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Lecture 9: Population genetics, first-passage problems. Outline: population genetics Moran model fluctuations ~ 1/ N but not ignorable effect of mutations effect of selection neurons: integrate-and-fire models interspike interval distribution no leak with leaky cell membrane - PowerPoint PPT Presentation
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Lecture 9: Population genetics, first-passage problems
Outline:• population genetics
• Moran model• fluctuations ~ 1/N but not ignorable• effect of mutations• effect of selection
• neurons: integrate-and-fire models• interspike interval distribution
•no leak• with leaky cell membrane
• evolution• traffic
Population genetics: Moran model
2 alleles, N haploid organisms
Population genetics: Moran model
2 alleles, N haploid organismschoose 2 individual at random: 1 dies, the other reproduces
Population genetics: Moran model
2 alleles, N haploid organismschoose 2 individual at random: 1 dies, the other reproducesIf there are n1 organisms of type 1 before this step, then afterwards there are
Population genetics: Moran model
2 alleles, N haploid organismschoose 2 individual at random: 1 dies, the other reproducesIf there are n1 organisms of type 1 before this step, then afterwards there are
n1 + 1 with probability x(1 – x) x = n1/N
Population genetics: Moran model
2 alleles, N haploid organismschoose 2 individual at random: 1 dies, the other reproducesIf there are n1 organisms of type 1 before this step, then afterwards there are
n1 + 1 with probability x(1 – x) x = n1/Nn1 – 1 with probability x(1 – x)
Population genetics: Moran model
2 alleles, N haploid organismschoose 2 individual at random: 1 dies, the other reproducesIf there are n1 organisms of type 1 before this step, then afterwards there are
n1 + 1 with probability x(1 – x) x = n1/Nn1 – 1 with probability x(1 – x)n1 with probability x2 + (1 – x)2.
Population genetics: Moran model
2 alleles, N haploid organismschoose 2 individual at random: 1 dies, the other reproducesIf there are n1 organisms of type 1 before this step, then afterwards there are
n1 + 1 with probability x(1 – x) x = n1/Nn1 – 1 with probability x(1 – x)n1 with probability x2 + (1 – x)2.
So
€
n1(t + δt) = n1(t)
Population genetics: Moran model
2 alleles, N haploid organismschoose 2 individual at random: 1 dies, the other reproducesIf there are n1 organisms of type 1 before this step, then afterwards there are
n1 + 1 with probability x(1 – x) x = n1/Nn1 – 1 with probability x(1 – x)n1 with probability x2 + (1 – x)2.
So
€
n1(t + δt) = n1(t)
n1(t + δt) − n1(t + δt)( )2
= 2n1(t)n2(t) = 2n1(t) 1− n1(t)( )
Population genetics: Moran model
2 alleles, N haploid organismschoose 2 individual at random: 1 dies, the other reproducesIf there are n1 organisms of type 1 before this step, then afterwards there are
n1 + 1 with probability x(1 – x) x = n1/Nn1 – 1 with probability x(1 – x)n1 with probability x2 + (1 – x)2.
So
€
n1(t + δt) = n1(t)
n1(t + δt) − n1(t + δt)( )2
= 2n1(t)n2(t) = 2n1(t) 1− n1(t)( )
or
€
x(t + δt) = x(t)
Population genetics: Moran model
2 alleles, N haploid organismschoose 2 individual at random: 1 dies, the other reproducesIf there are n1 organisms of type 1 before this step, then afterwards there are
n1 + 1 with probability x(1 – x) x = n1/Nn1 – 1 with probability x(1 – x)n1 with probability x2 + (1 – x)2.
