Spike Train Statistics from Empirical Facts to Theory: The Case of the Retina - Modeling in Computational Biology and Biomedicine - page 269

8.3.2.5 Matching Experimental Averages

Assume that an experimentalist observes rasters, and assume that all those rasters

are distributed according to an hidden probability distribution μ . Is to possible to
determine or, at least, to approach μ from those rasters? One possibility relies on
the maximal entropy principle described in the next sections. We assume for the
moment that statistics is stationary.
Fix

N

K observables O k , k =1 ,..., K , and compute their empirical average

π ( T )

[ O k ]. The remarks of the previous sections hold: since all rasters are dis-
tributed according to μ , π ( T )

ω

[ O k ] is a random variable with mean μ [ O k ] and

ω

Gaussian 1

1

N > 1 rasters
the experimentalist can estimate the order of magnitude of those fluctuations and
also analyze the raster-length dependence. In fine , he obtains an empirical average
value for each observable, π ( T )

fluctuations about its mean, of order

√ T .Ifthereare

. Now, to estimate the
hidden probability μ , by some approximated probability μ ap ,hehastoassume,asa
minimal requirement, that:

[ O k ]= C k ,k =1 ,..., K

ω

π ( T )
ω

[ O k ]= C k = μ ap [ O k ] , k =1 ,..., K , (8.13)

i.e., the expected average of each observable, computed with respect to μ ap is equal
to the average found in the experiment. This fixes a set of constraints to approach
μ . We call μ ap a statistical model .
Unfortunately, this set of conditions does not fix a unique solution but infinitely
many! As an example if we have only one neuron whose firing rate is

1

2 ,
then a straightforward choice for μ ap is the probability where successive spikes
are independent ( P [ ω k ( n ) ω k ( n

− 1)] = P [ ω k ( n )] P [ ω k ( n − 1)])andwhere

1

the probability of a spike is

2 . However, one can also take a one-step mem-
ory model where transition probabilities obey P [ ω k ( n )=0 | ω k ( n − 1) = 0] =

P [ ω k ( n )=1 | ω k ( n − 1) = 1] = p , P [ ω k ( n )=0 | ω k ( n − 1) = 1] =

P [ ω k ( n )=1 | ∈ [0 , 1]. In this case, indeed the invariant

probability of the corresponding Markov chain is μ ap [ ω k ( n )=0 , 1] =

ω k ( n − 1) = 0] = 1 − p , p

1

2 ,since

from Eq. ( 8.5 ),

μ ap [ ω k ( n )=0]= P [ ω k ( n )=0 | ω k ( n − 1)] μ ap [ ω k ( n − 1)] ,

ω k ( n− 1)=0 , 1

p

2 + 1 −

p

= 1
2

.

2

The same holds for μ ap [ ω k ( n )=1]. In this case, we match the constraint too but
with a model where successive spikes are not independent. Now, since p takes values

1 Fluctuations are not necessarily Gaussian, if the system undergoes a second order phase transition
where the topological pressure introduced in Sect. 8.3.1.5 is not twice differentiable.