Heston

Still WIP

Model:

\[ \frac{dS}{S}(t) = r(t) dt + \sqrt{\sigma(t)} dW_{t} \]

\[ d \sigma(t) = \alpha(t) (\theta(t) - \sigma(t)) dt + \sqrt{\sigma(t)} \psi(t) dV_t \]

\[ S(t)\sigma(t) = \rho(t) \]

After isolation and application:

\[ \int dS = \int S(t)(r(t) dt + \sqrt{\sigma(t)} dW_{t}) \]

\[ \int d \sigma(t) = \int \alpha(t) (\theta(t) - \sigma(t)) dt + \sqrt{\sigma(t)} \psi(t) dV_t \]

After resolution - in the traditional Riemannian approach but what about the ito calculus?: we may need to re-interpret the boundary condition \(S(0)\), \(\sigma(0)\) differently

\[ S(t) - S(0) = \int S(t)(r(t) dt + \sqrt{\sigma(t)} dW_{t}) \]

\[ \sigma(t) - \sigma(0) = \int \alpha(t) (\theta(t) - \sigma(t)) dt + \sqrt{\sigma(t)} \psi(t) dV_t \]

and final re-write:

\[ S(t) = S(0) + \int S(t)(r(t) dt + \sqrt{\sigma(t)} dW_{t}) \]

\[ \sigma(t) = \sigma(0) + \int \alpha(t) (\theta(t) - \sigma(t)) dt + \sqrt{\sigma(t)} \psi(t) dV_t \]

and then linearity of the integral:

\[ S(t) = S(0) + \int S(t)r(t) dt + \int S(t)\sqrt{\sigma(t)} dW_{t} \]

\[ \sigma(t) = \sigma(0) + \int \alpha(t) (\theta(t) - \sigma(t)) dt + \int \sqrt{\sigma(t)} \psi(t) dV_t \]

Now Stratonovich translation:

\[ S(t) = S(0) + \int S(t)r(t) + \frac{1}{2}S(t)\sqrt{\sigma(t)}\frac{d}{dS}(S(t)\sqrt{\sigma(t)})dt + \int S(t)\sqrt{\sigma(t)} \circ dW_{t} \]

\[ \sigma(t) = \sigma(0) + \int \alpha(t) (\theta(t) - \sigma(t)) + \frac{1}{2} \sqrt{\sigma(t)} \psi(t) \frac{d}{dS} \sqrt{\sigma(t)} \psi(t) dt + \int \sqrt{\sigma(t)} \psi(t) \circ dV_t \]

THEN CONSTRUCT CIRCUIT AND DO STRATONOVICH expansion

does the usual anti-deriv work? i.e. if we do not resolve an integral can we still just shift the stream for the integrator or do we need something more? Need to look at the stratonovich-taylor expansion to figure this out..

equally: when we assume that the usual multiplication/addition/scalar rules work, i.e. the usual semantics of these operators work the same if so the convolution, powers(square root) should also work the same most likely the problems come with deriv, inter and resolution of inter, i.e. boundary conditions

Note that we may have to multiply the fixed-functions: we can probably get away with doing a scalar mult of some sort BUT! when they are not constant we have to do something tricksy... since they do not necessarily have a Taylor expansion but are inside the integration so cannot just be integrated away without some split integration

It looks like we will be staying in the Stratonovich world!! If the Stratonovich Taylor expansion works out we will stay and will only have to figure out the function mult/term function and how it works with the square root etc.... Term structures we do by the integral linearity: i.e. we split the integral into small pieces at each singularity of the term structure