So
€
n1(t + δt) = n1(t)
n1(t + δt) − n1(t + δt)( )2
= 2n1(t)n2(t) = 2n1(t) 1− n1(t)( )
or
€
x(t + δt) = x(t)
x(t + δt) − x(t + δt)( )2
=2x(t) 1− x(t)( )
N 2
continuum limit: FP equation
€
δt = 1N (N steps/generation)
continuum limit: FP equation
€
δt = 1N
⇒∂P(x, t)
∂t= 1
2 σ 2 ∂ 2
∂x 2x(1− x)P(x, t)[ ], σ 2 =
2
N
(N steps/generation)
continuum limit: FP equation
€
δt = 1N
⇒∂P(x, t)
∂t= 1
2 σ 2 ∂ 2
∂x 2x(1− x)P(x, t)[ ], σ 2 =
2
N
(N steps/generation)
boundary conditions: P(0,t) = P(1,t) = 0
continuum limit: FP equation
€
δt = 1N
⇒∂P(x, t)
∂t= 1
2 σ 2 ∂ 2
∂x 2x(1− x)P(x, t)[ ], σ 2 =
2
N
(N steps/generation)
boundary conditions: P(0,t) = P(1,t) = 0(once an allele dies out, it can not come back)
continuum limit: FP equation
€
δt = 1N
⇒∂P(x, t)
∂t= 1
2 σ 2 ∂ 2
∂x 2x(1− x)P(x, t)[ ], σ 2 =
2
N
(N steps/generation)
boundary conditions: P(0,t) = P(1,t) = 0(once an allele dies out, it can not come back)
stochastic differential equation:
€
dx = σ x(1− x)dW (t)
continuum limit: FP equation
€
δt = 1N
⇒∂P(x, t)
∂t= 1
2 σ 2 ∂ 2
∂x 2x(1− x)P(x, t)[ ], σ 2 =
2
N
(N steps/generation)
boundary conditions: P(0,t) = P(1,t) = 0(once an allele dies out, it can not come back)
stochastic differential equation:
notice
€
dx = σ x(1− x)dW (t)
dx = 0
heterozygocityEventually P(x,t) gets concentrated at one boundary,
heterozygocityEventually P(x,t) gets concentrated at one boundary, i.e., one allele completely dominates the other.
heterozygocityEventually P(x,t) gets concentrated at one boundary, i.e., one allele completely dominates the other. But it is randomwhich one.
heterozygocity
€
H(t) = 2x(t) (1− x(t)( )
Eventually P(x,t) gets concentrated at one boundary, i.e., one allele completely dominates the other. But it is randomwhich one. Measure this by the heterozygocity
heterozygocity
€
H(t) = 2x(t) (1− x(t)( )
Eventually P(x,t) gets concentrated at one boundary, i.e., one allele completely dominates the other. But it is randomwhich one. Measure this by the heterozygocityuse Ito’s lemma:
€
dF =∂F
∂xu(x) +
∂F
∂t+ 1
2 σ 2 ∂ 2F
∂x 2G2(x)
⎛
⎝ ⎜
⎞
⎠ ⎟dt + σ
∂F
∂xG(x)dW
heterozygocity
€
H(t) = 2x(t) (1− x(t)( )
Eventually P(x,t) gets concentrated at one boundary, i.e., one allele completely dominates the other. But it is randomwhich one. Measure this by the heterozygocityuse Ito’s lemma:
€
dF =∂F
∂xu(x) +
∂F
∂t+ 1
2 σ 2 ∂ 2F
∂x 2G2(x)
⎛
⎝ ⎜
⎞
⎠ ⎟dt + σ
∂F
∂xG(x)dW
G(x) = x(1− x), u(x) = 0
heterozygocity
€
H(t) = 2x(t) (1− x(t)( )
Eventually P(x,t) gets concentrated at one boundary, i.e., one allele completely dominates the other. But it is randomwhich one. Measure this by the heterozygocityuse Ito’s lemma:
€
dF =∂F
∂xu(x) +
∂F
∂t+ 1
2 σ 2 ∂ 2F
∂x 2G2(x)
⎛
⎝ ⎜
⎞
⎠ ⎟dt + σ
∂F
∂xG(x)dW
G(x) = x(1− x), u(x) = 0
d x(1− x)[ ] = 12 σ 2(−2)x(1− x)dt + σ (1− 2x) x(1− x)dW ⇒
heterozygocity
€
H(t) = 2x(t) (1− x(t)( )
Eventually P(x,t) gets concentrated at one boundary, i.e., one allele completely dominates the other. But it is randomwhich one. Measure this by the heterozygocityuse Ito’s lemma:
€
dF =∂F
∂xu(x) +
∂F
∂t+ 1
2 σ 2 ∂ 2F
