<p>The distribution is not Gaussian, it's Poisson, as noted in your question. However, outside of certain edge cases (e.g. when there are very few spikes, or very few trials), a Poisson Distribution with rate parameter $\lambda$ looks very like a Gaussian with both mean and variance equal to $\lambda$ (standard deviation $= \sqrt{\lambda}$). This is useful, because if you treat the distribution as a Gaussian, you can apply methods such as the t-test or linear regression, without being too far wrong.</p>

<p>Here's some R code that shows the correspondance between the Poisson and the Gaussian.</p>

<pre><code>library(tidyverse)
x = seq(0, 30, 1) # Possible spike counts
poisson.density = dopois(x, 10)
gaussian.density = dnorm(x, 10, sqrt(10))
d = data.frame(x = x, Poisson=poisson.density, Gaussian=gaussian.density) %&gt;%
  gather(Distribution, Density, -x)
ggplot(d, aes(x, Density, color=Distribution, linetype=Distribution)) +
  geom_path() +
  theme(legend.position=c(1,1), legend.justification=c(1,1)) +
  labs(x='Spike count')
</code></pre>

<p><a href="https://i.sstatic.net/wZmD6.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/wZmD6.png" alt="enter image description here"></a></p>

CC BY-SA 4.0
2018-08-21T15:54:19.257Z
2018-08-21T15:54:19.257Z
Eoin
<p>I'm reading "Principles of Neural Information Theory" by James V Stone and in section 3.5 he says that the distribution of firing rates (of a single neuron) is generally assumed to be approximately Gaussian. He proceeds to give a mathematical argument for this.</p>

<p>If we measure the activity during a period of $T$ seconds over intervals $\Delta t$, there are $N=T/\Delta t$ possible positions for spikes to occur. The probability of $n$ spikes occurring  this period is
$$p(n)=\frac{N!}{n!(N-n)!}P^nQ^{N-n}$$
Where $P$ is the probability fo a spike occurring and $Q$ is the probability of a spike not occuring. If $N$ and $T$ increase while firing rate $r$ stays constant, $p(n)$ approaches the <a href="https://en.wikipedia.org/wiki/Poisson_distribution" rel="nofollow noreferrer">Poisson distribution</a>. If $T$ is held constant then a simple change of variables can be used to obtain a distribution for firing rate. If $N$ is large and $P$ is small, apparently this distribution is approximated by a Gaussian distribution.</p>

<p>I don't really understand this derivation. It would be nice if someone could provide an in-depth version of the argument with all steps full fleshed-out. I'm particularly confused about how one goes from p(n) to a Poisson distribution.</p>

CC BY-SA 4.0
2018-08-17T17:44:34.640Z
This is a question for Math SE. There's nothing specific to neuroscience in your actual question (how that formula gives a Gaussian distribution under some parameter assumption.)
got trolled too much this week
CC BY-SA 4.0
2018-08-22T15:47:00.487Z
@Fizz but the answer is relevant for neuroscience.
StrongBad
<neuroscience>
Why is the distribution of mean firing rates assumed to be Gaussian?
Eben Kadile
# Why is the distribution of mean firing rates assumed to be Gaussian?

(Asked by: username_0 on 2018-08-16)

I'm reading "Principles of Neural Information Theory" by James V Stone and in section 3.5 he says that the distribution of firing rates (of a single neuron) is generally assumed to be approximately Gaussian. He proceeds to give a mathematical argument for this.

If we measure the activity during a period of $T$ seconds over intervals $\Delta t$, there are $N=T/\Delta t$ possible positions for spikes to occur. The probability of $n$ spikes occurring this period is
$$p(n)=\frac{N!}{n!(N-n)!}P^nQ^{N-n}$$
Where $P$ is the probability fo a spike occurring and $Q$ is the probability of a spike not occuring. If $N$ and $T$ increase while firing rate $r$ stays constant, $p(n)$ approaches the [Poisson distribution](https://en.wikipedia.org/wiki/Poisson_distribution). If $T$ is held constant then a simple change of variables can be used to obtain a distribution for firing rate. If $N$ is large and $P$ is small, apparently this distribution is approximated by a Gaussian distribution.

I don't really understand this derivation. It would be nice if someone could provide an in-depth version of the argument with all steps full fleshed-out. I'm particularly confused about how one goes from p(n) to a Poisson distribution.

**Tags:** neuroscience

> **Comments:**
> **username_1**: This is a question for Math SE. There's nothing specific to neuroscience in your actual question (how that formula gives a Gaussian distribution under some parameter assumption.)  
> **username_2**: @Fizz but the answer is relevant for neuroscience.  

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## 1 answers

### Answer by username_3 on 2018-08-21. Score: 3

The distribution is not Gaussian, it's Poisson, as noted in your question. However, outside of certain edge cases (e.g. when there are very few spikes, or very few trials), a Poisson Distribution with rate parameter $\lambda$ looks very like a Gaussian with both mean and variance equal to $\lambda$ (standard deviation $= \sqrt{\lambda}$). This is useful, because if you treat the distribution as a Gaussian, you can apply methods such as the t-test or linear regression, without being too far wrong.
Here's some R code that shows the correspondance between the Poisson and the Gaussian.
library(tidyverse)
x = seq(0, 30, 1) # Possible spike counts
poisson.density = dopois(x, 10)
gaussian.density = dnorm(x, 10, sqrt(10))
d = data.frame(x = x, Poisson=poisson.density, Gaussian=gaussian.density) %>%
  gather(Distribution, Density, -x)
ggplot(d, aes(x, Density, color=Distribution, linetype=Distribution)) +
  geom_path() +
  theme(legend.position=c(1,1), legend.justification=c(1,1)) +
  labs(x='Spike count')


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