y****e
发帖数: 13 | 1 random walk x_i = 2 or -1 with probability 1/2 for each case.
S_n = x_1 + x_2 +.. x_n
Pr( S_n = a before -b) ?
除了用recursion 有什么快的方法么？ |
s*******n
发帖数: 66 | 2 define y = x-0.5, use martingale
【在 y****e 的大作中提到】
: random walk x_i = 2 or -1 with probability 1/2 for each case.
: S_n = x_1 + x_2 +.. x_n
: Pr( S_n = a before -b) ?
: 除了用recursion 有什么快的方法么？
|
l******i
发帖数: 1404 | 3 X_i = 2 or -1 with probability 1/2 for each case, S_n = X_1 + X_2 +.. X_n.
Define Y_i=(2/3)*X_i-(1/3), T_n = Y_1 + Y_2 +.. Y_n = (2/3)*S_n-(1/3),
T_n is a simple symmetric random walk.
P(S_n = a before -b) = P(T_n = (2a-1)/3) before -(2b+1)/3 ) = (2b+1)/(2a+2b). |
c**********s
发帖数: 295 | 4 In your approach,
T_n = (2/3)S_n -(1/3)n
2b).
【在 l******i 的大作中提到】
: X_i = 2 or -1 with probability 1/2 for each case, S_n = X_1 + X_2 +.. X_n.
: Define Y_i=(2/3)*X_i-(1/3), T_n = Y_1 + Y_2 +.. Y_n = (2/3)*S_n-(1/3),
: T_n is a simple symmetric random walk.
: P(S_n = a before -b) = P(T_n = (2a-1)/3) before -(2b+1)/3 ) = (2b+1)/(2a+2b).
|
i******t
发帖数: 370 | 5 You cannot get anything from here since n is variable...
Here is a sketch for Martingale approach:
assume exp(t*S) is martingale, then
(1/2)*[exp(t*(S+2))+exp(t*(S-1))]=exp(t*S)
=> 1/2*[exp(2t)+exp(-t)]=1
=> t=log((sqrt(5)-1)/2)
then solve p such that p*exp(t*a)+(1-p)exp(t*(-b))=1
【在 c**********s 的大作中提到】
: In your approach,
: T_n = (2/3)S_n -(1/3)n
:
: 2b).
|
r**a
发帖数: 536 | 6
What is this? S is not driftless, how can you assume this?
Actually S_n can be viewed as a Brownian motion W(t) with drift 0.5*t.
【在 i******t 的大作中提到】
: You cannot get anything from here since n is variable...
: Here is a sketch for Martingale approach:
: assume exp(t*S) is martingale, then
: (1/2)*[exp(t*(S+2))+exp(t*(S-1))]=exp(t*S)
: => 1/2*[exp(2t)+exp(-t)]=1
: => t=log((sqrt(5)-1)/2)
: then solve p such that p*exp(t*a)+(1-p)exp(t*(-b))=1
|
c**********s
发帖数: 295 | 7 this is how you transform a drifting process to a martingale process
【在 r**a 的大作中提到】
:
: What is this? S is not driftless, how can you assume this?
: Actually S_n can be viewed as a Brownian motion W(t) with drift 0.5*t.
|
r**a
发帖数: 536 | 8 what are you talking about? Can you show me that assumption is reasonable if
S is a BM with drift? i.e assume S(n)=W(n)+a*n, where a is a constant, then
can you show me exp(t*S(n)) is a martingale?
Rather than exp(t*S(n)), actually exp[S(n)- a * n - n/2] is a martingale if
S(n) = W(n) + a * n.
【在 c**********s 的大作中提到】
: this is how you transform a drifting process to a martingale process
|
i******t
发帖数: 370 | 9 didn't I show there exists t such that exp(t*S(n)) is martingale previously?
if
then
if
【在 r**a 的大作中提到】
: what are you talking about? Can you show me that assumption is reasonable if
: S is a BM with drift? i.e assume S(n)=W(n)+a*n, where a is a constant, then
: can you show me exp(t*S(n)) is a martingale?
