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In the book "Stochastic Differential Equations" by Bernt Øksendal, the Kolmogorov's backward equation is stated as following:

Let $X_t$ be an Ito diffusion and $A$ is the generator of $X_t$. Define $u(t,x)=E^x(f(X_t))$ where $E^x$ is the expectation with $X_0=x$. Then we have $$ \begin{array}{c}{\frac{\partial u}{\partial t}=A u, \quad t>0, x \in {R}^{n}} \\ {u(0, x)=f(x) ; \quad x \in {R}^{n}}\end{array}. $$ My question is why this equation is called "Backward"? Sicne now we have the initial condition $u(x,0)=f(x)$.

Besides, it seems that the definition of Kolmogorov's backward equation is in another form on Wiki: Kolmogorov backward equations (diffusion).

Q-Y
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1 Answers1

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Although it is suggestive to think of forward and backward as the direction of time in the equation, this is slightly misleading. The idea to have in mind should be:

  • Backward $\leftrightarrow$ initial state $\leftrightarrow$ generator $A$ of the Markov process $\leftrightarrow$ Feynman-Kac.

  • Forward $\leftrightarrow$ terminal state $\leftrightarrow$ adjoint $A^*$ of the generator $\leftrightarrow$ Fokker-Planck.

Indeed, you can see that both Oksendal and Wikipedia the operator appearing is $A$. Indeed, if you let $T_{\cdot}$ be the semigroup generated by $A$, you see that:

$$ \mathbb{E}_{0,x}[f(X_t)] = T_t f(x), \qquad \mathbb{E}_{t,x}[f(X_T)] = T_{T-t}f(x)$$

so Wikipedia is just writing the equation for the time-reversed process presented by Oskendal. If you take the probability density of $X_t$ started in $x$ to be $p_t(x,y)$, the above tells you:

$$ \partial_t p_t(x,y) = A_x p_t(x, y), \qquad p_0(x,y)= \delta(x {-} y)$$

Instead, the forward equation tells you that:

$$\partial_t p_t(x,y) = A^*_y p_t(x,y), \qquad p_0(x,y) = \delta(x{-}y).$$

Kore-N
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  • Would you mind explaining a bit more about why we have $ \partial_{t} p_{t}(x, y)=A_{x} p_{t}(x, y)$? And why do you say that Forward ↔ terminal condition? It seems that the forward equation also needs initial condition. – Q-Y Apr 02 '19 at 01:04
  • When I write $A_x$ I mean the operator applied to $p_t(x,y)$ as a function of $x$. For example, if $A$ is the laplacian, this would mean second derivative w.r.t. $x$. If you look at your equation, it's essentially the same as the first one I wrote integrated against $f$, that is for $\int p_t(x,y)f(y) dy$. When I say initial and terminal condition I do not mean of the PDE (I leave it to you to figure this one out), but of the SDE. I mean: $x$ is the initial state and $y$ is the terminal state. – Kore-N Apr 02 '19 at 08:48
  • Okay, I see. As for the $A_x$, I mean do you say that $ A_{y}^{*} p_{t}(x, y)=\partial_{t} p_{t}(x, y)=A_{x} p_{t}(x, y) $? – Q-Y Apr 02 '19 at 18:35
  • Yes, that is what I mean – Kore-N Apr 02 '19 at 19:07
  • But If $A$ is the generator, we have $\langle A_{y}^{} p_{t}(x, y),f\rangle=\langle p_t(x,y),Af\rangle$, why we have $A_{y}^{} p_{t}(x, y)=A_{x} p_{t}(x, y)$? – Q-Y Apr 02 '19 at 19:12
  • I am not claiming a proof of that. We have it, because it's a theorem (actually 2 of the most important ones, namely the Feynman-Kac and the Fokker-Planck formula). – Kore-N Apr 02 '19 at 19:15
  • I got it. Thank you very much! – Q-Y Apr 02 '19 at 19:49