t*****k
发帖数: 390 | 1 请问\int_0^1 f(t)B_t dt的variance是多少？ thanks |
t**o
发帖数: 64 | 2 版面上有的，不出前50贴
【在 t*****k 的大作中提到】
: 请问\int_0^1 f(t)B_t dt的variance是多少？ thanks
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t*****k
发帖数: 390 | 3 我如果用定义求没看出来后面怎么做？
X = \int_0^1 f(t)B_tdt
EX = E\int_0^1 f(t)B_tdt = \int_0^1 E(f(t)B_t)dt = 0
Var X = EX^2 - (EX)^2=EX^2 = E(\int_0^1 f(t)B_tdt)^2
=E(\int_0^1f(t)B_tdt \int_0^1f(u)B_udu)
=E(\int_0^1 \int_0^1 f(t)f(u)B_tB_udtdu)
=\int_0^1\int_0^1 f(t)f(u)E(B_tB_u)dtdu
=\int_0^1\int_0^1 f(t)f(u)min(u, t)dtdu
=2\int_0^1\int_0^u f(t)f(u)udtdu
Then, it seems a little complicated to move on... Thanks!
【在 t**o 的大作中提到】
: 版面上有的，不出前50贴
|
b*****t
发帖数: 10 | 4 可以 对 \int_0^t f(s)ds*B(t)用 ito lemma |
p*****k
发帖数: 318 | 5 trthank, your approach seems fine, just a minor typo:
2\int_0^1 dt f(t)*[\int_0^t du f(u)*u]
or as BITPITT suggested, one gets:
\int_0^1 dt [\int_t^1 du f(u)]^2
they are equivalent |
