Biomedical Engineering Reference
In-Depth Information
r
=1
π
(
T
)
empirical average by
1
N
ω
(
r
)
[
f
]. This also allows the computation of error
bars as well as more elaborated statistical estimation techniques [
48
].
What if the stationarity assumption is violated? Then, the average of
f
depends
on time and one computes the empirical average from the
N
rasters. We denote
π
(
N
)
[
f
(
n
)] the average of
f
at time
n
, performed over
N
rasters. For example
r
=1
ω
(
r
k
(
n
) is the sample-averaged
probability that neuron
k
fires at time
n
. If all rasters are described by the same
probability (the Gibbs distribution which is also defined in the non-stationary case),
then
π
(
N
)
when
f
(
ω
)=
ω
k
(
n
),
π
(
N
)
1
N
[
f
(
n
)] =
[
f
(
n
)]
→
μ
[
f
(
n
)]as
N→
+
∞
.
8.3.2.4
Example of Empirical Average: Estimating Instantaneous
Pairwise Correlations
Assume that spikes are distributed according to an hidden probability
μ
supposed to
be stationary for simplicity. The instantaneous pairwise correlations of neurons
k,j
with respect to
μ
is:
C
(
k,j
)=
μ
[
ω
k
(0)
ω
j
(0) ]
−
μ
[
ω
k
(0) ]
μ
[
ω
j
(0) ]
.
(8.11)
Since
μ
is stationary the index 0 can be replaced by any time index (time-translation
invariance of statistics).
Assume now that we have a raster
ω
distributed according to
μ
. An estimator of
C
(
k,j
) is:
C
(
T
)
ω
(
k,j
)=
π
(
T
)
ω
π
(
T
)
ω
[
ω
k
(0) ]
π
(
T
)
[
ω
k
(0)
ω
j
(0) ]
−
[
ω
j
(0) ]
.
(8.12)
ω
It converges to
C
(
k,j
) as
T
.
The events “neuron
k
fires at time 0”(
ω
k
(
n
)=1) and “neuron
j
fires at time 0”
(
ω
j
(
n
)=1)are
independent
if
μ
[
ω
k
(0)
ω
j
(0) ] =
μ
[
ω
k
(0) ]
μ
[
ω
j
(0) ], thus
C
(
k,j
)=0. (Note that independence implies vanishing correlation but the reverse
is not true in general. Here the two properties are equivalent thanks to the binary 0
,
1
form of the random variables
ω
k
(0)
,ω
j
(0)).
Assume now that the observed raster has been drawn from a probability where
these events are independent, but the experimentalist who analyzes this raster does
not know it. To check independence he computes
C
(
T
)
→
+
∞
ω
(
k,j
) from the experimental
raster
ω
. However, since
T
is finite,
C
(
T
ω
(
k,j
) will not be exactly 0.More
precisely, from the central limit theorem the following holds. The probability that
the random variable
C
(
T
)
(
k,j
)
is larger than
, is well approximated (for large
ω
T
and small
)by
e
−
2
T
2
K
.
K
can be exactly computed (Sect.
8.3.1.5
). In the simplest
case where spikes are drawn independently with a probability
p
of having a spike,
K
is equal to
p
2
p
2
(1
−
). Th
us,
fluctuations are Gaussian and their mean-square
deviation decay with
T
as
T
. As a consequence, even if neuron
j
and
k
are
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