∂x 2G2(x)
⎛
⎝ ⎜
⎞
⎠ ⎟dt + σ
∂F
∂xG(x)dW
G(x) = x(1− x), u(x) = 0
d x(1− x)[ ] = 12 σ 2(−2)x(1− x)dt + σ (1− 2x) x(1− x)dW ⇒
dH
dt= −σ 2H
heterozygocity
€
H(t) = 2x(t) (1− x(t)( )
Eventually P(x,t) gets concentrated at one boundary, i.e., one allele completely dominates the other. But it is randomwhich one. Measure this by the heterozygocityuse Ito’s lemma:
€
dF =∂F
∂xu(x) +
∂F
∂t+ 1
2 σ 2 ∂ 2F
∂x 2G2(x)
⎛
⎝ ⎜
⎞
⎠ ⎟dt + σ
∂F
∂xG(x)dW
G(x) = x(1− x), u(x) = 0
d x(1− x)[ ] = 12 σ 2(−2)x(1− x)dt + σ (1− 2x) x(1− x)dW ⇒
dH
dt= −σ 2H i.e., diversity dies out in about N generations
fluctuations of x
€
x(t) − x(t)( )2
= x 2(t) − x(t)2
=
fluctuations of x
€
x(t) − x(t)( )2
= x 2(t) − x(t)2
=
= x 2(t) − x(t) + x(t) − x(t)2
fluctuations of x
€
x(t) − x(t)( )2
= x 2(t) − x(t)2
=
= x 2(t) − x(t) + x(t) − x(t)2
= − 12 H(t) + x(t) 1− x(t)( )
fluctuations of x
€
x(t) − x(t)( )2
= x 2(t) − x(t)2
=
= x 2(t) − x(t) + x(t) − x(t)2
= − 12 H(t) + x(t) 1− x(t)( )
= − 12 H(t) + x(0) 1− x(0)( )
fluctuations of x
€
x(t) − x(t)( )2
= x 2(t) − x(t)2
=
= x 2(t) − x(t) + x(t) − x(t)2
= − 12 H(t) + x(t) 1− x(t)( )
= − 12 H(t) + x(0) 1− x(0)( ) t →∞
⏐ → ⏐ ⏐ x(0) 1− x(0)( )
fluctuations of x
€
x(t) − x(t)( )2
= x 2(t) − x(t)2
=
= x 2(t) − x(t) + x(t) − x(t)2
= − 12 H(t) + x(t) 1− x(t)( )
= − 12 H(t) + x(0) 1− x(0)( ) t →∞
⏐ → ⏐ ⏐ x(0) 1− x(0)( )
So mean-square fluctuations of x grow initially linearly in t and then saturate
with mutation:Mutation induces a drift term in the FP and sd equation
€
n1(t + δt) = n1(t) + 1N μ12 n2 − μ21 n1[ ] = n1(t) + 1
N μ12 N − n1( ) − μ21 n1⇒[ ]
with mutation:Mutation induces a drift term in the FP and sd equation
€
n1(t + δt) = n1(t) + 1N μ12 n2 − μ21 n1[ ] = n1(t) + 1
N μ12 N − n1( ) − μ21 n1⇒[ ]
⇒
dx = μ12 − μ12 + μ21( )x[ ]dt + σ x(1− x)dW (t)
with mutation:Mutation induces a drift term in the FP and sd equation
€
n1(t + δt) = n1(t) + 1N μ12 n2 − μ21 n1[ ] = n1(t) + 1
N μ12 N − n1( ) − μ21 n1⇒[ ]
⇒
dx = μ12 − μ12 + μ21( )x[ ]dt + σ x(1− x)dW (t)
∂P
∂t= −
∂
∂xμ12 − μ12 + μ21( )x( )P[ ] + 1
2 σ 2 ∂ 2
∂x 2x(1− x)P( )
with mutation:Mutation induces a drift term in the FP and sd equation
€
n1(t + δt) = n1(t) + 1N μ12 n2 − μ21 n1[ ] = n1(t) + 1
N μ12 N − n1( ) − μ21 n1⇒[ ]
⇒
dx = μ12 − μ12 + μ21( )x[ ]dt + σ x(1− x)dW (t)
∂P
∂t= −
∂
∂xμ12 − μ12 + μ21( )x( )P[ ] + 1
2 σ 2 ∂ 2
∂x 2x(1− x)P( )
stationary solution:
with mutation:Mutation induces a drift term in the FP and sd equation
€
n1(t + δt) = n1(t) + 1N μ12 n2 − μ21 n1[ ] = n1(t) + 1
N μ12 N − n1( ) − μ21 n1⇒[ ]
⇒
dx = μ12 − μ12 + μ21( )x[ ]dt + σ x(1− x)dW (t)
∂P
∂t= −
∂
∂xμ12 − μ12 + μ21( )x( )P[ ] + 1
2 σ 2 ∂ 2
∂x 2x(1− x)P( )
stationary solution:
€
dx = 0 ⇒ x =μ12
μ12 + μ21
fluctuationsUse Ito’s lemma on F(x) = x2:
fluctuationsUse Ito’s lemma on F(x) = x2:
€
d x 2 = 2 μ12 − μ12 + μ21( )x[ ]xdt + σ 2x(1− x)dt
fluctuationsUse Ito’s lemma on F(x) = x2:
at steady state:
€
d x 2 = 2 μ12 − μ12 + μ21( )x[ ]xdt + σ 2x(1− x)dt
€
μ12 + 12 σ 2
( ) x = μ12 + μ21 + 12 σ 2
( ) x 2
fluctuationsUse Ito’s lemma on F(x) = x2:
at steady state:
€
