Fokker Plank Equation

For a drift diffusion process

d \vec{z} = \vec{A} (t, \vec{z}) d t + 𝐁 (t, \vec{z}) d {\vec{W}}_{t}

is in differential form. The in integral form

\vec{z} (t_{b}) - \vec{z} (t_{a}) = \int_{t_{a}}^{t_{b}} \vec{A} (t, \vec{z} (t)) d t + \int_{t_{a}}^{t_{b}} 𝐁 (t, \vec{z} (t)) d {\vec{W}}_{t}

is defined by a regular Reinmann integral and an Ito integral with respect to a Weiner process. Expanding their definitions is

= \lim_{Δ t_{max} \to 0} \sum_{(t_{i}, t_{i + 1}) \in π (t_{a}, t_{b})} \vec{A} (t_{i}, \vec{z} (t_{i})) (t_{i + 1} - t_{i}))

+ \lim_{Δ t_{max} \to 0} \sum_{(t_{i}, t_{i + 1}) \in π (t_{a}, t_{b})} 𝐁 (t_{i}, \vec{z} (t_{i})) (\vec{W} (t_{i + 1}) - \vec{W} (t_{i})))

Let $Δ_{i} t = t_{i + 1} - t_{i}$ and $Δ_{i} \vec{W} = \vec{W} (t_{i + 1}) - \vec{W} (t_{i})$ .

= \lim_{Δ t_{max} \to 0} \sum_{Δ_{i} t \in π (t_{a}, t_{b})} \vec{A} (t_{i}, \vec{z} (t_{i})) Δ_{i} t + \lim_{Δ t_{max} \to 0} \sum_{Δ_{i} t \in π (t_{a}, t_{b})} 𝐁 (t_{i}, \vec{z} (t_{i})) Δ_{i} \vec{W}

Now consider an infinitelly differentiable, square integrable function $f (t, \vec{z} (t))$ that is a function of the vector.

f (t + k, \vec{z} (t + k)) - f (t, \vec{z} (t))

We can put in a dummy value equal to zero

= f (t + k, \vec{z} (t + k)) + \sum_{j = 1}^{N} (- f (t_{j}, \vec{z} (t_{j})) + f (t_{j}, \vec{z} (t_{j}))) - f (t_{a}, \vec{z} (t_{a}))

And move the 2 endpoints into the summation

= \sum_{i = 0}^{N} (f (t_{i + 1}, \vec{z} (t_{i + 1})) - f (t_{i}, \vec{z} (t_{i})))

where the endpoints are $t_{0} = t$ and $t_{N + 1} = t + k$ , and all the other indices are copied $t_{i} = t_{j}$ . Then make

= \sum_{i = 0}^{N} (f (t_{i + 1} + Δ_{i + 1} t, \vec{z} (t_{i + 1} + Δ_{i + 1} t)) - f (t_{i} + Δ_{i} t, \vec{z} (t_{i} + Δ_{i} t)))

And taking the limit it is still valid

= \lim_{N \to \infty} \sum_{i = 0}^{N} (f (t_{i + 1} + Δ_{i + 1} t, \vec{z} (t_{i + 1} + Δ_{i + 1} t)) - f (t_{i} + Δ_{i} t, \vec{z} (t_{i} + Δ_{i} t)))

We can make the sum a partition

= \lim_{Δ t_{max} \to 0} \sum_{Δ_{i} t \in π (t, t + k)} (f (t_{i + 1} + Δ_{i + 1} t, \vec{z} (t_{i + 1} + Δ_{i + 1} t)) - f (t_{i} + Δ_{i} t, \vec{z} (t_{i} + Δ_{i} t)))

And we the test function $f$ is chosen so that the radius of convergence for the Taylor series is nonzero. So we can expand the difference of the inside of the sum to a Taylor series.

