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I am studying computational neuroscience, particularly the modeling of neuronal spikes. Abstractly, we may think of a spike plainly as some event that either occurs or fails to occur in time. It has been established that the probability that $n$ events (spikes) occur in a trial of time $T$ follows a Poisson distribution with

$$ P_{T}(N = n) = \frac{(rT)^{n}}{n!}\exp(-rT) $$

Here, $r$ is the firing rate of the neuron (or the constant rate at which events occur in time). It was also established early in the book that $r\Delta t$ is the probability that any spike occurs at some time interval $[t, t + \Delta t]$. As a note, the equation above in the book is numbered $1.29$ - I clarify this for it is referenced in the fragment I need help with.

This is all the context required to understand the following fragment of the book, with which I need some help.


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I can fully comprehend the paragraph above up to equation $1.31$. What I don't understand is what immediately follows it. Namely, that the PDF of interspike intervals is equation $1.31$ without the factor $\Delta t$, or in other words, that the probability that an interval $[t + \tau, t + \tau + \Delta t]$ is an interspike interval is

$$ r\exp(-r\tau) $$

I did observe that such expression is the derivative with respect to $\Delta t$ of equation $1.31$. However, equation $1.31$ is not a CDF, and hence its derivative is not the PDF we are interested in. What am I missing?

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    $\begingroup$ It is not a cumulative distribution function. But if you divide both sides by $\Delta t$ and then take the limit as $\Delta t \to 0_+$ then you get the probability density (the derivative of the CDF) $r \exp(-r\tau)$ $\endgroup$
    – Henry
    Jul 27, 2023 at 10:27
  • $\begingroup$ Thanks for the comment. I considered that possibility, but after dividing by $\Delta t$, shouldn't the left hand side tend to infinity as $\Delta t$, the denominator, approaches $0$? $\endgroup$
    – lafinur
    Jul 27, 2023 at 10:32
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    $\begingroup$ Expanding a bit on @Henry's comment: The denominator is also approaching 0, as $\Delta t$ goes to $0$. So you have a ratio that goes to "0/0". Then you can replace both numerator and denominator by their derivative, here with respect to $\Delta t$. This is a consequence of de l'Hôpital's rule. Do you need a complete answer? $\endgroup$
    – Ute
    Jul 27, 2023 at 11:48
  • $\begingroup$ Hello @Ute, thank you for the comment. I can see how l'Hopital applies in theory, but in practice we do not know the derivative of $P[\tau \leq t_{i+1} - t_{i} < \tau + \Delta t]$ with respect to $\Delta t$, do we? $\endgroup$
    – lafinur
    Jul 27, 2023 at 11:56
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    $\begingroup$ It is a bit difficult to understand that mixed notation. Does it help if you denote by $D=t_{i+1}-t_i$ the random interspike length and then have $P[\tau \leq D\le \tau+\Delta t]$? $\endgroup$
    – Ute
    Jul 27, 2023 at 12:06

1 Answer 1

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Your difficulties seem to stem from translating the original form of equation (1.31) into the "probability that an interval $[t+\tau, t+\tau + \Delta t]$ is an interspike interval". This is due to the authors using the word "interval" synonymously with "length of an interval".

I rewrite the book paragraph using a bit more formalism and hope it becomes easier to understand. I also find it much easier to work with the cdf instead of the pdf.

Rewrite of book paragraph:

The probability density of length of time intervals between adjacent spikes is called the interspike interval length distribution, and it is a useful statistic for characterizing spiking patterns. Let $L$ denote this random length, and $F$ be its cumulative density function.

Suppose that a spike occurs at a time $t_i$ for some value of $i$. The following interspike interval is at least of length $\tau$, if there is no new spike in the interval $(t_i, t_i+\tau)$. From equation 1.29, with $n=0$, the probability of not firing a spike in this period, i.e., the probability that the interspike length exceeds $\tau$ is $$ P(L > \tau) = \exp (-r\tau). $$ From this we get the cdf of $L$, namely $$ F(\tau) = P(L\leq \tau) = 1 - \exp(-r\tau). $$ This is the cdf of an exponential distribution with rate parameter $r$.

New interpretation of book text:

Now look at equation (1.31). Using the symbol $L$ for length of interval again, we read $$ P(\tau\leq L < \tau+\Delta t) = r \Delta t \exp (-r\tau). $$ This probability can be written as difference of cdf in $\tau$ and $\tau+\Delta t$, which gives us $$ \frac{F(\tau+\Delta t)-F(\tau)}{\Delta t} = r\exp(-r\tau). $$ Now we can take the limit on the left hand side for $\Delta t \to 0$ and get the pdf of an exponential distribution.

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    $\begingroup$ There you go - and ask your money back :-D (for the book). Sometimes formal would have been easier than informal... $\endgroup$
    – Ute
    Jul 27, 2023 at 16:39
  • $\begingroup$ Jesus... To be honest I wouldn't have been able to reconstruct the statements in a way that made the logic as clear as you just did. $\endgroup$
    – lafinur
    Jul 27, 2023 at 20:43
  • $\begingroup$ Yeah the notation of the book is not excellent :( $\endgroup$
    – lafinur
    Jul 27, 2023 at 20:43
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    $\begingroup$ tx - :-) hehe, taught this more than once $\endgroup$
    – Ute
    Jul 27, 2023 at 20:47
  • $\begingroup$ The pedagogical skills do show. Have a great rest of your day $\endgroup$
    – lafinur
    Jul 27, 2023 at 21:34

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