A note on manuscript by S.E.Odom and A. Borisyuk “Estimating three synaptic
conductances in a stochastic neural model”
In this note we show that our reconstruction method will, at every time step generate at least one solution
(out of two) with at least one of the conductances taking a negative value.
We start by assuming that all solutions for the conductances g1 , g2 , and g3 at the ith time step are positive
real values and show that this leads to a contradiction, implying that at least one of them must be negative.
For convenience define ∆1 = (µVi −E1 ), ∆2 = (µVi −E2 ), and ∆3 = (µVi −E3 ). One can write the equation
relating the variances of Vm at successive time steps as
σV2 i+1
=
σV2 i
2
3
∆t
∆t X 2 2 2
1−
g σ (σ + ∆2j )
(gL + g1 + g2 + g3 ) + 2
C
C j=1 j ξj Vi
(1)
(same as equation (18) in the manuscript).
For ∆t sufficiently small one has that σV2 i ≈ σV2 i+1 allowing replacement of (1) by
2
3
∆t
∆t X 2 2 2
1−
g σ (σ + ∆2j )
(gL + g1 + g2 + g3 ) + 2
C
C j=1 j ξj Vi
(2)
3
∆t X 2 2 2
∆t2
∆t 2
2
g σ (σ + ∆2j )
σVi (gL + g1 + g2 + g3 ) + 2 σV2 i (gL + g1 + g2 + g3 ) + 2
C
C
C j=1 j ξj Vi
(3)
σV2 i
=
σV2 i
Expanding (2) gives
σV2 i = σV2 i − 2
The common σV2 i terms may be cancelled on both sides. We may also ignore the O(∆t2 ) term for small
∆t. Multiplying the resulting equation by C 2 /∆t gives
0 = −2CσV2 i (gL + g1 + g2 + g3 ) +
3
X
gj2 σξ2j (σV2 i + ∆2j )
(4)
j=1
We may now express (4) as a quadratic equation in terms of g1 ,
ag12 + bg1 + c = 0
(5)
where
a = σξ21 (σV2 i + ∆21 )
b=
c=
−2CσV2 i
g22 σξ22 (σV2 i
(6)
(7)
+
∆22 )
+
g32 σξ23 (σV2 i
+
∆23 )
−
2CσV2 i (gL
+ g2 + g3 ).
If both solutions for g1 are positive real values, then by the quadratic formula,
√
−b ± b2 − 4ac
g1 =
2a
1
(8)
it follows that −4ac < 0. Since (6) shows that a is always positive, this is equivalent to the condition that
c > 0. By (5) this implies that ag12 + bg1 < 0 or that
σξ21 (σV2 i + ∆21 )g12 − 2CσV2 i g1 < 0
(9)
Symmetric arguments can be made for g2 and g3 assuming all solutions are positive real values. These
yield,
σξ22 (σV2 i + ∆22 )g22 − 2CσV2 i g2 < 0
σξ23 (σV2 i
+
∆23 )g32
−
2CσV2 i g3
(10)
< 0.
(11)
gj2 σξ2j (σV2 i + ∆2j ).
(12)
Equation (4) may now be rewritten as
2CσV2 i gL = −2CσV2 i (g1 + g2 + g3 ) +
3
X
j=1
Adding the inequalities (9)-(11) shows that the right hand side of (12) is negative. However, since gL > 0,
it must be the case that the left hand side of (12) is positive. This contradiction implies that at least one
of the solutions for g1 , g2 , or g3 must be negative and extraneous. Moreover, the two associated solutions
for the other conductances will also be extraneous.
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