= \lim_{Δ t_{max} \to 0} \sum_{i} (\frac{\partial f}{\partial t} Δ t + \frac{\partial f}{\partial \vec{z}} \cdot (\vec{z} (t_{i + 1}) - \vec{z} (t_{i})) + \frac{1}{2} \frac{\partial^{2} f}{\partial^{2} t} Δ t^{2} + \dots)

Let

Δ_{i} \vec{z} = \vec{z} (t_{i + 1}) - \vec{z} (t_{i})

= \lim_{Δ t_{max} \to 0} \sum_{i} (\frac{\partial f}{\partial t} Δ t + \frac{\partial f}{\partial \vec{z}} \cdot Δ_{i} \vec{z} + \frac{1}{2} \frac{\partial^{2} f}{\partial^{2} t} Δ t^{2} + \frac{1}{2} \frac{\partial^{2} f}{\partial^{2} \vec{z}} Δ_{i} \vec{z} \cdot Δ_{i} \vec{z} + \frac{\partial^{2} f}{\partial t \partial \vec{z}} Δ t Δ_{i} \vec{z} + \dots)

Here $\frac{\partial f}{\partial \vec{z}}$ is the gradient and $\frac{\partial^{2} f}{\partial^{2} \vec{z}}$ is the Hessian and so on. If we expand the gradients and Hessians to a sum

= \lim_{Δ t_{max} \to 0} \sum_{i} (\frac{\partial f}{\partial t} Δ t + \sum_{k}^{U} \frac{\partial f}{\partial {\vec{z}}_{k}} Δ_{i} {\vec{z}}_{k} + \frac{1}{2} \frac{\partial^{2} f}{\partial^{2} t} Δ t^{2} + \frac{1}{2} \sum_{k}^{U} \sum_{l}^{U} \frac{\partial^{2} f}{\partial {\vec{z}}_{k} \partial {\vec{z}}_{l}} Δ_{i} {\vec{z}}_{k} Δ_{i} {\vec{z}}_{l} + \sum_{k}^{U} \frac{\partial^{2} f}{\partial t \partial {\vec{z}}_{k}} Δ t Δ_{i} {\vec{z}}_{k} + \dots)

one can see that

Δ_{i} {\vec{z}}_{k} = \int_{t_{i}}^{t_{i + 1}} \vec{A} (t, \vec{z} (t))_{k} d t + \int_{t_{i}}^{t_{i + 1}} 𝐁 (t, \vec{z} (t))_{k} d {\vec{W}}_{t}

and so

Δ_{i} {\vec{z}}_{k} Δ_{i} {\vec{z}}_{l} = (\int_{t_{i}}^{t_{i + 1}} {\vec{A}}_{k} (t, \vec{z} (t)) d t) (\int_{t_{i}}^{t_{i + 1}} {\vec{A}}_{l} (t, \vec{z} (t)) d t)

+ (\int_{t_{i}}^{t_{i + 1}} 𝐁_{k} (t, \vec{z} (t)) d {\vec{W}}_{t}) (\int_{t_{i}}^{t_{i + 1}} 𝐁_{l} (t, \vec{z} (t)) d {\vec{W}}_{t})

+ (\int_{t_{i}}^{t_{i + 1}} {\vec{A}}_{k} (t, \vec{z} (t)) d t) (\int_{t_{i}}^{t_{i + 1}} 𝐁_{l} (t, \vec{z} (t)) d {\vec{W}}_{t})

+ (\int_{t_{i}}^{t_{i + 1}} {\vec{A}}_{l} (t, \vec{z} (t)) d t) (\int_{t_{i}}^{t_{i + 1}} 𝐁_{k} (t, \vec{z} (t)) d {\vec{W}}_{t})

Get rid of the first term

For the first part let

M_{k} = {sup}_{t \in [t_{i}, t_{i + 1}]} {\vec{A}}_{k} (t, z (t))

M_{l} = {sup}_{t \in [t_{i}, t_{i + 1}]} {\vec{A}}_{l} (t, z (t))

Then

\int_{t_{i}}^{t_{i + 1}} {\vec{A}}_{k} (t, \vec{z} (t)) d t \leq M_{k} Δ_{i} t

(\int_{t_{i}}^{t_{i + 1}} {\vec{A}}_{k} (t, \vec{z} (t)) d t) (\int_{t_{i}}^{t_{i + 1}} {\vec{A}}_{l} (t, \vec{z} (t)) d t) \leq M_{k} M_{l} Δ_{i} t^{2}

