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Suppose that $\mathbf{s}(t)\in S$ denotes the spatial location of a process at time $t$. Further, let $\mathbf{x}(\mathbf{s}(t))$ denote covariates at the location $\mathbf{s}(t)$. My goal is to write an SDE for the process that can be discretized in the following way: $$\mathbf{s}(t) - \mathbf{s}(t-dt) = \mathbf{a}'(\mathbf{x}(\mathbf{s}(t)) - \mathbf{x}(\mathbf{s}(t-dt))) + \text{noise},$$ where $\mathbf{a}$ are advection coefficients constant with respect to time. How may I formulate the SDE such that I can get the desired discretization? Right now, I'm considering $$\frac{d\mathbf{s}(t)}{dt} = \frac{\mathbf{a}'d\mathbf{x}(\mathbf{s}(t))}{dt} + \frac{\text{noise}}{dt},$$ but I'm not sure that I'm using the correct approach/order when deriving the advection component to get the desired approach.

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    $\begingroup$ This might also fit better in physics.stackexchange.com since I think you are looking for a physics-based derivation. $\endgroup$ Feb 22, 2023 at 20:32
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    $\begingroup$ ignoring the noise for a second which actually forces $s(t)$ to be non-differentiable, by chain rule you get $$d(x(s(t)))/dt=x'(s(t))ds/dt.$$ $\endgroup$ Feb 22, 2023 at 20:34
  • $\begingroup$ @ThomasKojar Thank you for your suggestion about the physics stack exchange. I'll give that a go. Just to clarify on your second comment: $\mathbf{x}(\mathbf{s}(t))$ are covariates of the process at time $t$, so we wouldn't need to apply chain rule, right? I suppose the notation of $\mathbf{x}(\mathbf{s}(t))$ is a bit confusing, but I'm treating $\mathbf{y}(t)$ as an index for the covariates $\mathbf{x}$. $\endgroup$
    – Ron Snow
    Feb 22, 2023 at 21:55
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    $\begingroup$ I did the chain rule because we then get the sde $$\frac{ds(t)}{dt}=\frac{1}{1-a'x'(s(t))}dB_{t}.$$ But this is just a suggestion because I don't know what your particular situation is. $\endgroup$ Feb 22, 2023 at 22:08
  • $\begingroup$ I see. Thank you for all of your help, Thomas! $\endgroup$
    – Ron Snow
    Feb 22, 2023 at 22:29

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