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Notation: Here $\mathcal Y_t$ denotes the natural filtration of the process $Y_t$, and $\{\cdot\}$ denotes the fractional part of a real number.

This question concerns detecting the presence (or otherwise) of a brief, periodic impulse signal in the presence of noise.

Let $X$ be a uniform random variable on $[0, 1]$, and $0 < \delta < 1$ a real number. Consider the solution $Y$ to the SDE

$$dY_t = A \, \mathbf 1_{[0, \delta]} (\{t - X\}) \ dt + \sigma \,dW_t,$$

with $W$ a standard one dimensional Brownian motion independent of $X$, and $A, \sigma > 0$ constants. We assume the starting condition $Y_0 = 0$ a.s.

In this setup, $\delta$ is the duration of the impulse signal, $\sigma$ is the strength of the noise, and $A$ is the amplitude of the signal.

Question:

I believe that given enough time, we can ascertain with almost certainty whether or not the signal is present. More precisely, we should have $\mathbb E[X| \mathcal Y_t] \to X$ almost surely and in $L^1$ as $t \to \infty$, but this seems quite difficult to prove.

In any case, if this is true, I would like to quantify the effects of the parameters on the result. Can we estimate $\mathbb E[|\mathbb E[X | \mathcal Y_t] - X|]$ in terms of the parameters $\delta, A, \sigma$?

References: The ideas in the paper Seperating signal from noise (Liv, Peled, Peres) may be helpful.

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  • $\begingroup$ As soon as t > 2 $Y_t = A \delta + \sigma W_t$ so I don’t think you get any information about X after that time. $\endgroup$ Commented Aug 19, 2024 at 22:01
  • $\begingroup$ @mike I don’t think that holds, perhaps something in the problem set up is not clear? $\endgroup$ Commented Aug 20, 2024 at 1:00
  • $\begingroup$ I believe explicitly we have $$Y_t = A\delta ([t] - 1)_+ + \max(\{t\}, X) + \max(\{t\}, X + \delta - 1) - X + \sigma W_t.$$ Which is messier than I expected! Maybe there is a nicer way to write it. $\endgroup$ Commented Aug 20, 2024 at 1:18
  • $\begingroup$ Although I don’t think the explicit formula is of much help anyway to answer the problem. $\endgroup$ Commented Aug 20, 2024 at 1:19
  • $\begingroup$ … sorry typo, it is $$Y_t = A[\delta ([t] - 1)_+ +\max(\{t\}, X) + \max(\{t\}, X + \delta - 1) - X] + \sigma W_t.$$ $\endgroup$ Commented Aug 20, 2024 at 1:25

1 Answer 1

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Does the following make any sense?

For $n = 1, 2, \ldots$ and $t \in [0, 1)$ let $$ Z_n(t) = \frac{1}{n} \sum_{k = 0}^{n - 1} (Y_{k + t} - Y_k) . $$ Then $Z_n$ is measurable with respect to $\mathcal Y_{n + t}$ and $$ Z_n(t) = \frac{1}{n} \sum_{k = 0}^{n - 1} \biggl(\sigma W_{k + t} - \sigma W_k + A \int_0^t \mathbb 1_{[0,\delta]}(\{n + s - X\}) ds\biggr) . $$ Since the integral does not depend on $n$, in fact we have $$ Z_n(t) = \frac{\sigma}{n} \sum_{k = 0}^{n - 1} (W_{k + t} - W_k) + A \int_0^t \mathbb 1_{[0,\delta]}(\{s - X\}) ds . $$ The first term almost surely converges to zero (in fact uniformly with respect to $t \in [0, 1)$). Thus, $$ Z(t) = \lim_{n \to \infty} Z_n(t) = \int_0^t \mathbb 1_{[0,\delta]}(\{s - X\}) ds $$ is measurable with respect to $\mathcal Y_\infty$.

Knowing $Z(t)$, one easily reconstructs $X$ (e.g. the unique $t$ such that $Z'(t^-) = 0$ and $Z'(t^+) = 1$ is equal to $X$), so it turns out that $X$ is measurable with respect to $\mathcal Y_\infty$. In particular, $\mathbb E[X | \mathcal Y_t]$ is a martingale which converges to $\mathbb E[X | \mathcal Y_\infty] = X$ almost surely and in $L^1$.

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    $\begingroup$ Hm, I will check this in detail soon, interesting idea. $\endgroup$ Commented Oct 14, 2024 at 21:56
  • $\begingroup$ I assume you meant $W_{k+t} - W_k$ everywhere, so I am modifying it, but please let me know if I am mistaken! $\endgroup$ Commented Oct 14, 2024 at 23:03
  • $\begingroup$ The solution checks out IMO, very nicely done. $\endgroup$ Commented Oct 14, 2024 at 23:05
  • $\begingroup$ Wait a minute, there is still another question open, the one about the rate of convergence. Time permitting, I will get back to it later this week. Oh, and thanks for the edit, that is what I meant. $\endgroup$ Commented Oct 15, 2024 at 8:00
  • $\begingroup$ Yep, you could still answer that if you want. I will unaccept the answer so the problem will display as still open, but enjoy the +500 imaginary points as my show of gratitude! $\endgroup$ Commented Oct 15, 2024 at 8:55

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