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I am trying to solve an backwards integrated Ito Stochastic Differential equation of the form $$ dx = a dt + bdW \tag{1}$$ where dW is the random variable associated with a Wiener process. However I am confused about the corrosponding PDE equation it represents. Different sources show differences by a minus sign in the time derivative. It comes down to

$$\partial f/\partial t = v \partial f/\partial x + D \partial^2f / \partial x ^2 \tag{2}$$

versus

$$-\partial f/\partial s = v \partial f/\partial x + D \partial^2f / \partial x ^2 \tag{3}$$

Which one here is the correct one? I noticed that one is labled as "s" but I assume that can be simply relabeled to "t" or am I wrong?

In summary: I am wondering if a=v or a=-v, when trying to represent the diffusion equation as a backwards integrated SDE.

I don't have very thorough understanding of the mathmatics behind this concept so please keep the answers simple. I would very much appreicate it thanks!

patrick7
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  • What is a backwards integrated SDE? Please study this post first. Considering the right functions and not making sign mistakes one and the same ordinary SDE (forget backwards SDEs) gives rise to a Kolmogorov forward and a backward PDE. – Kurt G. Jan 22 '24 at 17:31
  • Backwards integrated SDE (perhaps I am using the wrong language) is where you start at the final position, and integrate backwards in time until you reach inital time. Judging by your post and comment here, you are suggesting that there is no difference in the PDE representation when you integrate the SDE forwards in time versus backwards in time? Thanks! – patrick7 Jan 22 '24 at 17:47
  • There exist BSDEs (backward SDEs) but we do not need (as my post shows) them to understand KBE and KFE. Reading your last comment (and to make use of acronyms worse) you seem to mix up PDE and SDE. Advice: concentrate and be patient. – Kurt G. Jan 22 '24 at 17:50
  • I see, thanks for pointing me in a better direction. In light of what I've just said regarding "Backwards integrated SDE", then are you able to provide me with an answer to my post? If I want to solve the diffusion equation (2nd equation in my post), by integrating the SDE backwards in time, should by coefficent a be v or -v? – patrick7 Jan 22 '24 at 18:06
  • I will not write a formal answer before we have eliminated the term "Backwards SDE". Secondly: \tag{2} is the magic mathJax code that will number your equations. Please go a head and clarify exactly what that 2nd equation is. – Kurt G. Jan 22 '24 at 18:11
  • Before I eliminate the term "backwards integrated SDE", can you confirm that BSDE is the concept of integrating a SDE backwards in time starting from a terminal position, where the resulting probability distribution solution f(t=T, x=xf) is weighted by the inital condition f(t=0)? Again please correct me if I am using any wrong teminologies here. – patrick7 Jan 22 '24 at 18:19
  • *No.* We integrate the ordinary SDE forwards in time by which we get a stochastic process that starts at $X_0$ and has conditional expectations $p(t,x)=\mathbb E[u(X_T)|X_t=x]$ where $T>t$ is a final time (see linked post). That $p(t,x)$ solves the Kolmogorov backwards PDE. Had you patiently read the other post you would realise that I am only repeating stuff here. – Kurt G. Jan 22 '24 at 19:47
  • Exercise: is your equation (2) a Kolmogorov backwards PDE or is it (3)? – Kurt G. Jan 22 '24 at 19:50
  • I think we are on the wrong track here. My intent is to numerically intergrate an SDE backwards in time (e.g., from final time T and a final position xf) back to the inital time (t=0) with a known initial condition p(0,x). Then the resulting probability density at p(T,xf) is constructed by a weighted average of the inital condition f(0,x). Is this clear? From sources I've read, integrating an SDE forward in time versus backwards in time (my intent) corrosponds to different PDE equations. I would like to know, which PDE equation (2 or 3 in post) is the correct one by solving a SDE this way. – patrick7 Jan 22 '24 at 20:11

1 Answers1

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I found the answer to my question in https://doi.org/10.1007/s11214-017-0351-y

I was wrong on my first assumption where I state "t" and "s" are interchangable. They are not! They are related by $$s = T - t$$ where T is the final time (or "starting time" for integrating the SDE backwards in time).

As a result by change of variables, equations 2 and 3 are identical, which the answer to my question is: $$a = v$$ I hope this helps.

patrick7
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