u( x)= Pr X() t hits C before 0 X( 0)= x ( ) 2 AMS 216 Stochastic Differential Equations Lecture #2

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1 AMS 6 Stochastic Differential Equations Lecture # Gambler s Ruin (continued) Question #: How long can you play? Question #: What is the chance that you break the bank? Note that unlike in the case of deterministic equations, for stochastic differential equations, it is not enough just to calculate X(t). For stochastic differential equations, we also want to know quantities that are not readily determined from X(t), such as answers to questions above. We answer question # first. Let u( )= Pr X() t hits C before 0 X( 0)= { } We have u(c) = and u(0) = 0. Xdt = + For fied 0 < < C and when dt is small enough, the probability of X(t) hits 0 or C in time interval [0, dt] is eponentially small. u( )= E { u( + )}+ odt = E u( )+ u + u = u + u dt + odt + odt ==> u ( )= 0 differential equation u( 0)= 0, uc = boundary conditions ==> u = C The probability of breaking the bank is proportional to your initial cash and inversely proportional to the total cash. Now we answer question #. Let T( )= E{ time until X()= t C or X()= t 0 X( 0)= }. We have T(0) = 0 and T(C) = 0. Xdt = + - -

2 ==> T( )= dt + E { T( + )}+ odt = dt + E T( )+ T + T = dt + T( ) + T dt + odt T ( )= differential equation T ( 0)= 0, T( C)= 0 boundary conditions ==> T( ) = C ( ) + odt But the average time does not give us the full picture! T() is the average of the time until going bankrupt or breaking the bank. However, this average does not give us a full picture of how long we can play with initial cash. Notice that T( )= C ( ) increases with C. In particular, when C =, we have T( ) = for > 0. This certainly does not mean we can play forever with initial cash. To have a more detailed picture of how long we can play when C is very large, we look at the probability that we can play longer than a certain time. Assume C =. Consider { [ ] X ( 0)= } P(,t)= Pr X ( )> 0 for 0, t P(, t) is the probability of surviving (at least) to time t given that X(0) =. We have P,0 = and P( 0,t) = 0. Xdt = + ==> P(, t)= E { P( +, t dt) }+ odt = E P(,t)+ P t ( dt)+ P + P = P,t + P t ( dt) + P dt + odt P(, t) satisfies the initial boundary value problem (IBVP) P t = P P(,0)=, P( 0,t)= 0 Converting it to an initial value problem (IVP) by odd etension P(,t)= P(,t) The etended function P(, t) satisfies the IVP + odt - -

3 P t = P P,0 =, < 0, > 0 Solution of the IVP u t = au u(,0)= f is given by u(,t)= 4at ep 4at f ( )d Using this formula to calculate P(, t), we obtain f( )= P(,t)= = =, <, > t t 0 t ep t d ep t d t ep( s )ds, s = t = erf t where the error function is defined as erf 0 t We can use P,t = erf z 0 ep( s )ds t ep t d to calculate the probability. P(,.) = 0.5 means that with initial cash =, the probability that we can play longer than t =. is 50%. P(, 63) = 0. means that with initial cash =, the probability that we can play longer than t = 63 is 0%. P(, 55) = 0.05 means that with initial cash =, the probability that we can play longer than t = 55 is 5%

4 Note that when the game is fair and there are many players, the casino s cash is almost unchanged with respect to the time. In other words, the casino cannot make money with a fair game. (Now we consider a) Biased game: dx = mdt + It is a biased game because E{ dx} = mdt Question #: How long can you play? Question #: What is the chance that you break the bank? Let u = Pr Xt hits C before 0 X( 0) = { } We have uc = and u( 0) = 0. Again, for fied 0 < < C and when dt is small enough, the probability of X(t) hits 0 or C in time interval [0, dt] is eponentially small. Xdt = + dx where dx = mdt + satisfies dx = O( dt) E{ dx}= mdt E{ ( dx) }= E{ o( dt)+ ( ) }= dt + odt Function u() satisfies u( )= E { u( + dx) }+ odt = E u( )+ u dx + u dx = u( ) u mdt + u dt + odt + odt u ==> ( ) mu ( )= 0 differential equation u( 0)= 0, uc = boundary conditions (The characteristic equation of the ODE is m - 4 -

5 The two roots are = m, = 0 A general solution has the form u( )= c e m + c Using the boundary conditions, we obtain) ==> u( )= em ( e mc = e m )e mc e mc Suppose mc is moderately large (for eample, mc = 0). We have u( ) ( e m )e mc m C < e The probability of breaking the bank is eponentially small. Let us compare fair game vs biased game Fair game: u( )= C ==> u C = Biased game: u C ( e mc )e mc mc e u C is the probability of winning all the cash of the other player when the two players start with the equal amount of cash. Now we study the average time to the end of game. Let T( )= E{ time until X()= t C or X()= t 0 X( 0)= }. We have T ( 0) = 0 and T( C) = 0. Eercise #: Derive an ODE for T(). Solve the boundary value problem to get T( )= m C m em e mc Suppose mc is moderately large and is not close to C, then we have T( )= m C em e mc m - 5 -

6 This is consistent with the intuitive deterministic picture that if your cash decreases with a speed m, then your initial cash will last time = m. Now consider a discrete version of the gambler s ruin problem. t = 0,,, 3, discrete X = 0,,, 3, discrete X( t + )= X()+ t dx +, prob = µ dx =, prob = + µ N = sum of your initial cash and casino s cash n = your initial cash { } { ()= N or X()= t 0 X( 0)= n} un = Pr X() t hits N before 0 X( 0)= n T( n)= E time until X t Eercise P: For µ = 0 (fair game), derive equations for u(n) and T(n). Solve the equations for u(n) and T(n). Eercise P: For µ > 0 (biased game), derive equations for u(n) and T(n). Solve the equations for u(n) and T(n). White noise A short story ) Z() t dt ()Z( s) ==> ) E{ Z t }= ( t s) ==> 3) e it E{ Z()Z t ( 0) }dt = ==> 4) Z(t) is white noise. First, we point out that Z() t dt is not a regular function

7 = O( dt) dt = O dt dt = O dt lim = dt0 dt 0 = We need to fill in some details to eplain each step, especially from line 3 to line 4. A long story We start with some mathematical preparations. Delta function (Dirac's delta function): We give two equivalent definitions of the Delta function. Each definition is mathematically more convenient in some situations. Definition (limit of Gaussian distribution): ( )= lim n ep n n where lim n = 0, for eample, n = n n. Definition (limit of impulse function): ( )= lim ( ) n n n, for where n ( )= n, n 0, otherwise - 7 -