h********g
发帖数: 22 | 1 W(t) is standard BM, X follow GBM : dX=\mu X dt + \sigma X dW.
Define Y = \int_0^\infinty X(u)du.
Ask: Are there simple necessary and sufficient conditions on \mu and \sigma
that guarantee that P(Y=\infinity) =1? |
n****e
发帖数: 629 | 2 From SPM?
要独立完成面试题嘛 haha
sigma
【在 h********g 的大作中提到】
: W(t) is standard BM, X follow GBM : dX=\mu X dt + \sigma X dW.
: Define Y = \int_0^\infinty X(u)du.
: Ask: Are there simple necessary and sufficient conditions on \mu and \sigma
: that guarantee that P(Y=\infinity) =1?
|
h********g
发帖数: 22 | |
s*******s
发帖数: 1568 | 4 mu!=0??
consider discrete case
hehe
sigma
【在 h********g 的大作中提到】
: W(t) is standard BM, X follow GBM : dX=\mu X dt + \sigma X dW.
: Define Y = \int_0^\infinty X(u)du.
: Ask: Are there simple necessary and sufficient conditions on \mu and \sigma
: that guarantee that P(Y=\infinity) =1?
|
Q***5
发帖数: 994 | 5 Why is that?
X(t) = exp( (mu-1/2sigma^2)t + sigma W(t)), so a negative mu turns to push
X(t) goes to 0 faster -- make Y to be less. The condition should be
something like mu>= something_about_sigma
My guess is mu>= -1/2sigma^2
【在 s*******s 的大作中提到】
: mu!=0??
: consider discrete case
: hehe
:
: sigma
|
h********g
发帖数: 22 | 6 why not \mu > + 1/2 sigma^2? |
s*******s
发帖数: 1568 | 7 sorry, I miss the X term,
But exp( (-1/2sigma^2t + sigma W(t)) is a Martingale, thus I think mu>0 is
enough
push
【在 Q***5 的大作中提到】
: Why is that?
: X(t) = exp( (mu-1/2sigma^2)t + sigma W(t)), so a negative mu turns to push
: X(t) goes to 0 faster -- make Y to be less. The condition should be
: something like mu>= something_about_sigma
: My guess is mu>= -1/2sigma^2
|
o****b
发帖数: 31 | 8 maybe this question is related. If P[y=\infinity]=1, does it mean that
E[y]=\infinity?
\sigma
【在 h********g 的大作中提到】
: W(t) is standard BM, X follow GBM : dX=\mu X dt + \sigma X dW.
: Define Y = \int_0^\infinty X(u)du.
: Ask: Are there simple necessary and sufficient conditions on \mu and \sigma
: that guarantee that P(Y=\infinity) =1?
|
s*******s
发帖数: 1568 | 9 ... definition of Leb Integration....
【在 o****b 的大作中提到】
: maybe this question is related. If P[y=\infinity]=1, does it mean that
: E[y]=\infinity?
:
: \sigma
|
o****b
发帖数: 31 | 10 I guess if P[y=\infinity]=1 means E[y]=\infinity. E[y]=\int_0^{\infinity}E[X
(u)]du=\int_0^{\infinity}e^{\mu u}du. Then the condition is \mu>0. |
|
|
Q***5
发帖数: 994 | 11 Perhaps you are right. I was comparing the drifting term (\mu - 1/2sigma^2)t
with the variance sigma^2 t, but it seems that comparing with std sigma*
sqrt(t) makes more sense.
Are you sure the condition is not \mu >= 1/2 sigma^2? -- can someone
provides a strict proof?
【在 h********g 的大作中提到】
: why not \mu > + 1/2 sigma^2?
|
f****e
发帖数: 590 | 12 use the fact that W_t is dominated by t at large t, i.e. W_t/t -> 0
hence if \mu-1/2 \sigma^2 >0 Y is infinite a.s.
but if \mu - 1/2 \sigma^2 < 0 Y is finite a.s
\mu - 1/2 \sigma^2 = 0 is the tricky part though
)t
【在 Q***5 的大作中提到】
: Perhaps you are right. I was comparing the drifting term (\mu - 1/2sigma^2)t
: with the variance sigma^2 t, but it seems that comparing with std sigma*
: sqrt(t) makes more sense.
