Single spike can have dramatic consequences on population activity.

Terms

1. PSTH: peri-stimulus-time histogram, meaning the probability density of firing as a function of time, after the stimulus.

## One Input Spike

1. Some neuron takes constant input $I_0$ and noise $I_{\mathrm{noise}}$.
2. We inject an extra input on to this neuron.

The factors of importance are

1. amount of noise
2. time course of PSP caused by the injection.

Relation between PSP and PSTH. Basically all well expalained in Fig 7.12:

1. For large noise, PSTH is similar to PSP,
2. For small noise, PSTH is the derivatives of PSP.

Amazing but why?

Read Fig 7.11. Consider two scenarios,

1. with noise, basically noise will trigger a spike,
2. without noise: related to the derivatives of psp because spike can only occur when the derivative is positive.

Understand the significance using homogeneous population model. Linearized equation is applied

$$f_{\mathrm{PSTH}}(t) = \frac{d}{dt} \int_0^\infty \mathcal L (x) \epsilon_0(t-x) dx,$$

where

$$\mathcal L(x) \sim \begin{cases} \delta(x) & \qquad \text{low noise limit} \ \text{broad} & \qquad \text{high noise} \end{cases}.$$

## Reverse Correlation

Reverse correlation: Record the input of the neuron just before it spikes, then average many spikes.

$$C^{\mathrm{rev}}(s) = \langle \Delta I(t^{(f)-s}) \rangle_f,$$

where $\Delta I$ is the stimulus right before the spike at time $t^{(f)}$.

Reverse correlation is related to correlation function $C$ through

$$\nu C^{rev}(s) = C(s),$$

where $\nu$ is the firing rate, $\nu=A_0$.

We will find the relation between this reverse correlation and transfer properties of a single neuron, which is described by

$$\hat A(\omega) = \hat G(\omega) \hat I(\omega),$$

We derive the population activity using the transfer function $\hat G(\omega)$

$$A(t) = A_0 + \int_0^\infty G(s) \Delta I(t-s) ds.$$

Fourier transform of multiplications leads to a convolution.

With the expression of $A(t)$, we could calculate reverse correlation

\begin{align} C(s) =& \lim_{T\to\infty} \frac{1}{T} \int_0^T A(t+s)\Delta I(t) dt \ =& \lim_{T\to\infty} \frac{1}{T} \int_0^T \int_0^\infty G(s’) \Delta I(t+s-s’) ds’ \Delta I(t) dt\ =& \int_0^\infty ds’ G(s’) \lim_{T\to\infty}\frac{1}{T}\int_0^T \Delta I(t+s-s’) \Delta I(t) dt\ =& \int_0^\infty ds’ G(s’) \langle \Delta I(t+s-s’)\Delta I(t)\rangle \end{align}

The reason we dropped the term $A_0$ is because we assume the input is stochastic i.e., $\langle \Delta I(t)\rangle=0$.

For white noise, we have

$$\langle \Delta I(t’)\Delta I(t)\rangle=\sigma^2\delta(t’-t).$$

Then we find the relation between reverse correlation and transfer function,

$$C^{rev}(s) = \frac{1}{\nu}C(s) = \frac{\sigma^2}{\nu} G(s).$$

Planted: by ;