio q / p gives the odds agaist wiig, or the fair payoff ratio. If a = (q / p) b, the the casio breaks eve i the log ru. By choosig a smaller payoff for wiig, a < (q / p) b, the the casio makes a profit i the log ru. (b) A player s fial fortue X ca be writte i terms of the umber of wis W ad the umber of losses L = W i bets. From gaiig $a for each wi ad losig $b o each loss, we have X = X 0 + aw bl = X 0 + aw b( W), where W ~ b(, p) (a basic biomial distributio that couts the umber of wis i idepedet attempts). We ote that E[ X ] ca be re-derived by E[ X ] = E[ X 0 ] + a E[W ] be[ W]= E[X 0 ]+ a(p) b (q) = E[ X 0 ] + (ap bq).
We also ca recover the variace from this form of X because first X = X 0 + aw b( W) = X 0 + (a + b)w b. The because addig a costat does ot chage the variace ad X 0 is idepedet of the umber of wis W, we have Var( X ) = Var( X 0 ) + Var ((a + b)w) = Var( X 0 ) + (a + b) 2 Var( W ) = Var( X 0 ) + (a + b) 2 pq. We ca further describe the distributio of X by lookig at its rage ad pdf. The possible edig heights are X 0 b, X 0 + a ( 1)b, X 0 + 2a ( 2)b,... X 0 + ( 1)a b, X 0 + a which are of the form X 0 + ia ( i)b for 0 i, where i is the umber of wis. The probability of edig at such a height of the form X 0 + ia ( i)b is give by P( X = X 0 + ia ( i)b) = P W = i ( ) = i p i q i. (c) How does a player at least break eve? Now to have X X 0, we must have i wis so that X 0 + ia ( i)b X 0 or ia ( i)b 0 or i b a + b. (d) Thus, if we choose the smallest iteger I such that I b / (a + b), the the probability that X is at least as much as X 0 is P( X X 0 ) = P( W I) = p i q i. i i = I Example. While playig roulette, a group of players each start with radom amout from $500 to $1500 ad each make a sequece of ie $50 colum bets havig a 2:1 payoff. (a) What is the average fortue after ie bets? (b) Give the probability distributio of a player s fial fortue. (c) How may wis are eeded for a player to at least come out eve after ie bets, ad what is the probability of there beig at least this may wis?
Solutio. With o other coditios stated, we ca oly assume that the iitial amout X 0 is uiformly from $500 to $1500, so that E[ X 0 ] = (500 +1500) / 2 = $1000. A colum bet is a bet o a block of 12 umbers (out of 38). So i this case, we have p = 12 / 38, q = 26 / 38, b = $50, ad a = $100 (due to 2:1 payoff). (a) The players average fortue after ie bets is E[ X 9 ] = E[ X 0 ] + (a p b q) = 1000 + 9(100 12 / 38 50 26 / 38) $976.32. (b) The table below gives the possibilities for the umbers of wis, losses, (ad ties), alog with the fial fortue ad its probability. W, L X = X 0 + W 100 L 50 9 (12 / 38) W (26 / 38) 9 W W 9, 0 X = X 0 + 900 0.0000312305 8, 1 X = X 0 + 750 0.000608995 7, 2 X = X 0 + 600 0.00527795 6, 3 X = X 0 + 450 0.026683 5, 4 X = X 0 + 300 0.0867197 4, 5 X = X 0 + 150 0.187893 3, 6 X = X 0 0.271401 2, 7 X = X 0 150 0.252015 1, 8 X = X 0 300 0.136508 0, 9 X = X 0 450 0.032863 Note: For a biomial distributio b(, p), the most likely umber of successes is the largest iteger k satisfyig k ( + 1)p. If ( + 1) p is already a iteger, the both k = ( + 1) p ad the previous iteger k 1 are equally likely ad are both modes. I this case, we have ( +1) p = 10 12 / 38 3.158. So we most likely have 3 wis i 9 tries which meas we most likely ed up with a fial fortue of X = X 0. (c) To at least break eve, we fid that a player eeds I b / (a +b) = 9 50 / 150 = 3 wis out of ie bets. The the probability of at least comig out eve is P( X 9 X 0 ) = P( W 3) = 1 P( W 2) 0.578614. With this example, we see that there is a slight paradox occurs i gamblig. If a gambler decides to make oe strig of ie bets as above, the there is over a 57% chace that he will at least break eve. Could a degeerate gambler resist these chaces o a oe-time opportuity? However if he cotiues to make such strigs of bets, the he will lose moey i the log ru because his average fial fortue will be less that his average startig amout.
There are several special cases listed below for which E[ X c ] = E X 0 ad Var( X c ) = Var( X 0 ) + a 2 p + b 2 q ( a p bq) 2 ( ) simplify. [ ] + (a p bq) [ ] = E[ X 0 ] ad Case (a): Assume ap = bq, which is equivalet to a / b = q / p. The E X c Var( X c ) = Var( X 0 ) + a p(a + b). This case represets a fair game for which the payoff ratio a / b equals the odds agaist wiig q / p. The average fortue after ay umber of bets equals the average iitial fortue; hece, either the house or the player has a advatage i the log ru. But the variace icreases without boud as the umber of bets icreases; thus there is wide dispersio i the possible fial outcomes. Case (b): Assume a = b ad p = q = 1/2. The E[ X c ] = E[ X 0 ] ad Var( X c ) = Var( X 0 ) + 2pa 2. This case is a subset of Case (a) ad represets the symmetric radom walk that moves up or dow the same heights with equal probability. If a = b = 1, the it is the symmetric simple radom walk. Case (c): Assume r = 0. The Var( X c ) = Var( X 0 ) + (a + b) 2 pq. Case (d): Assume a = b = 1. The E X c Var( X 0 ) + (p + q ( p q) 2 ). [ ] = E[ X 0 ] +(p q) ad Var( X c ) = Case (e): Assume a = b = 1 ad r = 0. The p + q = 1 ad 1 p q E X c simple radom walk. ( ) 2 = 4 pq ; thus, [ ] = E[ X 0 ] +(p q) ad Var( X c ) = Var( X 0 ) + 4 pq. This case is the traditioal [ ] = E[ X 0 ] ad Case (f): Assume a = b = 1, r = 0, ad p = q. The E X c Var( X c ) = Var( X 0 ) +. This case is the traditioal symmetric simple radom walk. Exercise. The probability of wiig a casio bet is p = 2/9. Bettors start with radom iteger amouts from $1000 to $20,000 ad each makes a sequece of 8 bets of $100. The actual payoff ratio is 3 : 1. Let X 8 be the bettor s fortue after these 8 bets. (a) What is the theoretical average startig amout E[ X 0 ]? (b) What are the average umber ad most likely umber of wis i 8 bets? (c) What is the fair payoff for a $100 bet? The actual payoff? (d) What is the rage of edig fortues after 8 bets of $100 each? (e) Compute P(X 8 = X 0 + 400). (f) At least how may wis are eeded so that X 8 X 0? (g) Compute P(X 8 X 0 ). (h) Compute E[ X 8 ], ( X 8 ), ad the most likely fial fortue after 8 bets.