: Are you sure the condition is not \mu >= 1/2 sigma^2? -- can someone
: provides a strict proof?
|
Q***5
发帖数: 994 | 13 For \mu - 1/2 \sigma^2 = 0, X(t) has the distribution of exp( \sigma W(t)),
so with probability 0.5, X(t)> 1 -- so it should be true that Y is infinity
a.e., one thing we need to worry about is the correlation between W(t)'s,
but this does not seem to be an issue...
【在 f****e 的大作中提到】
: use the fact that W_t is dominated by t at large t, i.e. W_t/t -> 0
: hence if \mu-1/2 \sigma^2 >0 Y is infinite a.s.
: but if \mu - 1/2 \sigma^2 < 0 Y is finite a.s
: \mu - 1/2 \sigma^2 = 0 is the tricky part though
:
: )t
|
z****e
发帖数: 2024 | 14 弱问，
X(u)du, 这个积分变量u是t还是W？
就是说是：
X(t,W(t))dt, t从0到infinity
还是X(t,W(t))dw(t), t从0到infinity?
有什么区别么？
谢谢。 |
z****i
发帖数: 406 | 15 SPM是什么啊？
多谢！！
【在 n****e 的大作中提到】
: From SPM?
: 要独立完成面试题嘛 haha
:
: sigma
|
z****i
发帖数: 406 | 16 SPM是什么啊？
多谢！！
【在 n****e 的大作中提到】
: From SPM?
: 要独立完成面试题嘛 haha
:
: sigma
|
f****e
发帖数: 590 | 17 so u r saying bm stay above zero for half of the time, a.s. in all the paths?
,
infinity
【在 Q***5 的大作中提到】
: For \mu - 1/2 \sigma^2 = 0, X(t) has the distribution of exp( \sigma W(t)),
: so with probability 0.5, X(t)> 1 -- so it should be true that Y is infinity
: a.e., one thing we need to worry about is the correlation between W(t)'s,
: but this does not seem to be an issue...
|
Q***5
发帖数: 994 | 18 I mean:
Y can be approximately written as (consider time step 1, but we can make the
step approaches 0 to get the real Y)
X(0)+X(1)+X(2)....
Now, IF all X(k) are independent, and for each X(k), Prob(X(k)>1)= 1/2,
then the sum will be infinite a.s.
The tricky part is, X(k)'s are not really independent, but this should not
be a problem -- we can for example pick a sub-sequence X(2^k), where the time
step grows up exponentially -- to make sure that the random variables are `
almost' independent.
p
【在 f****e 的大作中提到】
: so u r saying bm stay above zero for half of the time, a.s. in all the paths?
:
: ,
: infinity
|
f****e
发帖数: 590 | 19 well..
intuitively, ok..
but does not sound like a proof..
the
time
`
【在 Q***5 的大作中提到】
: I mean:
: Y can be approximately written as (consider time step 1, but we can make the
: step approaches 0 to get the real Y)
: X(0)+X(1)+X(2)....
: Now, IF all X(k) are independent, and for each X(k), Prob(X(k)>1)= 1/2,
: then the sum will be infinite a.s.
: The tricky part is, X(k)'s are not really independent, but this should not
: be a problem -- we can for example pick a sub-sequence X(2^k), where the time
: step grows up exponentially -- to make sure that the random variables are `
: almost' independent.
|
Q***5
发帖数: 994 | 20 Or you can show, as you suggested "bm stay above zero for half of the time"
a.e. -- is that a known result?
【在 f****e 的大作中提到】
: well..
: intuitively, ok..
: but does not sound like a proof..
:
: the
: time
: `
|
a******6
发帖数: 78 | 21 yes, I see the case mu>0.
But how about the mu =0 ?
If a X(t) is a brownian motion , then y(t) is N(0 , t^3/3) .
here dx = \sigma * x dW , if mu = 0, which makes the case harder to see.
is
【在 s*******s 的大作中提到】
: sorry, I miss the X term,
: But exp( (-1/2sigma^2t + sigma W(t)) is a Martingale, thus I think mu>0 is
: enough
:
: push
|
