Converting an Ito integral into a Backward Ito Integral

Question

In "Reverse Time Diffusion Equation Models", Brian D.O. Anderson begins with a multidimensional Ito SDE $$ dx_t = f(x_t,t) dt + g(x_t, t) dw_t,$$ with some initial condition $p(x_0, t_0)$. So $f: \mathbb{R}^n \times [0, \infty)$ and $g: \mathbb{R}^{n } \times [0, \infty) \rightarrow \mathbb{R}^{n \times m}$, and $w$ is the $m$-dimensional Wiener process.

The solution $x$ is assumed to have a probability distribution $p(x_t, t)$. So then Anderson defines an auxiliary process

$$ d \bar w_t^k = dw^k + \frac{1}{p(x_t, t)} \sum_j \frac{\partial}{\partial x_j} [p(x_t, t) g^{ik}(x_t, t)] dt$$

with initial condition $\bar w_0 = 0$.

Using the above equation, $dw_t$ can be put in terms of $d \bar w_t$ so that we can write $$(*) \ dx_t = \hat f(x_t, t) dt + g(x_t, t) d\bar{w_t},$$ where $\hat f^i(x_t, t) = f^i(x_t, t) - \frac{1}{p(x_t, t)} \sum_k g^{ik}g(x_t, t) \sum_j \frac{\partial}{\partial x_j}[p(x_t, t) g^{ik}(x_t, t)]$.

Now Anderson says something I find confusing.

In order to convert this equation $(*)$ to one to be understood as using backward Ito integration, we must make an adjustment if $g$ depends explicitly on $x_t$ - double in fact that required to convert this equation to a Stratonovich equation, or double that required to obtain a symmetrized integral with respect to the usual Wiener martingale, and by extension, with respect to $\bar w_t$, regarded as a semimartingale.

He proceeds to rewrite $(*)$ as $$dx_t = \bar f(x_t, t) dt + g(x_t, t) d\bar w_t,$$ where $\bar f$ is a modification of $\hat f$, and this is described as a "backward Ito integral." I take that to mean that, if we substitute $t$ with $-t$, this equation a standard Ito SDE on the reverse time filtration generated by $(x, \bar w)$, and that the solution matches in distribution, to the original $x$.

1. Is my interpretation correct?
2. What are the rules surrounding the modification $\hat f \rightarrow \bar f$? (What did he mean by the phrase "double that" in the quote above?)
3. Elsewhere, Anderson suggests that $\bar w$ is actually a Wiener process on the reverse filtration. Conditions (1,2,3) from his paper would almost suffice to show this (and also the reverse process would be a Brownian bridge?), but don't we additionally need that $\bar w$ has continuous sample paths?

Answer 1

I have deciphered what Anderson meant so I can answer my own questions.

The Answers to my Questions

1. The interpretation is correct (see the derivations below). Additionally the resulting process $x$ as a function of $\bar w$ should not only match the forward one in distribution, but also along sample paths!

2. See the explanations below.

3. If the preconditions of Anderson's theorem are satsified, then $\bar w$ is a forward time Wiener process since it has continuous sample paths and Brownian incremements. The reverse time $\bar w$ is a Brownian bridge.

Forward to Reverse Ito SDE (1D case)

First consider the one dimensional case of a semimartingale integrator $X$ and a stochastic integrand $Y$. The Ito integral is given $$\int_0^T Y_s dX_s = \lim_{\Delta t \rightarrow 0} \sum_{i=0}^{T/\Delta t} X_{i \Delta t/T} (Y_{(i+1)\Delta t/T} - Y_{i \Delta t/T})$$ $$= \lim_{\Delta t \rightarrow 0} \sum_{i=0}^{T/\Delta t} (X_{i \Delta t/T} - X_{(i+1)\Delta t/T }+ X_{(i+1)\Delta t/T}) (Y_{(i+1)\Delta t/T} - Y_{i \Delta t/T})$$ $$= \lim_{\Delta t \rightarrow 0} \sum_{i=0}^{T/\Delta t} (X_{i \Delta t/T} - X_{(i+1)\Delta t/T }) (Y_{(i+1)\Delta t/T} - Y_{i \Delta t/T})+ \lim_{\Delta t \rightarrow 0} \sum_{i=0}^{T/\Delta t} X_{(i+1)\Delta t/T} (Y_{(i+1)\Delta t/T} - Y_{i \Delta t/T})$$ $$= \langle X, Y \rangle_T + \int_T^0 X_s dY_s.$$

The stochastic integral term on the last line is taken in reverse time. In the conversion, one incurs a correction term due to the covariation $\langle X, Y \rangle$ process. These manipulations can be seen as a two step process, converting to the Stratonovich integral which incurs a $\frac {1}{2} \langle X, Y \rangle$ correction term, and then converting back -- in reverse time -- which incurs another $\frac{1}{2}\langle X, Y \rangle$ correction term.