d x 2 = 2 μ12 − μ12 + μ21( )x[ ]xdt + σ 2x(1− x)dt
€
μ12 + 12 σ 2
( ) x = μ12 + μ21 + 12 σ 2
( ) x 2
x 2 =μ12 + 1
2 σ 2( ) x
μ12 + μ21 + 12 σ 2
( )
fluctuationsUse Ito’s lemma on F(x) = x2:
at steady state:
€
d x 2 = 2 μ12 − μ12 + μ21( )x[ ]xdt + σ 2x(1− x)dt
€
μ12 + 12 σ 2
( ) x = μ12 + μ21 + 12 σ 2
( ) x 2
x 2 =μ12 + 1
2 σ 2( ) x
μ12 + μ21 + 12 σ 2
( )=
μ12 μ12 + 12 σ 2
( )
μ12 + μ21( ) μ12 + μ21 + 12 σ 2
( )
fluctuationsUse Ito’s lemma on F(x) = x2:
at steady state:
€
d x 2 = 2 μ12 − μ12 + μ21( )x[ ]xdt + σ 2x(1− x)dt
€
μ12 + 12 σ 2
( ) x = μ12 + μ21 + 12 σ 2
( ) x 2
x 2 =μ12 + 1
2 σ 2( ) x
μ12 + μ21 + 12 σ 2
( )=
μ12 μ12 + 12 σ 2
( )
μ12 + μ21( ) μ12 + μ21 + 12 σ 2
( )
€
x 2 − x2
=12 σ 2μ12μ21
μ12 + μ21( )2
μ12 + μ21 + 12 σ 2
( )
mean square fluctuations:
heterozygocity:
€
H = 2 x − x 2( ) = 2 x 1−
μ12 + 12 σ 2
( )
μ12 + μ21 + 12 σ 2
( )
⎡
⎣ ⎢ ⎢
⎤
⎦ ⎥ ⎥
heterozygocity:
€
H = 2 x − x 2( ) = 2 x 1−
μ12 + 12 σ 2
( )
μ12 + μ21 + 12 σ 2
( )
⎡
⎣ ⎢ ⎢
⎤
⎦ ⎥ ⎥
=2μ12μ21
μ12 + μ21( ) μ12 + μ21 + 12 σ 2
( )
heterozygocity:
€
H = 2 x − x 2( ) = 2 x 1−
μ12 + 12 σ 2
( )
μ12 + μ21 + 12 σ 2
( )
⎡
⎣ ⎢ ⎢
⎤
⎦ ⎥ ⎥
=2μ12μ21
μ12 + μ21( ) μ12 + μ21 + 12 σ 2
( )
small noise (large population):
heterozygocity:
€
H = 2 x − x 2( ) = 2 x 1−
μ12 + 12 σ 2
( )
μ12 + μ21 + 12 σ 2
( )
⎡
⎣ ⎢ ⎢
⎤
⎦ ⎥ ⎥
=2μ12μ21
μ12 + μ21( ) μ12 + μ21 + 12 σ 2
( )
small noise (large population):
€
H →2μ12μ21
μ12 + μ21( )2 = x 1− x ; x 2 − x
2∝σ 2 → 0
heterozygocity:
€
H = 2 x − x 2( ) = 2 x 1−
μ12 + 12 σ 2
( )
μ12 + μ21 + 12 σ 2
( )
⎡
⎣ ⎢ ⎢
⎤
⎦ ⎥ ⎥
=2μ12μ21
μ12 + μ21( ) μ12 + μ21 + 12 σ 2
( )
small noise (large population):
€
H →2μ12μ21
μ12 + μ21( )2 = x 1− x ; x 2 − x
2∝σ 2 → 0
large noise (small population):
heterozygocity:
€
H = 2 x − x 2( ) = 2 x 1−
μ12 + 12 σ 2
( )
μ12 + μ21 + 12 σ 2
( )
⎡
⎣ ⎢ ⎢
⎤
⎦ ⎥ ⎥
=2μ12μ21
μ12 + μ21( ) μ12 + μ21 + 12 σ 2
( )
small noise (large population):
€
H →2μ12μ21
μ12 + μ21( )2 = x 1− x ; x 2 − x
2∝σ 2 → 0
large noise (small population):
€
H →4μ12μ21
μ12 + μ21( )σ 2⇒
heterozygocity:
€
H = 2 x − x 2( ) = 2 x 1−
μ12 + 12 σ 2
( )
μ12 + μ21 + 12 σ 2
( )
⎡
⎣ ⎢ ⎢
⎤
⎦ ⎥ ⎥
=2μ12μ21
μ12 + μ21( ) μ12 + μ21 + 12 σ 2
( )
small noise (large population):
€
H →2μ12μ21
μ12 + μ21( )2 = x 1− x ; x 2 − x
2∝σ 2 → 0
large noise (small population):
€
H →4μ12μ21
μ12 + μ21( )σ 2⇒ usually one allele dominates, rare transitions
selectionLet the alleles chosen to reproduce do so with with probabilities
€
p1 =w1x
w1x + w2(1− x), p2 =
w2(1− x)
w1x + w2(1− x)
selectionLet the alleles chosen to reproduce do so with with probabilities
€
p1 =w1x
w1x + w2(1− x), p2 =
w2(1− x)
w1x + w2(1− x)
Now, if there are n1 organisms of type 1 before this step, then afterwards there are
selectionLet the alleles chosen to reproduce do so with with probabilities
€
p1 =w1x
w1x + w2(1− x), p2 =
w2(1− x)
w1x + w2(1− x)
Now, if there are n1 organisms of type 1 before this step, then afterwards there are
n1 + 1 with probability p1(1 – x)
selectionLet the alleles chosen to reproduce do so with with probabilities
€
p1 =w1x
w1x + w2(1− x), p2 =
w2(1− x)
w1x + w2(1− x)
