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I'm reading a book on theoretical neuroscience [1], in which the following definitions are given:

$\rho(t)=\sum_{i=1}^n \delta(t-t_i)$

where $\delta$ is Dirac's delta and the $t_i$ are timestamps at which some event occurs. Noting $<z>$ the expected value or the random variable $z$ over independent trials, the function $r(t)$ is defined as

$r(t)=\frac{1}{\Delta t}\int_t^{t+\Delta t}<\rho(u)>du$

with $\Delta t \rightarrow 0$.

The author then states that for any well-behaved function $h(t)$, convolution between $h$ and $r$ is equivalent to convolution between $h$ and $<\rho>$, i.e.

\begin{equation} \int h(\tau) r(t-\tau) d\tau=\int h(\tau)<\rho(t-\tau)>d\tau. \end{equation}

This statement (equation 1.6 in the book) is made without any proof, and I assume it must therefore be rather trivial. However, I haven't been able to prove it [2]. Any help would greatly be appreciated.

[1]: Dayan and Abbott, Theoretical neuroscience, https://boulderschool.yale.edu/sites/default/files/files/DayanAbbott.pdf

[2]: All I've tried is to, quite unimaginatively, plug in the definition of $r$ into the left hand side, but then I get stuck doing random variable changes. I think that I haven't understood that equivalence well on an intuitive level.

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    $\begingroup$ The assumption that omitted proofs are trivial is generally false. $\endgroup$
    – Jazzmaniac
    Nov 27, 2023 at 12:05
  • $\begingroup$ @Jazzmaniac That's a good prior to have, and we can even refine it based on the size of the margins! $\endgroup$
    – Ash
    Nov 27, 2023 at 13:36

1 Answer 1

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It's unclear to me what $r(t)$ is supposed to be doing here. Let's consider

$$r(t) = \frac{1}{\delta}\int_{t}^{t+\delta} f(x) dx, \, \delta \rightarrow 0 $$

If we are integrating over a small enough interval and $f(x)$ is reasonably smooth, we can approximate it through it's Taylor Expansion around $x$, i.e.

$$f(x+\gamma) \approx f(x) + f'(x)\cdot \gamma= a \cdot \gamma+ b$$

The integral over a straight line is easy enough

$$\int_{x_1}^{x_2} (ax+b)dx = (x_2-x_1) \cdot \frac{1}{2}(f(x_2)+f(x_1))$$

Dividing by the integration interval simply yields the mean of the function values at the integration limits

$$\frac{1}{x_2-x_x1}\int_{x_1}^{x_2} (ax+b)dx = \frac{1}{2}(f(x_2)+f(x_1)) $$

If we plug this into the original integral we get

$$r(t) = \frac{1}{\delta}\int_{t}^{t+\delta} (f'(x) \cdot \gamma+ f(x))d\gamma = \\ \frac{1}{2}[f(t+\delta)+f(t)]$$

And for $\delta \rightarrow 0$ this simply becomes

$$r(t) = \frac{1}{2}[f(t)+f(t)] = f(t)$$

This makes intuitively sense: Integrating over an infinitely small interval just yields the function value at this point. There is no other information going into this integral, so it can't be anything else.

So the questions here: what is the purpose of $r(t)$ how would it be different from using $<p(t)>$ directly ? Maybe it has to do with properties $<p(t)>$ when it's NOT smooth as it is the ensemble average of pulse trains which are definitely not smooth.

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  • $\begingroup$ Thanks a lot, that was really interesting. Regarding your questions, I think that intuitively, we can think of $<\rho(t)>$ as defining a density, and that $r(t)$ gives us a probability over that interval. This is as you pointed out yourself, different from the case where the function under the integral is smooth. $\endgroup$
    – Ash
    Oct 8, 2023 at 8:45
  • $\begingroup$ I suppose that $r(t)$ is supposed to be the firing rate as function of time. However in this case $\Delta t$ needs to "suitable", i.e. small enough to give you decent time resolutions but big enough so that a meaningful number of events happened inside the interval, which is definitely NOT $\Delta t \rightarrow 0$. In this case I would have swapped the order of $<>$ and $\int$. Applying the integral first simply counts the number of events inside the interval. Averaging that over multiple trials gives you the average firing rate. $\endgroup$
    – Hilmar
    Oct 8, 2023 at 12:23
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    $\begingroup$ Since both operations are linear you can probably swap the order anyway. If my assumption of a finite $\Delta t$ is correct, than the integral becomes a moving average filter, i.e. a lowpass (with a lot of sidelobes). In that case the statement about the convolution feels wrong. I'll update my answer, when I have a chance $\endgroup$
    – Hilmar
    Oct 8, 2023 at 12:23

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