And inside the summation is

\lim_{Δ t_{max} \to 0} \sum_{i} \frac{\partial f}{\partial {\vec{z}}_{k} \partial {\vec{z}}_{l}} (\int_{t_{i}}^{t_{i + 1}} {\vec{A}}_{k} (t, \vec{z} (t)) d t) (\int_{t_{i}}^{t_{i + 1}} {\vec{A}}_{l} (t, \vec{z} (t)) d t) = \lim_{Δ t_{max} \to 0} \sum_{i} \frac{\partial f}{\partial {\vec{z}}_{k} \partial {\vec{z}}_{l}} \frac{}{} M_{k} M_{l} Δ_{i} t^{2} \leq \lim_{Δ t_{max} \to 0} Δ_{i} t_{max} \sum_{i} \frac{\partial f}{\partial {\vec{z}}_{k} \partial {\vec{z}}_{l}} \frac{}{} M_{k} M_{l} Δ_{i} t = 0

and doing the same thing with a lower bound we know the first part goes to zero

The second term

Looking at the second term

(\int_{t_{i}}^{t_{i + 1}} 𝐁_{k} (t, \vec{z} (t)) d {\vec{W}}_{t}) (\int_{t_{i}}^{t_{i + 1}} 𝐁_{l} (t, \vec{z} (t)) d {\vec{W}}_{t})

We can expand the dot product to

= (\sum_{s}^{U} \int_{t_{i}}^{t_{i + 1}} 𝐁_{k, s} (t, \vec{z} (t)) d {\vec{W}}_{t, s}) (\sum_{q}^{U} \int_{t_{i}}^{t_{i + 1}} 𝐁_{l, q} (t, \vec{z} (t)) d {\vec{W}}_{t, q})

= \sum_{s}^{U} \sum_{q}^{U} (\int_{t_{i}}^{t_{i + 1}} 𝐁_{k, s} (t, \vec{z} (t)) d {\vec{W}}_{t, s}) (\int_{t_{i}}^{t_{i + 1}} 𝐁_{l, q} (t, \vec{z} (t)) d {\vec{W}}_{t, q})

If we take the expected value then

= \sum_{s}^{U} \sum_{q}^{U} 𝔼 [(\int_{t_{i}}^{t_{i + 1}} 𝐁_{k, s} (t, \vec{z} (t)) d {\vec{W}}_{t, s}) (\int_{t_{i}}^{t_{i + 1}} 𝐁_{l, q} (t, \vec{z} (t)) d {\vec{W}}_{t, q})]

Swap in the definition of Ito integrals

= \sum_{s}^{U} \sum_{q}^{U} 𝔼 [(\lim_{Δ t_{max} \to 0} \sum_{Δ_{a} t \in π (t_{i}, t_{i + 1})} 𝐁_{k, s} (t, \vec{z} (t)) Δ_{a} {\vec{W}}_{t, s}) (\lim_{Δ t_{max} \to 0} \sum_{Δ_{b} t \in π (t_{i}, t_{i + 1})} 𝐁_{l, q} (t, \vec{z} (t)) Δ_{b} {\vec{W}}_{t, q})]

Factor out the limit and swap it with the expected value (I think this works but not sure)

= \lim_{Δ t_{max} \to 0} 𝔼 [\sum_{s}^{U} \sum_{q}^{U} (\sum_{Δ_{a} t \in π (t_{i}, t_{i + 1})} 𝐁_{k, s} (t, \vec{z} (t)) Δ_{a} {\vec{W}}_{t, s}) (\sum_{Δ_{b} t \in π (t_{i}, t_{i + 1})} 𝐁_{l, q} (t, \vec{z} (t)) Δ_{b} {\vec{W}}_{t, q})]

= \lim_{Δ t_{max} \to 0} 𝔼 [\sum_{s}^{U} \sum_{q}^{U} \sum_{Δ_{a} t \in π (t_{i}, t_{i + 1})} \sum_{Δ_{b} t \in π (t_{i}, t_{i + 1})} (𝐁_{k, s} (t, \vec{z} (t)) Δ_{a} {\vec{W}}_{t, s}) (𝐁_{l, q} (t, \vec{z} (t)) Δ_{b} {\vec{W}}_{t, q})]

then rearange

= \lim_{Δ t_{max} \to 0} \sum_{s}^{U} \sum_{q}^{U} \sum_{Δ_{a} t \in π (t_{i}, t_{i + 1})} \sum_{Δ_{b} t \in π (t_{i}, t_{i + 1})} 𝔼 [(𝐁_{k, s} (t, \vec{z} (t)) 𝐁_{l, q} (t, \vec{z} (t))) (Δ_{a} {\vec{W}}_{t, s} Δ_{b} {\vec{W}}_{t, q})]