Now, if there are n1 organisms of type 1 before this step, then afterwards there are
n1 + 1 with probability p1(1 – x)n1 – 1 with probability p2x
selectionLet the alleles chosen to reproduce do so with with probabilities
€
p1 =w1x
w1x + w2(1− x), p2 =
w2(1− x)
w1x + w2(1− x)
This leads to a drift in x proportional to x(1 - x):
Now, if there are n1 organisms of type 1 before this step, then afterwards there are
n1 + 1 with probability p1(1 – x)n1 – 1 with probability p2x
selectionLet the alleles chosen to reproduce do so with with probabilities
€
p1 =w1x
w1x + w2(1− x), p2 =
w2(1− x)
w1x + w2(1− x)
This leads to a drift in x proportional to x(1 - x):
Now, if there are n1 organisms of type 1 before this step, then afterwards there are
n1 + 1 with probability p1(1 – x)n1 – 1 with probability p2x
€
dx = μ12 − μ12 + μ21( )x[ ]dt + sx(1− x)dt + σ x(1− x)dW (t);
s =2(w1 − w2)
(w1 + w2)(s <<1)
selection: large population limit
with selection but no mutations:
€
dx
dt= sx(1− x)
selection: large population limit
with selection but no mutations:
€
dx
dt= sx(1− x)
logx
1− x⋅1− x0
x0
⎛
⎝ ⎜
⎞
⎠ ⎟= stsolution:
selection: large population limit
with selection but no mutations:
€
dx
dt= sx(1− x)
logx
1− x⋅1− x0
x0
⎛
⎝ ⎜
⎞
⎠ ⎟= st
x =1
1+1− x0
x0
exp(−st)
solution:
selection: large population limit
with selection but no mutations:
€
dx
dt= sx(1− x)
logx
1− x⋅1− x0
x0
⎛
⎝ ⎜
⎞
⎠ ⎟= st
x =1
1+1− x0
x0
exp(−st)
t >>1
slog
1− x0
x0
⎛
⎝ ⎜
⎞
⎠ ⎟: x →1
solution:
Neurons
Neurons receive synaptic input from other neurons
Neurons
Neurons receive synaptic input from other neurons~ injected current
Neurons
Neurons receive synaptic input from other neurons~ injected current
€
CdV
dt= I(t) V measured from resting potential
Neurons
Neurons receive synaptic input from other neurons~ injected current
€
CdV
dt= I(t)
CdV
dt= −gV + I(t)
V measured from resting potential
with leak g = membrane conductance
Neurons
Neurons receive synaptic input from other neurons~ injected current
€
CdV
dt= I(t)
CdV
dt= −gV + I(t)
V measured from resting potential
with leak g = membrane conductance
(experimental fact:) input current is noisy, very small τc compared tomembrane time constant τ = C/g
Neurons
Neurons receive synaptic input from other neurons~ injected current
€
CdV
dt= I(t)
CdV
dt= −gV + I(t)
V measured from resting potential
with leak g = membrane conductance
(experimental fact:) input current is noisy, very small τc compared tomembrane time constant τ = C/g
V(t) is described by Wiener process (g = 0) or Brownian motion (g ≠ 0)
SpikesThe above is approximately true as long as V stays below a criticalvalue VT.
SpikesThe above is approximately true as long as V stays below a criticalvalue VT. Above this threshold, active ion channels amplify incomingcurrents and produce an action potential (spike).
SpikesThe above is approximately true as long as V stays below a criticalvalue VT. Above this threshold, active ion channels amplify incomingcurrents and produce an action potential (spike). After a few ms, V returns to its sub-threshold equilibrium level.
