t*****e
发帖数: 53 | 1 u = exp(sigma *sqrt(delta_t))
d = exp(-sigma *sqrt(delta_t))
p = (expt(r * delta_t) - d) / (u - d).
S_0 = 1;
At step n, whats the variance?
I follow the suggestion of this board and use the variance definition:
a node at nth is u^kd^(n-k). the probability is C_n_k * p^k * (1-p)^(n-k)
the expected stock price is exp(r * n *delta_t)
variance is sum over 0..n (C_n_k *p^k *(1-p)^(n-k) * (u^kd^(n-k)^2 - expt(-2
*n * delta_t) and I fursther simply this to:
(p*u^2 + (1-p) * d^2)^2 - expt(2*r*n*delta_t) |
m******n
发帖数: 354 | 2 no support for latex?
delta_t = T/n,
S_T = S_0 * (u^k)*d^(n-k)
= S_0 * [(exp(sigma *sqrt(delta_t)))^k] * [exp(-sigma *sqrt(delta_t))^(n
-k)]
= S_0* [exp(k*sigma*sqrt(delta_t))]*[exp((n-k)*(-sigma)*sqrt(delta_t))]
= S_0*exp[(2k-n)* sigma*sqrt(delta_t)]
= S_0*exp[(2k-n)* sigma*sqrt(T/n)]
= S_0*exp[sigma*sqrt(T)((2k-n)/sqrt(n))] (1)
where k is the number of up-moves, which has binomial distribution that is
k~bin_{n,1/2}, so E(k) = np, V(k)= np(1-p),here p =1/2 is th |
m******n
发帖数: 354 | 3 if you use martingale measure, just do the same way, k is still binomial
distributed, but with p = (exp(r * delta_t) - d) / (u - d).
to further simply, you should use the approximation:
e^x = 1 + x/1 + (x^2)/2!+ ..., and ignore the terms of delta_t of order >= 2, you will have p = 1/2(1 + sqrt(delta_t)*(r-1/2*(delta_t^2))/sigma).
The reat is easy, you will find the Variance of the stock return should be the same, but the mean changed. |
t*****e
发帖数: 53 | 4 k is a binomial distribution, but k~bin(n, p) and p = exp(sigma * sqrt(delta
_t)). this one is correct, I found on a lot of books and I can prove it.
My understanding is from the variance definition, then use tailor expansion.
But I cannot get it right? |
t*****e
发帖数: 53 | 5 I cannot see why the central limit theorum is useful or appropriate here? |
m******n
发帖数: 354 | 6 you can ignore the central limit theorem part, i just want show you the
relationship with the continuous case.
do you want direct use
E(ln(S_T/S_0)^2) - E(ln(S_t/S_0))^2 tu calculate the variance of time T? why
this then:
"variance is sum over 0..n (C_n_k *p^k *(1-p)^(n-k) * (u^kd^(n-k)^2 - expt(-
2*n * delta_t) and I fursther simply this to:
(p*u^2 + (1-p) * d^2)^2 - expt(2*r*n*delta_t)" ? |
t*****e
发帖数: 53 | 7 I just used the definition of the variance, the expected stock price at nth
step is exp(r*delta_t*n). So, the variance is:
sum over 0..n [C_n_k *p^k *(1-p)^(n-k) * u^k * d^(n-k)^2] - exp(-
2*n * delta_t).
do you agree with this? If yes,then this is equivalent to
[p*u^2+(1-p)*d^2]^n - exp(2*n*delta_t).
Let me know if you can get to this step? |
t*****e
发帖数: 53 | 8 BTW, I want to compute the stock price return variance. but not the log of
it. |
t*****e
发帖数: 53 | 9 and for one step, John Hull's book page 225 explained clearly. |
m******n
发帖数: 354 | 10 then the variance should not be sigma^2*T
【在 t*****e 的大作中提到】
: BTW, I want to compute the stock price return variance. but not the log of
: it.
|
|
|
t*****e
发帖数: 53 | 11 Oh...
so, you think LN(S_T/S_0) is sigma^2*T? I guess that might be why I cannot
continue the derivation.
Can you show it using the variance definition directly? |
m******n
发帖数: 354 | 12 I really don't get it, have u mistyped sth.?
nth
【在 t*****e 的大作中提到】
: I just used the definition of the variance, the expected stock price at nth
: step is exp(r*delta_t*n). So, the variance is:
: sum over 0..n [C_n_k *p^k *(1-p)^(n-k) * u^k * d^(n-k)^2] - exp(-
: 2*n * delta_t).
: do you agree with this? If yes,then this is equivalent to
: [p*u^2+(1-p)*d^2]^n - exp(2*n*delta_t).
: Let me know if you can get to this step?
|
t*****e
发帖数: 53 | 13 you said the variance of the stock price return is not sigma^2*T, but the
variance of Ln(stock price return) is sigma^2 *T? right?
Can you prove the Ln(stock price return) is sigma^2 * T by variance
definition? |
m******n
发帖数: 354 | 14 actually i have done it already, see the formel (1) in my first post,
you can deduce that
Ln(S_T/S_0) = sigma*sqrt(T)*((2k-n)/sqrt(n))
k is binomial distributed,
so E(k) = np, V(k)= np(1-p), you know that fact?
i think you are actually intrested in how to show V(k) = np(1-p), you can
find it in any probability text book
then you can see direct see V(Ln(S_T/S_0)) = sigma^2*T,
the trick is to use the features of binomial distribution. |
t*****e
发帖数: 53 | 15 Yes, I got this step: V(Ln(S_T/S_0)) = 4 * sigma^2 * T * p(1-p).
You used the tailor expansion for p in your previous posting. May I ask if
you have more strict proof that the limit(p*(1-p)) = 4 as delta_t -> 0.
thanks. |
w**********y
发帖数: 1691 | 16 P[S_T=u^a*d^(n-a)]=C(n,a)*p^a*(1-p)^(n-a) a=0,1,2,.....n
then E[X^2]-(EX)^2 |
