c**a
发帖数: 316 | 1 【 以下文字转载自 Mathematics 讨论区 】
发信人: ccca (cc), 信区: Mathematics
标 题: a simple stochastic process problem that I do not get
发信站: BBS 未名空间站 (Sat Apr 4 17:47:33 2015, 美东)
Please help!
X(t) is the stochastic exponential of a standard Brownian motion,
i.e. X(t) = exp(w(t) - 0.5 t), w(t) being a standard Brownian motion.
What is the probability that X(T) is greater than H (>1)?
My solution:
log(X(T)) > log (H)
=>
w(T) - 0.5 T > log(H)
=>
w(T) > log(H) + 0.5 T
=>
y > log(H)/sqrt(T) + 0.5*sqrt(T) where y is a standard normal r.v.
The PROBLEM is as T approaches +inf
[log(H)/sqrt(T) + 0.5*sqrt(T)] approaches +inf as well.
Hence, we have y > +inf.
Hence, the probability that X(T) is greater than H is zero.
Which does not make ANY sense. |
L*******t
发帖数: 2385 | 2 please dont forget the mean of your random variable approaches inf as well,
that's why
【在 c**a 的大作中提到】
: 【 以下文字转载自 Mathematics 讨论区 】
: 发信人: ccca (cc), 信区: Mathematics
: 标 题: a simple stochastic process problem that I do not get
: 发信站: BBS 未名空间站 (Sat Apr 4 17:47:33 2015, 美东)
: Please help!
: X(t) is the stochastic exponential of a standard Brownian motion,
: i.e. X(t) = exp(w(t) - 0.5 t), w(t) being a standard Brownian motion.
: What is the probability that X(T) is greater than H (>1)?
: My solution:
: log(X(T)) > log (H)
|
L*******t
发帖数: 2385 | 3 居然还发到大数学版去了，
以后有stochastic的问题来我们大金工版问。
【在 c**a 的大作中提到】
: 【 以下文字转载自 Mathematics 讨论区 】
: 发信人: ccca (cc), 信区: Mathematics
: 标 题: a simple stochastic process problem that I do not get
: 发信站: BBS 未名空间站 (Sat Apr 4 17:47:33 2015, 美东)
: Please help!
: X(t) is the stochastic exponential of a standard Brownian motion,
: i.e. X(t) = exp(w(t) - 0.5 t), w(t) being a standard Brownian motion.
: What is the probability that X(T) is greater than H (>1)?
: My solution:
: log(X(T)) > log (H)
|
c**a
发帖数: 316 | 4
,
But y = w(T)/sqrt(T) is standard Gaussian. Its mean does not approach inf.
【在 L*******t 的大作中提到】
: please dont forget the mean of your random variable approaches inf as well,
: that's why
|
c**a
发帖数: 316 | 5 Still do not understand. :(
y = w(T)/sqrt(T) is standard Gaussian. Its mean does not approach inf. |
L*******t
发帖数: 2385 | 6 对了你知道这个事实么，你的Ｘ（Ｔ）在Ｔ趋向于Ｉｎｆ的时候是趋向于零的。
这么以来，就清楚了
【在 c**a 的大作中提到】
: Still do not understand. :(
: y = w(T)/sqrt(T) is standard Gaussian. Its mean does not approach inf.
|
s*******1
发帖数: 20 | 7 are you sure? limit in what sense?
【在 L*******t 的大作中提到】
: 对了你知道这个事实么，你的Ｘ（Ｔ）在Ｔ趋向于Ｉｎｆ的时候是趋向于零的。
: 这么以来，就清楚了
|
L*******t
发帖数: 2385 | 8 almost surely, W(t)是以Sqrt(t)的速度增长的比-0.5t来的慢很多，可以这么理解一
下啊
【在 s*******1 的大作中提到】
: are you sure? limit in what sense?
|
c**a
发帖数: 316 | 9 是的 没错 x(T) -> 1 as T -> inf. making sense as x(t) is a martingale. 多谢
！ |
c**a
发帖数: 316 | 10 还是不明白。
For simplicity, S(0) = 1, \sigma = 1. (spot price = 1, vol = 1).
Define strike price K = H > 1, and Cn as the an vanilla European call
option with expiration T = n. Zero interest rate. So payoff is S(T_n)-K.
Then S(T_n) = exp(w(t) - 0.5 * t)
What does C_n look like as n -> +\inf.
If S(T_n) -> 1 a.s., the prob. the option Cn is in the money is zero.
But on the other hand, we know that C_n -> S_0.
Interesting. I think I get this. Let A be a set of zero measure, \int_{A}f(A
)dA can still be greater than zero, if f(A) is unbounded... |
|
|
x******a
发帖数: 6336 | 11 Interesting. I think I get this. Let A be a set of zero measure, int_{A}f(
A
This is simply not true. You really have to understand the different types
of convergence of random variables.
A |
L*******t
发帖数: 2385 | 12 sorry should be converge in prob
如果是almost surely的话，那么implyconverge in distribution
而你的process的expectation永远是1。。
不好意思造成了confusion
【在 L*******t 的大作中提到】
: almost surely, W(t)是以Sqrt(t)的速度增长的比-0.5t来的慢很多，可以这么理解一
: 下啊
|
c**a
发帖数: 316 | 13 So if X is r.v., and X -> 0 in probablity then E(X) = 0 may not be true,
but if X->0 almost surely, then E(X) = 0 is true? |
s**t
发帖数: 66 | 14 Why doesn't it make sense?
【在 c**a 的大作中提到】
: So if X is r.v., and X -> 0 in probablity then E(X) = 0 may not be true,
: but if X->0 almost surely, then E(X) = 0 is true?
|
x******a
发帖数: 6336 | 15 it is not true.
some basic knowledge of real analysis would be helpful.
【在 c**a 的大作中提到】
: So if X is r.v., and X -> 0 in probablity then E(X) = 0 may not be true,
: but if X->0 almost surely, then E(X) = 0 is true?
|
b***y
发帖数: 14281 | 16 It makes all the sense. When t->inf, X(t)->0*(a random gaussian), of course
the P(X>H)->0, as long as H>0. |
