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.

enter image description here

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?

  • 3
    $\begingroup$ 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
  • 2
    $\begingroup$ 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
  • 1
    $\begingroup$ 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


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.

  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.