t******m
发帖数: 255 | 1 We know that int_0^tW(u)du =N(0,t^3/3). But I need to calculate E[e^(iaW(t))
*the above integral]. Can anybody answer this question? Many Thanks. |
d*j
发帖数: 13780 | 2 a is a constant ?
把那个e^复数, 想想 normal 分布 cos 和 sin 的对称性
))
【在 t******m 的大作中提到】
: We know that int_0^tW(u)du =N(0,t^3/3). But I need to calculate E[e^(iaW(t))
: *the above integral]. Can anybody answer this question? Many Thanks.
|
t******m
发帖数: 255 | 3 a is a constant
【在 d*j 的大作中提到】
: a is a constant ?
: 把那个e^复数, 想想 normal 分布 cos 和 sin 的对称性
:
: ))
|
x******a
发帖数: 6336 | 4 It is exp(a^2t/2).
one can evaluate the expectation in many ways, in particular,
one can calculate directly by writing down the integral since the
distribution of W(t) is known.
E[e^(iaW(t))
【在 t******m 的大作中提到】
: We know that int_0^tW(u)du =N(0,t^3/3). But I need to calculate E[e^(iaW(t))
: *the above integral]. Can anybody answer this question? Many Thanks.
|
t******m
发帖数: 255 | 5 sorry. I am a little confused. distribution of W(t)is known but does the
integral N(0, t^3/3) has anything to do with W(t)? Thanks
【在 x******a 的大作中提到】
: It is exp(a^2t/2).
: one can evaluate the expectation in many ways, in particular,
: one can calculate directly by writing down the integral since the
: distribution of W(t) is known.
:
: E[e^(iaW(t))
|
p******5
发帖数: 138 | 6 If W(u) is the Brownian motion, \int^T_0 W(u) du = N(0, T^3/3) . |
p******5
发帖数: 138 | 7 My solution: assume w(s) is a Brownian motion,
E(\int^T_0 e^{i a w(T)} w(s) ds )
= \int^T_0 E(e^{i a w(T)} w(s)) ds
= \int^T_0 E(e^{i a (w(T) - w(s) + w(s))} w(s)) ds
= \int^T_0 E(e^{i a (w(s))} w(s)) ds for w(T)-w(s) independent of w(s)
I think it is not hard to get E(e^{i a (w(s))} w(s)) by properties of
characteristic function. The density function of w(s) is N(0, s) |
x******a
发帖数: 6336 | 8 sorry，did not see the product...
paul1985 is correct.
the
【在 t******m 的大作中提到】
: sorry. I am a little confused. distribution of W(t)is known but does the
: integral N(0, t^3/3) has anything to do with W(t)? Thanks
|
t******m
发帖数: 255 | 9 thx
【在 p******5 的大作中提到】
: My solution: assume w(s) is a Brownian motion,
: E(\int^T_0 e^{i a w(T)} w(s) ds )
: = \int^T_0 E(e^{i a w(T)} w(s)) ds
: = \int^T_0 E(e^{i a (w(T) - w(s) + w(s))} w(s)) ds
: = \int^T_0 E(e^{i a (w(s))} w(s)) ds for w(T)-w(s) independent of w(s)
: I think it is not hard to get E(e^{i a (w(s))} w(s)) by properties of
: characteristic function. The density function of w(s) is N(0, s)
|
i****k
发帖数: 39 | 10
Why E(e^{i a (w(T) - w(s)}) term is dropped? Cannot see the connection
between "w(T)-w(s) independent of w(s)" and dropping the term. Could you
please give more detailed explanation? Thanks a lot!
【在 p******5 的大作中提到】
: My solution: assume w(s) is a Brownian motion,
: E(\int^T_0 e^{i a w(T)} w(s) ds )
: = \int^T_0 E(e^{i a w(T)} w(s)) ds
: = \int^T_0 E(e^{i a (w(T) - w(s) + w(s))} w(s)) ds
: = \int^T_0 E(e^{i a (w(s))} w(s)) ds for w(T)-w(s) independent of w(s)
: I think it is not hard to get E(e^{i a (w(s))} w(s)) by properties of
: characteristic function. The density function of w(s) is N(0, s)
|
y********l
发帖数: 11 | 11 Agree with idrunk. Shouldn't the integrand be:
E[e^{i a (w(T) - w(s)}]*E[e^{i a w(s)} w(s)]
= e^{a^2(T-s)/2}*E[e^{i a w(s)} w(s)]?
Thanks~
【在 i****k 的大作中提到】
:
: Why E(e^{i a (w(T) - w(s)}) term is dropped? Cannot see the connection
: between "w(T)-w(s) independent of w(s)" and dropping the term. Could you
: please give more detailed explanation? Thanks a lot!
|
a*********0
发帖数: 613 | 12 弱问：why int_0^tW(u)du =N(0,t^3/3)... |
c********d
发帖数: 173 | 13 Shouldn't E[e^(iaW(t))=exp(-a^2t/2), rather than exp(a^2t/2)? |