SpikesThe above is approximately true as long as V stays below a criticalvalue VT. Above this threshold, active ion channels amplify incomingcurrents and produce an action potential (spike). After a few ms, V returns to its sub-threshold equilibrium level.
“integrate-and-fire” or “leaky integrate-and-fire” model of a neuron
SpikesThe above is approximately true as long as V stays below a criticalvalue VT. Above this threshold, active ion channels amplify incomingcurrents and produce an action potential (spike). After a few ms, V returns to its sub-threshold equilibrium level.
“integrate-and-fire” or “leaky integrate-and-fire” model of a neuron
our question here: if I(t) is white noise, what is the distribution ofinterspike intervals?
SpikesThe above is approximately true as long as V stays below a criticalvalue VT. Above this threshold, active ion channels amplify incomingcurrents and produce an action potential (spike). After a few ms, V returns to its sub-threshold equilibrium level.
“integrate-and-fire” or “leaky integrate-and-fire” model of a neuron
our question here: if I(t) is white noise, what is the distribution ofinterspike intervals?
This is a first-passage-time problem
with no leak:
€
CV ≡ x
dx
dt= I(t), x < θ = CVT ; I(t)I( ′ t ) = σ 2δ(t − ′ t )
with no leak:
Assume at t = 0, x = 0
€
CV ≡ x
dx
dt= I(t), x < θ = CVT ; I(t)I( ′ t ) = σ 2δ(t − ′ t )
with no leak:
Assume at t = 0, x = 0boundary condition at θ: P(θ) = 0.
€
CV ≡ x
dx
dt= I(t), x < θ = CVT ; I(t)I( ′ t ) = σ 2δ(t − ′ t )
with no leak:
Assume at t = 0, x = 0boundary condition at θ: P(θ) = 0.
€
CV ≡ x
dx
dt= I(t), x < θ = CVT ; I(t)I( ′ t ) = σ 2δ(t − ′ t )
We have solved this problem when there is no threshold:
with no leak:
Assume at t = 0, x = 0boundary condition at θ: P(θ) = 0.
€
CV ≡ x
dx
dt= I(t), x < θ = CVT ; I(t)I( ′ t ) = σ 2δ(t − ′ t )
We have solved this problem when there is no threshold:
€
P(x, t) =1
2πσ 2texp −
x 2
2σ 2t
⎛
⎝ ⎜
⎞
⎠ ⎟
with no leak:
Assume at t = 0, x = 0boundary condition at θ: P(θ) = 0.
€
CV ≡ x
dx
dt= I(t), x < θ = CVT ; I(t)I( ′ t ) = σ 2δ(t − ′ t )
We have solved this problem when there is no threshold:
€
P(x, t) =1
2πσ 2texp −
x 2
2σ 2t
⎛
⎝ ⎜
⎞
⎠ ⎟
But it does not satisfy the boundary condition.
solution with images:
Add an extra source, of opposite sign, at x = 2θ:
€
P(x, t) =1
2πσ 2texp −
x 2
2σ 2t
⎛
⎝ ⎜
⎞
⎠ ⎟− exp −
(x − 2θ)2
2σ 2t
⎛
⎝ ⎜
⎞
⎠ ⎟
⎡
⎣ ⎢
⎤
⎦ ⎥
solution with images:
Add an extra source, of opposite sign, at x = 2θ:
€
P(x, t) =1
2πσ 2texp −
x 2
2σ 2t
⎛
⎝ ⎜
⎞
⎠ ⎟− exp −
(x − 2θ)2
2σ 2t
⎛
⎝ ⎜
⎞
⎠ ⎟
⎡
⎣ ⎢
⎤
⎦ ⎥
cumulative probability of firing by t:
€
F(t) = 2dx
2πσ 2texp −
x 2
2σ 2t
⎛
⎝ ⎜
⎞
⎠ ⎟
θ
∞
∫
solution with images:
Add an extra source, of opposite sign, at x = 2θ:
€
P(x, t) =1
2πσ 2texp −
x 2
2σ 2t
⎛
⎝ ⎜
⎞
⎠ ⎟− exp −
(x − 2θ)2
2σ 2t
⎛
⎝ ⎜
⎞
⎠ ⎟
⎡
⎣ ⎢
⎤
⎦ ⎥
cumulative probability of firing by t:
€
F(t) = 2dx
2πσ 2texp −
x 2
2σ 2t
⎛
⎝ ⎜
⎞
⎠ ⎟
θ
∞
∫
€
f (t) =dF(t)
dt= 2
d
dt
du
2πe− 1
2 u2
θ
σ t
∞
∫ =θ
2πσ 2t 3exp −
θ 2
2σ 2t
⎛
⎝ ⎜
⎞
⎠ ⎟
interspike interval density:
solution with images:
Add an extra source, of opposite sign, at x = 2θ:
€
P(x, t) =1
2πσ 2texp −
x 2
2σ 2t
⎛
⎝ ⎜
⎞
⎠ ⎟− exp −
(x − 2θ)2
2σ 2t
⎛
⎝ ⎜
⎞
⎠ ⎟
⎡
⎣ ⎢
⎤
⎦ ⎥
cumulative probability of firing by t:
€
F(t) = 2dx
2πσ 2texp −
x 2
2σ 2t
⎛
⎝ ⎜
⎞
⎠ ⎟
θ
∞
∫
€
f (t) =dF(t)
dt= 2
d
dt
du
2πe− 1
2 u2
θ
σ t
∞
∫ =θ
2πσ 2t 3exp −
θ 2
2σ 2t
⎛
⎝ ⎜
⎞
⎠ ⎟
interspike interval density:
Levy distribution (one-sided stable distribution with α = ½
another way to get the answer:
The event rate is just the (diffusive) current
€
J(x) = −D∂P
∂x= − 1
2 σ 2 ∂P
∂x
evaluated at x = θ.