The $(Δ_{a} {\vec{W}}_{t, s} Δ_{b} {\vec{W}}_{t, q})$ are are independent, so their covariance is always zero if $s \neq q$ . So in the sume we only keep the diagonals

= \lim_{Δ t_{max} \to 0} \sum_{s}^{U} \sum_{Δ_{a} t \in π (t_{i}, t_{i + 1})} \sum_{Δ_{b} t \in π (t_{i}, t_{i + 1})} 𝔼 [(𝐁_{k, s} (t, \vec{z} (t)) 𝐁_{l, s} (t, \vec{z} (t))) (Δ_{a} {\vec{W}}_{t, s} Δ_{b} {\vec{W}}_{t, s})]

And the increments in $(Δ_{a} {\vec{W}}_{t, s} Δ_{b} {\vec{W}}_{t, s})$ are also independent by definition of a Weiner process, so the expected value goes to zero for different increments $a \neq b$

= \lim_{Δ t_{max} \to 0} \sum_{s}^{U} \sum_{Δ_{a} t \in π (t_{i}, t_{i + 1})} 𝔼 [(𝐁_{k, s} (t, \vec{z} (t)) 𝐁_{l, s} (t, \vec{z} (t))) (Δ_{a} {\vec{W}}_{t, s} Δ_{a} {\vec{W}}_{t, s})]

And the variance of a Weiner processes increment is by definition the increment of time, so $(Δ_{a} {\vec{W}}_{t, s} Δ_{a} {\vec{W}}_{t, s}) = Δ_{a} t$

= \lim_{Δ t_{max} \to 0} \sum_{s}^{U} \sum_{Δ_{a} t \in π (t_{i}, t_{i + 1})} 𝔼 [(𝐁_{k, s} (t, \vec{z} (t)) 𝐁_{l, s} (t, \vec{z} (t)))] Δ_{a} t

And now its just the definition of a Reinmann integral

= \sum_{s}^{U} \int_{t_{i}}^{t_{i + 1}} 𝔼 [(𝐁_{k, s} (t, \vec{z} (t)) 𝐁_{l, s} (t, \vec{z} (t)))] d t

Third and Fourth terms

So with

(\int_{t_{i}}^{t_{i + 1}} {\vec{A}}_{k} (t, \vec{z} (t)) d t) (\int_{t_{i}}^{t_{i + 1}} 𝐁_{l} (t, \vec{z} (t)) d {\vec{W}}_{t})

inside the summation it looks like

\lim_{Δ t_{max} \to 0} \sum_{i} (\int_{t_{i}}^{t_{i + 1}} {\vec{A}}_{k} (t, \vec{z} (t)) d t) (\int_{t_{i}}^{t_{i + 1}} 𝐁_{l} (t, \vec{z} (t)) d {\vec{W}}_{t})

If we use Cauchy Schwarz inequality where $\sum_{i} | A_{i} B_{i} | \leq \sqrt{\sum_{i} A_{i}^{2}} \sqrt{\sum_{i} B_{i}^{2}}$ then

\lim_{Δ t_{max} \to 0} | \sum_{i} (\int_{t_{i}}^{t_{i + 1}} {\vec{A}}_{k} (t, \vec{z} (t)) d t) (\int_{t_{i}}^{t_{i + 1}} 𝐁_{l} (t, \vec{z} (t)) d {\vec{W}}_{t}) | \leq \lim_{Δ t_{max} \to 0} \sum_{i} (\int_{t_{i}}^{t_{i + 1}} {\vec{A}}_{k} (t, \vec{z} (t)) d t)^{2} (\int_{t_{i}}^{t_{i + 1}} 𝐁_{l} (t, \vec{z} (t)) d {\vec{W}}_{t})^{2}

And we've seen that the first term goes to zero, and the second term is finite so

= 0

and we don't have to worry about it at all

Taylor series substituted

So substituting what we have, where only 1 of the terms isn't zero gives

\lim_{Δ t_{max} \to 0} 𝔼 [Δ_{i} {\vec{z}}_{k} Δ_{i} {\vec{z}}_{l}] = \sum_{s}^{U} \int_{t_{i}}^{t_{i + 1}} 𝔼 [(𝐁_{k, s} (t, \vec{z} (t)) 𝐁_{l, s} (t, \vec{z} (t)))] d t