: Rather than exp(t*S(n)), actually exp[S(n)- a * n - n/2] is a martingale if
: S(n) = W(n) + a * n.
|
r**a
发帖数: 536 | 10 if you are right, then for the standard random walk S(n) without drift, then
you also can show that there exists a t such that exp(t*S(n)) is a
martingale. Do you think it is
reasonable?
previously?
【在 i******t 的大作中提到】
: didn't I show there exists t such that exp(t*S(n)) is martingale previously?
:
: if
: then
: if
|
|
|
i******t
发帖数: 370 | 11 这什么逻辑，无理取闹
then
【在 r**a 的大作中提到】
: if you are right, then for the standard random walk S(n) without drift, then
: you also can show that there exists a t such that exp(t*S(n)) is a
: martingale. Do you think it is
: reasonable?
:
: previously?
|
r**a
发帖数: 536 | 12 ?? 我的逻辑很清楚吧。是你没看明白吧？再说一遍！
如果你的方法是对的话，用你的方法我也可以证明对于simple random walk S(n)，存
在一个t使得exp(t*S(n))是martingale。这显然不合理。simple random walk 对应到
Brownian motion W(n)。你的结果的意思是说exp(t*W(n))是martingale。
我认为正确的思路是把random walk with drift对应到一个brownian motion with
drift，然后再去考虑exponential martingale。
如果你认为我的逻辑有错误，请指出。不要说一些无关的话。谢谢。
【在 i******t 的大作中提到】
: 这什么逻辑，无理取闹
:
: then
|
i******t
发帖数: 370 | 13 If there is no drift, t=0, the equation is still right. But you do not get
anything useful to solve the problem!
Look, I gave a solution. If you think it is useful, take it. If you think it
is wrong, prove it is wrong. If you are not happy, find your own solution (
there are other ways to solve this!). Anything else is wasting your time and
my time. |
r**a
发帖数: 536 | 14 Hi dude. Calm down. I am just arguing the academic problem with you, nothing
else.
I just dig deeper about your way. I am convinced. Sorry for the previous
confusion.
For the solution, I do have my own method and already mentioned. The random
walk can be mapped to a BM with drift. The theory underlying is similar to
the numerical tree method for the Black-Scholes model. But my way is much
harder than yours. The details are as follows:
Regarding S(n), we have S(n) = S(n-1) + 0.5 + 1.5*Z_{n-1}, where Z_n is a
simple random walk. Then S(n+1)= 0.5n+1.5sum_{i=1}^nZ_i. So asymptotically S
(n+1)to 0.5t+1.5W(t). But my way is only true asymptotically.
it
(
and
【在 i******t 的大作中提到】
: If there is no drift, t=0, the equation is still right. But you do not get
: anything useful to solve the problem!
: Look, I gave a solution. If you think it is useful, take it. If you think it
: is wrong, prove it is wrong. If you are not happy, find your own solution (
: there are other ways to solve this!). Anything else is wasting your time and
: my time.
|
i******t
发帖数: 370 | 15 OK, man. The process is BM with drift as you said. You can draw a binary
tree same way as Black-Scholes.
The difficulty here is in BS, boundary (at expiration) is at the same level
of the tree, while this problem has boundary (reaching a or -b) at different
levels. This makes it harder to compute. |
i******t
发帖数: 370 | 16 Actually there is no magic about martingale.
If you are interested, here is the other solution mentioned in the previous
post. Essentially we want to solve recurrence relation:
p(k)=1/2*p(k+2)+1/2*p(k-1), with p(a)=1, p(-b)=0
The characteristic equation of this recurrence relation is x=1/2 * x^3 + 1/2
. Roots are x1=1, x2=(sqrt(5)-1)/2, x3=(-sqrt(5)-1)/2. General solution is p
(n)=w1*x1^n+w2*x2^n+w3*x3^n+w0. You will need to solve w_i, and make p(n) in
[0, 1].
