n*********3
发帖数: 21 | 1 We all know the common symmetric random walk problem as below:
start from 0, move on axis by brownian motion, then we have probability of
first reaching a to be b/(a+b), and probability of first reaching -b to be a
/(a+b), also we know the expectation of reaching a or b is a*b.
P(a)+P(b) = 1
E[W_t] = a*P(a) + b*P(b) = 0 //martingale
E[W_t^2-t] = a^2*P(a) + b^2*P(b) - E[t] = 0 //martingale
Hence derive the solution (suppose t is stopping time).
The question is how to compute E[t^k], where k>=1, obviously we have k=1
solution which is a*b.
I have tried to think of the following, using moment generation function f:
function of m as f(m) = E[Exp(m*t)], then f'(m) = E[t*Exp(m*t)] => f'(0) = E
[t]
Hence, recall Exp(lamda*W_t-0.5*lamda^2*t] is martingale, we have
E[Exp(lamda*W_t-0.5*lamda^2*t]] = E[Exp(0)] = 1
= P(a)*Exp(lamda*a)*E[Exp(-0.5*lamda^2*t)]+P(b)*Exp(lamda*(-b))*E[Exp(-0.5*
lamda^2*t)]
simplify to derive:
E[Exp(-0.5*lamda^2*t)] = (a+b)/(b*Exp(lamda*a)+a*Exp(-lamda*b))
then take lamda = sqrt(-2m), we have E[Exp(m*t)]= ...
The problem is I can derive E[t] = a*b by using f'(m) where m = 0. Where it
goes wrong? | a********e
发帖数: 508 | 2 many mistakes...W_t and t are not independent
a
【在 n*********3 的大作中提到】
: We all know the common symmetric random walk problem as below:
: start from 0, move on axis by brownian motion, then we have probability of
: first reaching a to be b/(a+b), and probability of first reaching -b to be a
: /(a+b), also we know the expectation of reaching a or b is a*b.
: P(a)+P(b) = 1
: E[W_t] = a*P(a) + b*P(b) = 0 //martingale
: E[W_t^2-t] = a^2*P(a) + b^2*P(b) - E[t] = 0 //martingale
: Hence derive the solution (suppose t is stopping time).
: The question is how to compute E[t^k], where k>=1, obviously we have k=1
: solution which is a*b.
| n*********3
发帖数: 21 | 3 I don't fully understand,
1. can we not assume they are independent?
2. if dependent, how can we derive all moments of stopping time t? | a********e
发帖数: 508 | 4 1. E[W_t^2-t] = 0 can't be derived directly from optional sampling theory
2. E[Exp(lamda*W_t-0.5*lamda^2*t]] != E[Exp(lamda*W_t)]*E[Exp(-0.5*lamda^2*t
)]
obviously W_t=a or b is correlated with t
The MGF can't be derived explicitly, but the n-th moment can be calculated
by
taking the n-th derivative of E[Exp(lamda*W_t-0.5*lamda^2*t]] and let lambda
=0.
Then you can work E[t^n] recursively
【在 n*********3 的大作中提到】
: I don't fully understand,
: 1. can we not assume they are independent?
: 2. if dependent, how can we derive all moments of stopping time t?
| p*****k
发帖数: 318 | 5 nanfeng1213, seems to me your approach (via m.g.f. / Laplace transform of
the p.d.f.) more or less works, despite all the valid issues raised by
allthesame.
the extra step you need is to take lambda -> -lambda, which results the
following linear equations (tower rule):
E_a * e^(lambda*a) + E_b * e^(-lambda*b) = 1
E_a * e^(-lambda*a) + E_b * e^(lambda*b) = 1
where
E_a = P(W_T=a) * E[e^(-alpha*T)|W_T=a],
E_b = P(W_T=-b) * E[e^(-alpha*T)|W_T=-b]
with alpha = lambda^2/2
thus one gets:
E[e^(-alpha*T)] = E_a + E_b
= [sinh(lambda*a)+sinh(lambda*b)]/sinh[lambda*(a+b)]
= cosh[ lambda*(a-b)/2 ] / cosh[ lambda*(a+b)/2 ]
i believe its Laplace tranform agrees with the standard infinite sum
expression of the p.d.f.
(see either Karatzas & Shreve or some double barrier option papers) | c**********e
发帖数: 2007 | 6 You can also use martingale. The first 3 martingales are
for t: W_t^2 - t
for t^2: W_t^4 - 6t*W_t^4 + 3t^2
for t^3: W_t^6 - 15t*W_t^4 + 45t^2*W_t^2 - 15t^3
a
【在 n*********3 的大作中提到】
: We all know the common symmetric random walk problem as below:
: start from 0, move on axis by brownian motion, then we have probability of
: first reaching a to be b/(a+b), and probability of first reaching -b to be a
: /(a+b), also we know the expectation of reaching a or b is a*b.
: P(a)+P(b) = 1
: E[W_t] = a*P(a) + b*P(b) = 0 //martingale
: E[W_t^2-t] = a^2*P(a) + b^2*P(b) - E[t] = 0 //martingale
: Hence derive the solution (suppose t is stopping time).
: The question is how to compute E[t^k], where k>=1, obviously we have k=1
: solution which is a*b.
|
|