Looking back at the Taylor series if we take the expected value

𝔼 [f (t + k, \vec{z} (t + k)) - f (t, \vec{z} (t))] = 𝔼 [\lim_{Δ t_{max} \to 0} \sum_{i} (\frac{\partial f}{\partial t} Δ t + \sum_{k}^{U} \frac{\partial f}{\partial {\vec{z}}_{k}} Δ_{i} {\vec{z}}_{k} + \frac{1}{2} \frac{\partial^{2} f}{\partial^{2} t} Δ t^{2} + \frac{1}{2} \sum_{k}^{U} \sum_{l}^{U} \frac{\partial^{2} f}{\partial {\vec{z}}_{k} \partial {\vec{z}}_{l}} Δ_{i} {\vec{z}}_{k} Δ_{i} {\vec{z}}_{l} + \sum_{k}^{U} \frac{\partial^{2} f}{\partial t \partial {\vec{z}}_{k}} Δ t Δ_{i} {\vec{z}}_{k} + \dots)]

and the higher order terms other than these ones go to zero, pretty easy to prove similar to others but will skip it for now

= 𝔼 [\lim_{Δ t_{max} \to 0} \sum_{i} (\frac{\partial f}{\partial t} Δ t + \sum_{k}^{U} \frac{\partial f}{\partial {\vec{z}}_{k}} Δ_{i} {\vec{z}}_{k} + \frac{1}{2} \sum_{k}^{U} \sum_{l}^{U} \frac{\partial^{2} f}{\partial {\vec{z}}_{k} \partial {\vec{z}}_{l}} Δ_{i} {\vec{z}}_{k} Δ_{i} {\vec{z}}_{l})]

= \lim_{Δ t_{max} \to 0} \sum_{i} (\frac{\partial f}{\partial t} Δ t + \sum_{k}^{U} 𝔼 [\frac{\partial f}{\partial {\vec{z}}_{k}} Δ_{i} {\vec{z}}_{k}] + \frac{1}{2} \sum_{k}^{U} \sum_{l}^{U} 𝔼 [\frac{\partial^{2} f}{\partial {\vec{z}}_{k} \partial {\vec{z}}_{l}} Δ_{i} {\vec{z}}_{k} Δ_{i} {\vec{z}}_{l}])

The first part is just a Reinmann integral

= \int_{t}^{t + k} \frac{\partial f}{\partial t} d t \lim_{Δ t_{max} \to 0} \sum_{i} (\sum_{k}^{U} 𝔼 [\frac{\partial f}{\partial {\vec{z}}_{k}} Δ_{i} {\vec{z}}_{k}] + \frac{1}{2} \sum_{k}^{U} \sum_{l}^{U} 𝔼 [\frac{\partial^{2} f}{\partial {\vec{z}}_{k} \partial {\vec{z}}_{l}} Δ_{i} {\vec{z}}_{k} Δ_{i} {\vec{z}}_{l}])

Substitute

= \int_{t}^{t + k} \frac{\partial f}{\partial t} d t \lim_{Δ t_{max} \to 0} \sum_{i} (\sum_{k}^{U} 𝔼 [\frac{\partial f}{\partial {\vec{z}}_{k}} (\int_{t_{i}}^{t_{i + 1}} \vec{A} (t, \vec{z} (t))_{k} d t + \int_{t_{i}}^{t_{i + 1}} 𝐁 (t, \vec{z} (t))_{k} d {\vec{W}}_{t})] + \frac{1}{2} \sum_{k}^{U} \sum_{l}^{U} 𝔼 [\frac{\partial^{2} f}{\partial {\vec{z}}_{k} \partial {\vec{z}}_{l}} Δ_{i} {\vec{z}}_{k} Δ_{i} {\vec{z}}_{l}])

and adding the integrals just concatenates them so

= \int_{t}^{t + k} \frac{\partial f}{\partial t} d t + \sum_{k}^{U} (\int_{t}^{t + k} 𝔼 [\frac{\partial f}{\partial {\vec{z}}_{k}} \vec{A} (t, \vec{z} (t))_{k}] d t + \int_{t_{i}}^{t_{i + 1}} 𝔼 [𝐁 (t, \vec{z} (t))_{k} d {\vec{W}}_{t}]) + \lim_{Δ t_{max} \to 0} \sum_{i} (\frac{1}{2} \sum_{k}^{U} \sum_{l}^{U} 𝔼 [\frac{\partial^{2} f}{\partial {\vec{z}}_{k} \partial {\vec{z}}_{l}} Δ_{i} {\vec{z}}_{k} Δ_{i} {\vec{z}}_{l}])

