- #1
TroyElliott
- 59
- 3
I am trying to show that given the following stochastic differential equation: ##\dot{x} = W(x(\tau))+\eta(\tau),## we have
##det|\frac{d\eta(\tau)}{dx(\tau')}| = exp^{\int_{0}^{T}d\tau \,Tr \ln([\frac{d}{d\tau}-W'(x(\tau))]\delta (\tau - \tau'))} = exp^{\frac{1}{2}\int_{0}^{T}d\tau W'(x(\tau))}.##
Attempt at the solution: We can write the differential equation as ##\eta(\tau) = \dot{x} - W(x(\tau)),## and then take the function derivative with respect to ##x(\tau')## to get ##\frac{d\eta(\tau)}{dx(\tau')} = (\frac{d}{d\tau}-W')\delta(\tau - \tau')##. Next we can use the identity ##det(exp^{M}) = exp^{Tr(M)}## to write ##det|\frac{d\eta(\tau)}{dx(\tau')}| = exp^{Tr \, \ln(\frac{d\eta(\tau)}{dx(\tau')})} = exp^{Tr \, \ln[(\frac{d}{d\tau}-W')\delta(\tau - \tau')]}.##
This is where I am stuck at. I don't see how the integral arises in the first equality in the above equation, and furthermore I do not see how to evaluate this integral to end up with the second equality in the above equation. Any insight would be greatly appreciated. Thanks!
##det|\frac{d\eta(\tau)}{dx(\tau')}| = exp^{\int_{0}^{T}d\tau \,Tr \ln([\frac{d}{d\tau}-W'(x(\tau))]\delta (\tau - \tau'))} = exp^{\frac{1}{2}\int_{0}^{T}d\tau W'(x(\tau))}.##
Attempt at the solution: We can write the differential equation as ##\eta(\tau) = \dot{x} - W(x(\tau)),## and then take the function derivative with respect to ##x(\tau')## to get ##\frac{d\eta(\tau)}{dx(\tau')} = (\frac{d}{d\tau}-W')\delta(\tau - \tau')##. Next we can use the identity ##det(exp^{M}) = exp^{Tr(M)}## to write ##det|\frac{d\eta(\tau)}{dx(\tau')}| = exp^{Tr \, \ln(\frac{d\eta(\tau)}{dx(\tau')})} = exp^{Tr \, \ln[(\frac{d}{d\tau}-W')\delta(\tau - \tau')]}.##
This is where I am stuck at. I don't see how the integral arises in the first equality in the above equation, and furthermore I do not see how to evaluate this integral to end up with the second equality in the above equation. Any insight would be greatly appreciated. Thanks!