Comparing this to the martingale approach, exp(lambda)=x. This is just a
different way describing the same structure.
nothing
random
S
【在 r**a 的大作中提到】
: Hi dude. Calm down. I am just arguing the academic problem with you, nothing
: else.
: I just dig deeper about your way. I am convinced. Sorry for the previous
: confusion.
: For the solution, I do have my own method and already mentioned. The random
: walk can be mapped to a BM with drift. The theory underlying is similar to
: the numerical tree method for the Black-Scholes model. But my way is much
: harder than yours. The details are as follows:
: Regarding S(n), we have S(n) = S(n-1) + 0.5 + 1.5*Z_{n-1}, where Z_n is a
: simple random walk. Then S(n+1)= 0.5n+1.5sum_{i=1}^nZ_i. So asymptotically S
|
r**a
发帖数: 536 | 17 When using BM, we'd better view this problem as a continuous one. Thus, the
original problem becomes the following one:
Consider a process S(t):=1.5W(t)+0.5t, what is the probability that S(t) hit
b before hitting -a, where a, b are positive numbers.
Then we can use exponential martingale to solve it. Since I did not solve it
physically, so I
am not sure how different our final answers are. Naively, I think as long
as a, b are big enough, which implies that we need to take tons of steps
random walk to reach a or b, then our answers should be close enough.
level
different
【在 i******t 的大作中提到】
: OK, man. The process is BM with drift as you said. You can draw a binary
: tree same way as Black-Scholes.
: The difficulty here is in BS, boundary (at expiration) is at the same level
: of the tree, while this problem has boundary (reaching a or -b) at different
: levels. This makes it harder to compute.
|
D********e
发帖数: 8 | 18 Just to make sure the audiences follow, the solution by Iamrobot is right.
To make the discreet process martingale, it is just one equation, which you
solve for t. |
M****e
发帖数: 3715 | 19 dX=0.5dt+1.5dW
=>
solve 0.5P''+1.5P'=0 with bc P(a)=1 and P(-b)=0
【在 y****e 的大作中提到】
: random walk x_i = 2 or -1 with probability 1/2 for each case.
: S_n = x_1 + x_2 +.. x_n
: Pr( S_n = a before -b) ?
: 除了用recursion 有什么快的方法么？
|
f*******y
发帖数: 267 | 20 I guess iamrobot is correct.
Mahone, what is your approach based on? Can you explain in details? I am new
to Stochastic calculus...
Is this related to Fokker-planck? Thanks!
【在 M****e 的大作中提到】
: dX=0.5dt+1.5dW
: =>
: solve 0.5P''+1.5P'=0 with bc P(a)=1 and P(-b)=0
|
r**a
发帖数: 536 | 21 I do not understand his formula either. He offered a 2nd order ODE with a BC
. But the solution of his ode will be a function instead of a number.
If you follow my idea, you can got S(n) ~ 0.5n+1.5W(n). Then use the
expectation martingale to get the prob. But as i mentioned, using a BM with
drift only can give you the asymptotic solution, which implies that if a and
b are big enough, then it should be a good estimation.
new
【在 f*******y 的大作中提到】
: I guess iamrobot is correct.
: Mahone, what is your approach based on? Can you explain in details? I am new
: to Stochastic calculus...
: Is this related to Fokker-planck? Thanks!
|
M****e
发帖数: 3715 | 22 sorry made a mistake. I thought it was a Brownian motion problem.
For random walk problem, iamrobot's method is correct.
new
【在 f*******y 的大作中提到】
: I guess iamrobot is correct.
: Mahone, what is your approach based on? Can you explain in details? I am new
: to Stochastic calculus...
: Is this related to Fokker-planck? Thanks!
|