The expected value of just the drift term is zero so

= \int_{t}^{t + k} \frac{\partial f}{\partial t} d t + \sum_{k}^{U} \int_{t}^{t + k} 𝔼 [\frac{\partial f}{\partial {\vec{z}}_{k}} \vec{A} (t, \vec{z} (t))_{k}] d t + \lim_{Δ t_{max} \to 0} \sum_{i} (\frac{1}{2} \sum_{k}^{U} \sum_{l}^{U} 𝔼 [\frac{\partial^{2} f}{\partial {\vec{z}}_{k} \partial {\vec{z}}_{l}} Δ_{i} {\vec{z}}_{k} Δ_{i} {\vec{z}}_{l}])

Then if we substitute again

= \int_{t}^{t + k} \frac{\partial f}{\partial t} d t + \sum_{k}^{U} \int_{t}^{t + k} 𝔼 [\frac{\partial f}{\partial {\vec{z}}_{k}} \vec{A} (t, \vec{z} (t))_{k}] d t + \lim_{Δ t_{max} \to 0} \sum_{i} (\frac{1}{2} \sum_{k}^{U} \sum_{l}^{U} 𝔼 [\frac{\partial^{2} f}{\partial {\vec{z}}_{k} \partial {\vec{z}}_{l}} (\sum_{s}^{U} \int_{t_{i}}^{t_{i + 1}} (𝐁_{k, s} (t, \vec{z} (t)) 𝐁_{l, s} (t, \vec{z} (t))) d t)])

and the integrals just concatenate again so the limit doesn't matter

= \int_{t}^{t + k} \frac{\partial f}{\partial t} d t + \sum_{k}^{U} \int_{t}^{t + k} 𝔼 [\frac{\partial f}{\partial {\vec{z}}_{k}} \vec{A} (t, \vec{z} (t))_{k}] d t + \frac{1}{2} \sum_{k}^{U} \sum_{l}^{U} \int_{t}^{t + k} 𝔼 [\frac{\partial^{2} f}{\partial {\vec{z}}_{k} \partial {\vec{z}}_{l}} (\sum_{s}^{U} (𝐁_{k, s} (t, \vec{z} (t)) 𝐁_{l, s} (t, \vec{z} (t))))] d t

Then let

𝐃 = 𝐁^{T} 𝐁

so then

= \int_{t}^{t + k} \frac{\partial f}{\partial t} d t + \sum_{k}^{U} \int_{t}^{t + k} 𝔼 [\frac{\partial f}{\partial {\vec{z}}_{k}} \vec{A} (t, \vec{z} (t))_{k}] d t + \frac{1}{2} \sum_{k}^{U} \sum_{l}^{U} \int_{t}^{t + k} 𝔼 [\frac{\partial^{2} f}{\partial {\vec{z}}_{k} \partial {\vec{z}}_{l}} 𝐃_{k, l} (t, \vec{z} (t))] d t

So we get the formula expression

𝔼 [f (t + k, \vec{z} (t + k)) - f (t, \vec{z} (t))] = \int_{t}^{t + k} \frac{\partial f}{\partial t} d t + \sum_{k}^{U} \int_{t}^{t + k} 𝔼 [\frac{\partial f}{\partial {\vec{z}}_{k}} \vec{A} (t, \vec{z} (t))_{k}] d t + \frac{1}{2} \sum_{k}^{U} \sum_{l}^{U} \int_{t}^{t + k} 𝔼 [\frac{\partial^{2} f}{\partial {\vec{z}}_{k} \partial {\vec{z}}_{l}} 𝐃_{k, l} (t, \vec{z} (t))] d t

