v*****n
发帖数: 22 | 1 Find the linear correlation coefficient betweenX(T) and Y (T) whereX(t)=
W(t) and Y (t) = int_0_t{W(s)}ds. W(t) is a Brownian motion process and
assume W(0) = 0. | J*******g
发帖数: 267 | 2 var(X(t)) = t
var(Y(t)) = t^3/3
cov(X(t), Y(t)) = t^2/2
so, corr(X(t), Y(t)) = sqrt(3)/2
【在 v*****n 的大作中提到】
: Find the linear correlation coefficient betweenX(T) and Y (T) whereX(t)=
: W(t) and Y (t) = int_0_t{W(s)}ds. W(t) is a Brownian motion process and
: assume W(0) = 0.
| c******r
发帖数: 300 | 3 sorry wrong calculation of var(y(t))
d(X(t)Y(t)) = W(t)^2dt + Y(t)dW(t)
=>
X(t)Y(t) = \int_0^t W(s)^2ds + \int_0^tY(t)dW(t)
=>
E(X(t)Y(t)) = \int_0^t sds = t^2/2
=>
Cov(X(t),Y(t))=t^2/2
Var(X(t)) = t, Var(Y(t)) = E(Y(t)^2)
dY(t)^2 = 2Y(t)dY(t) + W(t)^2dt
= 2Y(t)W(t)dt + W(t)^2 dt
=>
Y(t)^2 = 2 \int_0^t Y(t)W(t)dt
=> E(Y(t)^2) = 2 \int_0^t E(Y(t)W(t))dt = t^3/3
=> Var(Y(t)) = t^3/3
Or you can use the definition of Riemann integral to solve the problem.
【在 v*****n 的大作中提到】
: Find the linear correlation coefficient betweenX(T) and Y (T) whereX(t)=
: W(t) and Y (t) = int_0_t{W(s)}ds. W(t) is a Brownian motion process and
: assume W(0) = 0.
| v*****n
发帖数: 22 | 4 Thanks for the detailed solution.
Uncleared on one equation.
sorry wrong calculation of var(y(t))
d(X(t)Y(t)) = W(t)^2dt + Y(t)dW(t)
=>
X(t)Y(t) = \int_0^t W(s)^2ds + \int_0^tY(t)dW(t)
=>
E(X(t)Y(t)) = \int_0^t sds = t^2/2
=>
Cov(X(t),Y(t))=t^2/2
Var(X(t)) = t, Var(Y(t)) = E(Y(t)^2)
dY(t)^2 = 2Y(t)dY(t) + W(t)^2dt
~~~~~~~~~~~
Where is that from? Shouldn't it be 1/2W(t)^2dtdt and
then=0?
= 2Y(t)W(t)dt + W(t)^2 dt
=>
Y(t)^2 = 2 \int_0^t Y(t)W(t)dt
=> E
【在 c******r 的大作中提到】
: sorry wrong calculation of var(y(t))
: d(X(t)Y(t)) = W(t)^2dt + Y(t)dW(t)
: =>
: X(t)Y(t) = \int_0^t W(s)^2ds + \int_0^tY(t)dW(t)
: =>
: E(X(t)Y(t)) = \int_0^t sds = t^2/2
: =>
: Cov(X(t),Y(t))=t^2/2
: Var(X(t)) = t, Var(Y(t)) = E(Y(t)^2)
: dY(t)^2 = 2Y(t)dY(t) + W(t)^2dt
| v*****n
发帖数: 22 | 5 Thanks.
Always right but not in detail. :)
【在 J*******g 的大作中提到】
: var(X(t)) = t
: var(Y(t)) = t^3/3
: cov(X(t), Y(t)) = t^2/2
: so, corr(X(t), Y(t)) = sqrt(3)/2
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