another way to get the answer:
The event rate is just the (diffusive) current
€
J(x) = −D∂P
∂x= − 1
2 σ 2 ∂P
∂x
evaluated at x = θ.
€
f (t) =d
dx
1
2πσ 2texp −
x 2
2σ 2t
⎛
⎝ ⎜
⎞
⎠ ⎟−
1
2πσ 2texp −
(x − 2θ)2
2σ 2t
⎛
⎝ ⎜
⎞
⎠ ⎟
⎡
⎣ ⎢
⎤
⎦ ⎥x=θ
another way to get the answer:
The event rate is just the (diffusive) current
€
J(x) = −D∂P
∂x= − 1
2 σ 2 ∂P
∂x
evaluated at x = θ.
€
f (t) =d
dx
1
2πσ 2texp −
x 2
2σ 2t
⎛
⎝ ⎜
⎞
⎠ ⎟−
1
2πσ 2texp −
(x − 2θ)2
2σ 2t
⎛
⎝ ⎜
⎞
⎠ ⎟
⎡
⎣ ⎢
⎤
⎦ ⎥x=θ
=2
2πσ 2t
d
dxexp −
x 2
2σ 2t
⎛
⎝ ⎜
⎞
⎠ ⎟
⎡
⎣ ⎢
⎤
⎦ ⎥x=θ
=θ
2πσ 2t 3exp −
θ 2
2σ 2t
⎛
⎝ ⎜
⎞
⎠ ⎟
a problem:The mean interspike interval is infinite:
a problem:The mean interspike interval is infinite:
€
t = tf (t)dt0
∞
∫ ~tdt
t 3 / 2∫ = ∞
a problem:The mean interspike interval is infinite:
€
t = tf (t)dt0
∞
∫ ~tdt
t 3 / 2∫ = ∞
so the firing rate (= 1/<t>) is zero!
adding a constant drift term:
€
dx
dt= μ + I(t), x < θ = CVT ; I(t)I( ′ t ) = σ 2δ(t − ′ t )
adding a constant drift term:
€
dx
dt= μ + I(t), x < θ = CVT ; I(t)I( ′ t ) = σ 2δ(t − ′ t )
solution with no boundary:
€
P(x, t) =1
2πσ 2texp −
(x − μt)2
2σ 2t
⎛
⎝ ⎜
⎞
⎠ ⎟
adding a constant drift term:
€
dx
dt= μ + I(t), x < θ = CVT ; I(t)I( ′ t ) = σ 2δ(t − ′ t )
solution with no boundary:
€
P(x, t) =1
2πσ 2texp −
(x − μt)2
2σ 2t
⎛
⎝ ⎜
⎞
⎠ ⎟
need a moving image:
€
P(x, t) =1
2πσ 2texp −
(x − μt)2
2σ 2t
⎛
⎝ ⎜
⎞
⎠ ⎟−
C
2πσ 2texp −
(x − 2θ − μt)2
2σ 2t
⎛
⎝ ⎜
⎞
⎠ ⎟
adding a constant drift term:
€
dx
dt= μ + I(t), x < θ = CVT ; I(t)I( ′ t ) = σ 2δ(t − ′ t )
solution with no boundary:
€
P(x, t) =1
2πσ 2texp −
(x − μt)2
2σ 2t
⎛
⎝ ⎜
⎞
⎠ ⎟
need a moving image:
€
P(x, t) =1
2πσ 2texp −
(x − μt)2
2σ 2t
⎛
⎝ ⎜
⎞
⎠ ⎟−
C
2πσ 2texp −
(x − 2θ − μt)2
2σ 2t
⎛
⎝ ⎜
