Variance of Ito Integral when the integrand is a function of a different Wienner process

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I need to find the expression for variance of the integral with the following form:



$int_0^T X_tdW_2t$



when $X_t=f(W_1t)$ and both $W_1t$ and $W_2t$ are defined on the same sample space and are correlated. More specifically, is it still possible to use (a version of) Ito Isometry that would take the correlation into account?




probability integration variance




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  • As long as the integrand $X_t$ is adapted to a filtration which is admissible for $W_2t$, you can apply Itô's isometry. Note that you need this measurability assumption in any case in order to make sense of the integral.
    – saz
    Aug 27 at 6:29










  • @saz, makes sense, completely. Just as a follow up. So, I do not need to include correlation coefficient anywhere then?
    – Ovi
    Aug 27 at 6:57














up vote
0
down vote

favorite












I need to find the expression for variance of the integral with the following form:



$int_0^T X_tdW_2t$



when $X_t=f(W_1t)$ and both $W_1t$ and $W_2t$ are defined on the same sample space and are correlated. More specifically, is it still possible to use (a version of) Ito Isometry that would take the correlation into account?




probability integration variance




share|cite|improve this question




















  • As long as the integrand $X_t$ is adapted to a filtration which is admissible for $W_2t$, you can apply Itô's isometry. Note that you need this measurability assumption in any case in order to make sense of the integral.
    – saz
    Aug 27 at 6:29










  • @saz, makes sense, completely. Just as a follow up. So, I do not need to include correlation coefficient anywhere then?
    – Ovi
    Aug 27 at 6:57












up vote
0
down vote

favorite









up vote
0
down vote

favorite











I need to find the expression for variance of the integral with the following form:



$int_0^T X_tdW_2t$



when $X_t=f(W_1t)$ and both $W_1t$ and $W_2t$ are defined on the same sample space and are correlated. More specifically, is it still possible to use (a version of) Ito Isometry that would take the correlation into account?




probability integration variance




share|cite|improve this question












I need to find the expression for variance of the integral with the following form:



$int_0^T X_tdW_2t$



when $X_t=f(W_1t)$ and both $W_1t$ and $W_2t$ are defined on the same sample space and are correlated. More specifically, is it still possible to use (a version of) Ito Isometry that would take the correlation into account?





probability integration variance





share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Aug 26 at 23:22









Ovi

12




12











  • As long as the integrand $X_t$ is adapted to a filtration which is admissible for $W_2t$, you can apply Itô's isometry. Note that you need this measurability assumption in any case in order to make sense of the integral.
    – saz
    Aug 27 at 6:29










  • @saz, makes sense, completely. Just as a follow up. So, I do not need to include correlation coefficient anywhere then?
    – Ovi
    Aug 27 at 6:57
















  • As long as the integrand $X_t$ is adapted to a filtration which is admissible for $W_2t$, you can apply Itô's isometry. Note that you need this measurability assumption in any case in order to make sense of the integral.
    – saz
    Aug 27 at 6:29










  • @saz, makes sense, completely. Just as a follow up. So, I do not need to include correlation coefficient anywhere then?
    – Ovi
    Aug 27 at 6:57















As long as the integrand $X_t$ is adapted to a filtration which is admissible for $W_2t$, you can apply Itô's isometry. Note that you need this measurability assumption in any case in order to make sense of the integral.
– saz
Aug 27 at 6:29




As long as the integrand $X_t$ is adapted to a filtration which is admissible for $W_2t$, you can apply Itô's isometry. Note that you need this measurability assumption in any case in order to make sense of the integral.
– saz
Aug 27 at 6:29












@saz, makes sense, completely. Just as a follow up. So, I do not need to include correlation coefficient anywhere then?
– Ovi
Aug 27 at 6:57




@saz, makes sense, completely. Just as a follow up. So, I do not need to include correlation coefficient anywhere then?
– Ovi
Aug 27 at 6:57















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