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It is known how to calculate stochastic integrals of the kind, e.g., $\int_0^T W_t \, dW_t$ or $\int_0^T W_t^2 \,dW_t$, where $W_t$ is the Wiener process, aka Brownian motion.

Question: How about the same but with white noise? Let the continuous time white noise $\dot{W}_t$ be the formal derivative of the Wiener process, where $d{W}_t=\dot{W}_t \, dt$ (see the note below). Is it possible to give a meaning to random variables $Y_\alpha(T)$ of the kind

$$ Y_\alpha(T) = \int_0^T (\dot{W}_t)^\alpha \, dW_t = \int_0^T (\dot{W}_t)^{\alpha+1} \, dt $$

for a fixed integer $\alpha>0$ and $T>0$? The trivial case $\alpha =0$ is $Y_0(T)=W_T$, given that $W_0 = 0$.

Note: By continuous time white noise, I mean the formal derivative of the Wiener process, $\dot{W}_t=dW_t/dt$. Its autocovariance is distributed according to a Dirac delta, $\langle \dot{W}_t \dot{W}_s \rangle = \delta(t-s)$, while for the Wiener process, we have $\langle {W}_t {W}_s\rangle = \min(t,s)$. Complementary approaches for the white noise may be found in this question, this, this, this or this. For reference, a discussion of integration with respect to the Brownian motion and white noise can be found here.

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    $\begingroup$ I believe in order to make sense of this, one would have to define what is meant by the product of (Schwarz) distributions, which in general is a nontrivial task. One case which we can work out formally is $\alpha = 1$ if we are to assume that $(dW_t)^2 = dt$. In this case $Y_1(T) = T$ (here we are writing $dW_t = \dot{W}_t$). Moreover since the product differentials of finite variation processes vanish (again, this is formal), this would imply that $Y_\alpha(T) \equiv 0$ for any $\alpha \geq 2$, and probably also for $\alpha > 1$. $\endgroup$ Commented Jul 3, 2023 at 15:04
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    $\begingroup$ The white noise is not a well defined random variable, but rather a stochastic distribution. The integral you are trying to compute, i.e. the Ito integral of a power of the white noise is ill defined. The only thing that comes to my mind is, that if you instead you use the Wick product (powers) then you can write $Y_\alpha(T) = \int_0^T (\dot{W}_t)^{\diamond\alpha} \, dW_t = \int_0^T (\dot{W}_t)^{\diamond(\alpha+1)} \, dt$ (wher the first expression is a bit an abuse of notation) $\endgroup$ Commented Jul 12, 2023 at 11:51
  • $\begingroup$ @Chaos thank you, If you had written your comment in response form, I would have accepted it. The idea of using Wick's product is interesting. For reference: en.wikipedia.org/wiki/Wick_product $\endgroup$ Commented Jul 12, 2023 at 20:51

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There is a nice way to interpret such stochastic integrals by using scale functions and time changes of Brownian Motion. The idea is as follows:

Suppose we have some "usual SDE", i.e. one where $b$ and $\sigma$ satisfy the usual conditions, so that there exists a solution to $$dX_t=b(X_t)dt+\sigma(X_t)dB_t$$ One can then show that a solution to the SDE given above can be represented as $$X_t=s^{-1}(B_{\gamma_t})$$ where $s$ is the scale function and $\gamma_t$ is a time change. The crucial part here is the scale function, which we can then generalize to "messier" coefficients $b,\sigma$:

The scale function of the SDE given above is defined as $$s(x)=\int_0^x \exp(-F(y))dy$$ where $$F(y)=\int_0^y \frac{2b(s)}{\sigma^2(s)}ds$$ (This definition may seem strange to you, but there is a nice derivation of why one would define a function like this. I refer you to these lecture notes, chapter 1, for a nice introduction)

Now the key insight is that if we can express any solution to the SDE as $s^{-1}(B_{\gamma_t})$, then why not take exactly this process as the solution in the first place, even if the coefficients of the SDE may be ill-behaved? Let's consider the following example:

Consider the (formal) SDE $$dX_t=W_t'dt+dB_t$$ where $W_t$ and $B_t$ are two Brownian Motions. Of course, $W'$ may be ill-defined, but using the scale functions as defined before we may compute $$s(x)=\int_0^x\exp\bigg(\int_0^y -2W'_tdt\bigg)dy = \int_0^x \exp(-2W_y)dy$$ where we assume that any interpretation of $W'$ would obey the fundamental theorem of calculus, i.e. $\int_0^t W_s'ds=W_t$.

Now note that even though $W'$ may be ill-defined, the scale function $s$ definitely is well-defined! It remains to construct the time change $\gamma_t$, but again i'd refer you to the lecture notes linked above to see the construction.

Further references for scale functions and time changes are Foundations of Modern Probability Chapter 33 or Continuous Martingales and Brownian Motion Chapter 7.

I know that this approach does not perfectly address your question, but perhaps it is a good starting point. You may find a similar approach for general $(W')^n$.

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