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We understand SDEs like "$dX_t = b(t,X_t)dt + \sigma(t,X_t)dB_t$" for Brownian process $B$ to be formally the same as "$\frac{dX_t}{dt} = b(t,X_t) + \sigma(t,X_t)W_t$" where $W$ is a ``white noise" which is not a stochastic process in the usual sense of the term.

I wanted to know as to how far does this go : Like can I define a product of derivatives of two stochastic processes $X_t$ above and another $Y_t$ which is suppose given as, $dY_t = b_1(X_t,t)dt + \sigma_1(X_t,t)dB_t$?

  • Like is there meaning to writing something like,

$$ \frac{dX_t}{dt} \cdot \frac{dY_t}{dt} = (b(t,X_t) + \sigma(t,X_t)W_t )\cdot (b_1(t,X_t) + \sigma_1(t,X_t)W_t ) $$

  • Further, is there a way to define $\mathbb{E} \left [ \frac{dX_t}{dt} \cdot \frac{dY_t}{dt}\right ]$ from the above and compute it? If yes, can someone kindly show/reference it?
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    $\begingroup$ Perhaps I misunderstood the question: If you define $dX_t/dt$ in the above way, then of course you can multiply these random random variables. Usually the Itô integral is a fundamental object, and "$dX_t = U_t dt + V_t dB_t$" is understood as $X_t = X_0 + \int_0^t U_t dt + \int_0^t V_t dB_t$. $\endgroup$
    – Mateusz Kwaśnicki
    Sep 25, 2020 at 16:20
  • $\begingroup$ If you see my first bullet point, it seems unclear what to do with $W_t$. How does one deal with squares of $W_t$ and such? How can one compute the expectation in the second bullet point? $\endgroup$
    – gradstudent
    Sep 25, 2020 at 16:23
  • $\begingroup$ Ah, I was reading too quickly and I thought $W_t$ is the Brownian motion, not its "derivative"; sorry! I do not think there is a natural way to define what could $(W_t)^2$ mean. Any approach I can think of leads to infinities — so perhaps one can define $(W_t)^2 = \infty$? On the other hand, it is quite common to denote $dB_t \cdot dB_t = dt$ (which makes the integration by parts formula easier to remember), so it may seem reasonable to write $W_t \cdot W_t = \frac{dB_t}{dt} \cdot \frac{dB_t}{dt} = \frac{1}{dt}$. This looks fun, but I fail to see any use of such a definition. :-) $\endgroup$
    – Mateusz Kwaśnicki
    Sep 25, 2020 at 23:01
  • $\begingroup$ Has it never naturally occured to try to take a product of the time derivatives of 2 solutions of 2 SDEs? $\endgroup$
    – gradstudent
    Sep 26, 2020 at 2:54
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    $\begingroup$ The problem here is that actually these derivatives do not exist. You can define them only in a distributional sense. However, if $W_{t}$ is just a distribution then products or expressions like $W_{t}^{2}$ are ill-posed. $\endgroup$
    – Tobsn
    Jan 3, 2021 at 17:12

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The above formulas are only meant as integrals because Brownian motion is nowhere differentiable. As mentioned in the comments the process $X_{t}$ is not differentiable because it inherits the regularity of Brownian motion. For example, see here (https://math.stackexchange.com/questions/3962360/derivative-of-a-stochastic-integral-with-respect-to-limit-with-respect-to-inte)

If $H$ is adapted, right-continuous, and bounded then $$\lim_{t \rightarrow 0} \frac{1}{W_t} \int_0^t H_s dW_s = H_0$$ in probability.

and then using that Brownian motion is nowhere differentiable.

Having said that, there have been many approaches to make sense/consistently-interpret of products of distributions (eg. derivative of Brownian motion is white noise)

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