Forward to Reverse Ito SDE (Multidimensional Case) Now I will rederive Anderson's reverse time formula.

Let $x$ be a multidimensional stochastic process defined by $$dx_t = f(x_t, t) dt + g(x_t, t) dw_t $$ and suppose that the joint pdf $p(x_t, t)$ exists for all $t \ge 0$. Anderson's SDE reads $$d \bar w_t^k = dw_t^k + \frac{1}{p(x_t, t)} \sum_{j} \frac{\partial}{\partial x_j}[p(x_t, t) g^{jk}(x_t, t)] dt.$$

We may rewrite this as $d \bar w_t^k - \frac{1}{p(x_t, t)} \sum_{j} \frac{\partial}{\partial x_j}[p(x_t, t) g^{jk}(x_t, t)] dt = dw_t^k$ and substituting into the SDE for $x$, we have

\begin{aligned} dx_t^l &= f^l(x_t, t) dt + \sum_k g^{lk}(x_t, t)d w_t^k \\ &= f^l(x_t, t) dt + \sum_k g^{lk}(x_t, t)\left(d \bar w_t^k - \frac{1}{p(x_t, t)} \sum_{j} \frac{\partial}{\partial x_j}[p(x_t, t) g^{jk}(x_t, t)] dt\right) \\ &= \left(f^l(x_t, t) - \sum_k \frac{g^{lk}(x_t, t)}{p(x_t, t)} \sum_{j} \frac{\partial}{\partial x_j}[p(x_t, t) g^{jk}(x_t, t)]\right) dt + \sum_k g^{lk}(x_t, t) d \bar w_t^k \\ &= \hat f(x_t, t) dt + g(x_t, t) d \bar w_t . \\ \end{aligned}

Write the diffusion term as $g(x_t, t) d\bar w_t = G_t d\bar w_t$, where $G_t$ is a matrix of stochastic processes. We have

\begin{aligned} \int_0^T G_t d\bar w_t &= \lim_{\Delta t \rightarrow 0} \sum_{i=0}^{T/\Delta t} G_{i T/\Delta t} (\bar w_{(i+1) T/\Delta t} - \bar w_{i T / \Delta t}) \\ &= \lim_{\Delta t \rightarrow 0} \sum_{i=0}^{T/\Delta t} (G_{i T/\Delta t} + G_{(i+1) T/\Delta t} - G_{(i+1) T/\Delta t}) (\bar w_{(i+1) T/\Delta t} - \bar w_{i T / \Delta t})\\ &= \lim_{\Delta t \rightarrow 0} \sum_{i=0}^{T/\Delta t} (G_{i T/\Delta t} - G_{(i+1) T/\Delta t}) (\bar w_{(i+1) T/\Delta t} - \bar w_{i T / \Delta t}) + \sum_{i=0}^{T/\Delta t} G_{(i+1) T/\Delta t} (\bar w_{(i+1) T/\Delta t} - \bar w_{i T / \Delta t}) \\ &= \lim_{\Delta t \rightarrow 0} \sum_{i=0}^{T/\Delta t} (G_{i T/\Delta t} - G_{(i+1) T/\Delta t}) (\bar w_{(i+1) T/\Delta t} - \bar w_{i T / \Delta t}) + \int_T^0 G_t d\bar w_t \\ \end{aligned}

Note the last integral is taken in reverse time, and I will denote the reverse time differential as $(d\bar w_t)^R$.

