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4 votes
Let X

t

be the geometric Brownian motion dX
t

=rX
t

dt+αX
t

dB
t

. Find E
x
[X
T

∣F
t

] for t≤T by a) using the Markov property and b) writing X
t

=xe
rt
M
t

, where M
t

=exp(αB
t


2
1

α
2
t) is a martingale

User Lemnar
by
7.8k points

1 Answer

5 votes

To find E[X_T | F_t] for t ≤ T, we can use two different approaches:

a) Using the Markov property:
The Markov property states that the future behavior of X_T, given F_t, depends only on the current value X_t.

To find E[X_T | F_t], we can write it as E[X_T | X_t]. This means we want to find the expected value of X_T given the current value X_t.

Since X_t follows a geometric Brownian motion, we can use the solution for the stochastic differential equation (SDE) dX_t = rX_t dt + αX_t dB_t.

By solving this SDE, we find that X_t = x * exp((r - α^2/2)t + αB_t), where x is the initial value of X and B_t is a Brownian motion.

Using this solution, we can calculate E[X_T | X_t].

b) Writing X_t = xe^(rt)M_t, where M_t = exp(αB_t - (1/2)α^2t) is a martingale:
We can rewrite X_t as xe^(rt)M_t, where x is the initial value of X, r is the constant drift term, t is the time, and M_t is a martingale.

Since M_t is a martingale, we have E[M_T | F_t] = M_t.

Therefore, E[X_T | F_t] = E[xe^(rT)M_T | F_t] = xe^(rt)E[M_T | F_t] = xe^(rt)M_t.

In this approach, we use the fact that the expectation of a martingale is itself at any given time.

So, to find E[X_T | F_t] for t ≤ T, we have two methods:
a) Using the Markov property: E[X_T | X_t] = E[x * exp((r - α^2/2)T + αB_T) | X_t]
b) Writing X_t = xe^(rt)M_t: E[X_T | F_t] = xe^(rt)M_t

Please let me know if you need further clarification or have any other questions!

User Ahmed Fouad
by
8.4k points