And since its deterministic you can do

\frac{d}{d t} 𝔼 [f (t, \vec{z} (t))] = 𝔼 [\frac{\partial f}{\partial t} + \sum_{k}^{U} \frac{\partial f}{\partial {\vec{z}}_{k}} \vec{A} (t, \vec{z} (t))_{k} + \frac{1}{2} \sum_{k}^{U} \sum_{l}^{U} \frac{\partial^{2} f}{\partial {\vec{z}}_{k} \partial {\vec{z}}_{l}} 𝐃_{k, l} (t, \vec{z} (t))]

final part

We also know that

𝔼 [f (\vec{z} (t))] = \int_{V} f (\vec{z}) p (t, \vec{z}) d \vec{z}

so taking the derivative of both sides

\frac{d}{d t} 𝔼 [f (\vec{z} (t))] = \int_{V} f (\vec{z}) \frac{d p (t, \vec{z})}{d t} d \vec{z}

and substituting

\int_{V} (\sum_{k}^{U} \frac{\partial f}{\partial {\vec{z}}_{k}} \vec{A} (t, \vec{z} (t))_{k} + \frac{1}{2} \sum_{k}^{U} \sum_{l}^{U} \frac{\partial^{2} f}{\partial {\vec{z}}_{k} \partial {\vec{z}}_{l}} 𝐃_{k, l} (t, \vec{z} (t))) p (t, \vec{z}) d \vec{z} = \int_{V} f (\vec{z}) \frac{d p (t, \vec{z})}{d t} d \vec{z}

Divergence theorem

Split the integral up

= \int_{V} \sum_{k}^{U} \frac{\partial f}{\partial {\vec{z}}_{k}} \vec{A} (t, \vec{z} (t))_{k} p (t, \vec{z}) d \vec{z} + \int_{V} \frac{1}{2} \sum_{k}^{U} \sum_{l}^{U} \frac{\partial^{2} f}{\partial {\vec{z}}_{k} \partial {\vec{z}}_{l}} 𝐃_{k, l} (t, \vec{z} (t)) p (t, \vec{z}) d \vec{z}

first part

focusing on the first part

= \int_{V} \sum_{k}^{U} \frac{\partial f}{\partial {\vec{z}}_{k}} \vec{A} (t, \vec{z} (t))_{k} p (t, \vec{z}) d \vec{z}

We can write it like this

= \int_{V} \sum_{k}^{U} ((\nabla_{z} f) \cdot (\vec{A} p)) d \vec{z}

and remember that a scalar and a vector

\nabla \cdot (a \vec{b}) = (\nabla a \cdot \vec{b}) + (a \nabla \vec{b})

And with the divercence theorem

\int_{V} \nabla_{z} \cdot \vec{c} (\vec{z}) d \vec{z} = \oint_{Ω (V)} (\vec{c} (\vec{z}) \cdot \vec{n}) d \vec{z}

And if its two of then dotted then

\int_{V} (\nabla_{z} a \cdot \vec{b}) d \vec{z} + \int_{V} (a \nabla_{z} \cdot \vec{b}) d \vec{z} = \oint_{Ω (V)} ((a \vec{b}) \cdot \vec{n}) d \vec{z}

= \oint_{Ω (V)} ((p \vec{A} f) \cdot \vec{n}) d \vec{z} - \int_{V} f \nabla_{z} \cdot (p \vec{A}) d \vec{z}

And if we assume the probability goes to zero at the edge (or in the limit to infinity) then the boundary condition dissapears

= - \int_{V} f \nabla_{z} \cdot (p \vec{A}) d \vec{z}

second part

TODO, just the same thing but twice

result

So we have

\int_{V} \frac{\partial p}{\partial t} f d \vec{z} = \int_{V} (- \sum_{k}^{U} \frac{\partial}{\partial {\vec{z}}_{k}} (p \vec{A} (t, \vec{z} (t))_{k}) + \frac{1}{2} \sum_{k}^{U} \sum_{l}^{U} \frac{\partial^{2}}{\partial {\vec{z}}_{k} \partial {\vec{z}}_{l}} (p 𝐃_{k, l} (t, \vec{z} (t)))) f (t, \vec{z}) d \vec{z}

and since it works for arbitrary function f it must be that

\frac{\partial p}{\partial t} = - \sum_{k}^{U} \frac{\partial}{\partial {\vec{z}}_{k}} (p \vec{A} (t, \vec{z} (t))_{k}) + \frac{1}{2} \sum_{k}^{U} \sum_{l}^{U} \frac{\partial^{2}}{\partial {\vec{z}}_{k} \partial {\vec{z}}_{l}} (p 𝐃_{k, l} (t, \vec{z} (t)))