Define the correction term $C := \lim_{\Delta t \rightarrow 0} \sum_{i=0}^{T/\Delta t} (G_{i T/\Delta t} - G_{(i+1) T/\Delta t}) (\bar w_{(i+1) T/\Delta t} - \bar w_{i T / \Delta t})$. In what follows, I will use $A^l$ and $B^l$ to be a family (indexed by $l$) of nuisance functions arising from the multidimensional Ito formula, whose time integral will be killed when the covariation (e.g. $\langle \int A^l(X_t, t) dt, \cdot \rangle$) is taken with any other process.

\begin{aligned} C^k &= \lim_{\Delta t \rightarrow 0} \sum_{i=0}^{T/\Delta t} \sum_l (G_{i T/\Delta t} - G_{(i+1) T/\Delta t})^{kl} (\bar w_{(i+1) T/\Delta t} - \bar w_{i T / \Delta t})^l \\ &= -\sum_l \lim_{\Delta t \rightarrow 0} \sum_{i=0}^{T/\Delta t} (G_{(i+1) T/\Delta t} - G_{i T/\Delta t})^{kl} (\bar w_{(i+1) T/\Delta t} - \bar w_{i T / \Delta t})^l \\ &= -\sum_l \int_0^T dG_t^{kl} d\bar w_t^l \\ &= -\sum_l \int_0^T d(g^{kl}(x_t, t)) d\bar w_t^l \\ &= -\sum_l \int_0^T d(g^{kl}(x_t, t))(A^l(x_t, t) dt + dw_t^l) \\ &= -\sum_{l} \int_0^T (B^l(x_t, t)dt + \nabla g^{kl}(x_t, t)^T g(x_t, t) dw_t)(A^l(x_t, t) dt + dw_t^l) \\ &= -\sum_{lm} \int_0^T \frac{\partial g^{kl}}{\partial x_m} (x_t, t) g^{mn}(x_t, t) dw_t^n)(dw_t^l) \\ &= -\sum_{lm} \int_0^T \frac{\partial g^{kl}}{\partial x_m}(x_t, t)^m g^{ml}(x_t, t) dt \\ \end{aligned}

Then $dC_t^k = -\sum_{lm} \frac{\partial g^{kl}}{\partial x_m} (x_t, t)^m g(x_t, t)^{ml} dt$, so that

\begin{aligned} dx_t^k &= dC_t^k + \hat f^k(x_t, t)dt + \sum_lg^{kl}(x_t, t) (d\bar{w}_t^l)^R \\ &= \left( \hat f(x_t, t) -\sum_{lm} \frac{\partial g^{kl}}{\partial x_m} (x_t, t) g^{ml}(x_t, t) \right)dt + \sum_l g^{kl}(x_t, t) (d \bar w_t^l)^R \\ \end{aligned}

Answer 2

Too long for a comment. I had a quick look at Anderson's paper and find it indeed very confusing. I suggest to read instead Haussmann and Pardoux's paper.

So here is a simple example to illustrate idea how the time reversal works: $$ dx_t = dw_t $$ with initial condition $x_0=0$. The solution is, naturally, $x_t = w_t$ with density $p(t,x) = \frac{e^{-\frac{x^2}{2t}}}{\sqrt{2\pi t}}.$ Since $\frac{1}{p(t,x)}\cdot \frac{\partial}{\partial x} p(t,x) = - \frac{x}{t}$, we have $$\overline w_t = w_t - \int_{0}^t \frac{1}{p(s,x_s)}\cdot \frac{\partial}{\partial x} p(s, x_s)ds = w_t - \int_0^t \frac{w_s}{s}ds.$$ Now let us reverse the time, which means replacing $t$ by $T-t$: $$\widetilde w_t = \overline{w}_{T-t} = w_{T-t} - \int_0^{T-t} \frac{w_s}{s}ds = w_{T-t} - \int_{0}^T \frac{w_s}s ds + \int_{T-t}^T \frac{w_s}{s}ds\\ =w_{T-t} - \xi + \int_0^t \frac{w_{T-t}}{T-t}dt = b_t - \xi +\int_0^t \frac{b_t}{T-t}dt,$$ where $\xi = \int_0^T \frac{w_s}{s}ds$ and $b_t = w_{T-t}$ is the time-reversed Wiener process. Now differentiate and rewrite it a bit: $$ db_t = -\frac{b_t}{T-t} + d\widetilde w_t. $$
Looks familiar? Yes, this looks like the SDE solved by a Brownian bridge. But $b_t$ is a Brownian bridge started at $w_T$! So $\widetilde w_t$ (or $\widetilde w_t -\widetilde w_0$ if we want it to start from zero) is a Wiener process. Perhaps, this is what is meant by $\overline w_t$ being a Wiener process in the reverse time?

Anyway, I suggest you to read the article by Haussmann and Pardoux, to me it seems a lot clearer.