CNS 221 - Spring 2006 Lecture 8 (2006-Apr-20) Wolfgang Einhäuser Treyer

CNS 221 - Spring 2006 Lecture 8 (2006-Apr-20) Wolfgang Einhäuser Treyer

Even for constant stimulus, neurons fire irregularly

Sources of noise At individual neuron (“intrinsic”) thermal noise: <V2> ~ R kBT finite number of channels in membrane - channels are either open or closed - H&H-model (e.g.) considers average number of open channels at given potential At network level (“extrinsic”) stochasticity of synaptic transmission network effects (random connectivity)

Renewal system We know: - the input current I(t) - time the neuron last spiked t0 E.g. Integrate & Fire neuron: In the absence of noise, we can predict the time of the next spike as the first time after t0 at which V(t) crosses the threshold. In contrast, if we have noise, we can just provide a probability for the next spike to occur at time t. Renewal system (input-dependent): The probability of the next “event” (spike) is given by the time of the last event (t-t0 - the “age”) and I(t’) (for t0<t’<t). Example:

Firing rate spike raster response in trial j trial number t 1.5s 0

Firing rate spike raster response in trial j trial number 100 DT=20ms <f(t)> 0 t 1.5s 0

Firing rate trial number 100 DT=200ms <f(t)> 0 t 1.5s 0

Firing rate trial number 100 DT=2ms (DT->0 instanteous rate) <f(t)> 0 t 1.5s 0

Inter-spike interval (ISI) stationary input: Histogram of inter-spike-intervals 9 non-bursting MT-cell (moving random dot pattern at medium coherence) % 0 t (ms)

Poisson process • Poission process with mean rate µ: • Random process {N(t), t>0} for which • Let ti , tj timepoints with ti<tj for i < j • (1) N(tp)-N(tp-1) independent from N(tq)-N(tq-1) for any p, q with p≠q • (2) Average number of events between t1 and t2:(t2- t1)µ • P{N(t2)-N(t1)=k} = (µ(t2-t1)k)exp(-µ(t2-t1)) / k! • P{N(t+Dt)-N(t)=k} = ((µDt) k exp(-µDt)) / k! For small time windows (i.e. µDt<<1): P{N(t+Dt)-N(t)=0} = exp(-µDt) -> 1-µDt P{N(t+Dt)-N(t)=1} = (µDt) exp(-µDt) -> µDt

Poisson process For small time windows (i.e. µDt<<1): P{N(t+Dt)-N(t)=0} = 1-µDt P{N(t+Dt)-N(t)=1} = µDt So: - choose Dt small, that maximally 1 spike happens in Dt, - bin your time into intervals of Dt - for each bin draw a random variable 0<r<1 from a uniform distr. - spike if r ≤ µDt

Poisson process P{N(t+Dt)-N(t)=k} = (µDt)exp(-µDt) / k! If spike has ocurred at t=0 the probability that no spike has ocuered by time t is given: P{N(t)-N(0)=0} = exp(-µt) Hence the probability that at least one spike has ocurred is 1- exp(-µt) And the probability density p(t) = µexp(-µt) So in terms of ISIs:Given we had a spike at t0 the probability to have another spike at t is p(t|t0) = µexp(-µ(t-t0))

Poisson process Mean firing rate: Mean interval between spikes: So the mean firining rate is indeed µ, as intended

Inter-spike interval (ISI) 9 % p 0 refractory period ! t (ms) 3/µ 0 1/µ t

Autocorrelation with

Noise spectrum Powerspectrum: Relation to autocorrelation: Fourier transform of Cii(s) is P(w)

Normalized Autocorrelation • Mean firing rate: µ • probability to find spike in [t,t+Dt]: µ Dt • for large inter-spike distances, expectation for spike independent of spike long time ago, hence define • transform to Fourier space • divergence for w->0 in unnormalized case!

Autocorrelation So far: p(t) (see interspike interval) 0 t t Given we had a spike at t=0, what is the probability for the nextspike at t? s Now: t0 t1 t What is the probability to find two spikes at distance s (independent of the spikes between them)?

Autocorrelation Assume we had a spike at time t (which we expect with rate µ), the probability for a spike at t+s is given by Cii(s) Q(s), hence we can express the probability to find any two spikes at distance s as C+(s) by µC+(s) = Cii(s) Q(s) where µ is the (prior) probability for having a spike at t. Hence Or recursively: 2nd spike at s 3rd spike at s 1st spike at s

Autocorrelation with symmetry and trivial autocorrelation at s=0 we have: Now Fourier transform and obtain (convolution -> multiplication in Fourier space)

Autocorrelation

Example: Poisson process By definition, the probability Hence: As one would expect, the autocorrelation is flat, with delta peak at DC

Poisson Process with refractory period Add absoulte refractory period to Poisson procees Using one obtains

Poisson Process with refractory period for large w as without refractory period, for small w however noise is decreased =>refractoriness makes spiking more regular for finite DT the mean firing rate is bound, even for r->inf

Coefficient of variation (CV) Quantification of variation: standard deviation of ISI divided by mean Regular spiking: CV=0 Poisson process (without refractory period): CV=1 with refractory period: CV<1.

Random walk model Perfect integrator, receiving synaptic input from a random process Ne(t): V(t) = aeNe(t) nth events are needed to cross threshold: nth= [Vthresh/ae]+1 If nth=1, the spiking probability (of our postsynaptic cell) is given by the probability of one presynaptic event to occur, 1-exp(-µt). The average inter-spike interval is given by nth/µ, where µ is again the presynaptic rate. And the variance (as the presynaptic spikes are independent) by nth/µ2. So the CV scales with 1/sqrt(nth), i.e. (given our Vthresh ) we spike more regularly for many small inputs (small ae => large nth), than for few large ones (large ae, => small nth)

Random walk model Now add inhibitory input V(t) = aeNe(t) - aiNi(t) with µ=aeµe - aiµi we have <V(t)> = µt and with s2 = a2eµe + a2iµi we have var(V(t)) = s2 t probability to cross threshold in finite time: 1 if excitation > inhibition (µ>0) (aeµe/aiµ)vthrehs otherwise (µ<0)

Random walk model • V(t) = aeNe(t) + aiNi(t); excitation>inhibition (µ>0) • postsynaptic ISI: • mean: <Tth>=Vth/µ • variance: var(Tth)= Vth(aeµe + aiµi)/µ3 • Hence • CV = sqrt((aeµe + aiµi)/ (Vth(aeµe - aiµi))) • adding inhibition increases variability (small drift, high jitter) • “balanced” excitation and inhibition Adding a leak stabilizes the membrane potential <V(t)>~Rµ

CNS 221 - Spring 2006 Lecture 8 (2006-Apr-20) Wolfgang Einhäuser Treyer