⎞
⎠ ⎟
€
P(θ, t) = 0 ⇒ exp −(θ − μt)2
2σ 2t
⎛
⎝ ⎜
⎞
⎠ ⎟= C exp −
(θ + μt)2
2σ 2t
⎛
⎝ ⎜
⎞
⎠ ⎟
adding a constant drift term:
€
dx
dt= μ + I(t), x < θ = CVT ; I(t)I( ′ t ) = σ 2δ(t − ′ t )
solution with no boundary:
€
P(x, t) =1
2πσ 2texp −
(x − μt)2
2σ 2t
⎛
⎝ ⎜
⎞
⎠ ⎟
need a moving image:
€
P(x, t) =1
2πσ 2texp −
(x − μt)2
2σ 2t
⎛
⎝ ⎜
⎞
⎠ ⎟−
C
2πσ 2texp −
(x − 2θ − μt)2
2σ 2t
⎛
⎝ ⎜
⎞
⎠ ⎟
€
P(θ, t) = 0 ⇒ exp −(θ − μt)2
2σ 2t
⎛
⎝ ⎜
⎞
⎠ ⎟= C exp −
(θ + μt)2
2σ 2t
⎛
⎝ ⎜
⎞
⎠ ⎟ ⇒ C = exp
2μθ
σ 2
⎛
⎝ ⎜
⎞
⎠ ⎟
adding a constant drift term:
€
dx
dt= μ + I(t), x < θ = CVT ; I(t)I( ′ t ) = σ 2δ(t − ′ t )
solution with no boundary:
€
P(x, t) =1
2πσ 2texp −
(x − μt)2
2σ 2t
⎛
⎝ ⎜
⎞
⎠ ⎟
need a moving image:
€
P(x, t) =1
2πσ 2texp −
(x − μt)2
2σ 2t
⎛
⎝ ⎜
⎞
⎠ ⎟−
C
2πσ 2texp −
(x − 2θ − μt)2
2σ 2t
⎛
⎝ ⎜
⎞
⎠ ⎟
€
P(θ, t) = 0 ⇒ exp −(θ − μt)2
2σ 2t
⎛
⎝ ⎜
⎞
⎠ ⎟= C exp −
(θ + μt)2
2σ 2t
⎛
⎝ ⎜
⎞
⎠ ⎟ ⇒ C = exp
2μθ
σ 2
⎛
⎝ ⎜
⎞
⎠ ⎟
€
P(x, t) =1
2πσ 2texp −
(x − μt)2
2σ 2t
⎛
⎝ ⎜
⎞
⎠ ⎟−
1
2πσ 2texp
2μθ
σ 2
⎛
⎝ ⎜
⎞
⎠ ⎟exp −
(x − 2θ − μt)2
2σ 2t
⎛
⎝ ⎜
⎞
⎠ ⎟
solution:
ISI distribution:
€
J(x) = −D∂P
∂x= − 1
2 σ 2 ∂P
∂xfrom
ISI distribution:
€
J(x) = −D∂P
∂x= − 1
2 σ 2 ∂P
∂xfrom
€
f (t) = −12 σ 2
2πDσ 2t
d
dxexp −
(x − μt)2
2σ 2t
⎛
⎝ ⎜
⎞
⎠ ⎟− C exp −
(x − μt − 2θ)2
2σ 2t
⎛
⎝ ⎜
⎞
⎠ ⎟
⎡
⎣ ⎢
⎤
⎦ ⎥x=θ
ISI distribution:
€
J(x) = −D∂P
∂x= − 1
2 σ 2 ∂P
∂xfrom
€
f (t) = −12 σ 2
2πDσ 2t
d
dxexp −
(x − μt)2
2σ 2t
⎛
⎝ ⎜
⎞
⎠ ⎟− C exp −
(x − μt − 2θ)2
2σ 2t
⎛
⎝ ⎜
⎞
⎠ ⎟
⎡
⎣ ⎢
⎤
⎦ ⎥x=θ
=θ
2πσ 2 t 3 / 2exp −
(θ − μt)2
2σ 2t
⎛
⎝ ⎜
⎞
⎠ ⎟
ISI distribution:
€
J(x) = −D∂P
∂x= − 1
2 σ 2 ∂P
∂xfrom
€
f (t) = −12 σ 2
2πDσ 2t
d
dxexp −
(x − μt)2
2σ 2t
⎛
⎝ ⎜
⎞
⎠ ⎟− C exp −
(x − μt − 2θ)2
2σ 2t
⎛
⎝ ⎜
⎞
⎠ ⎟
⎡
⎣ ⎢
⎤
⎦ ⎥x=θ
=θ
2πσ 2 t 3 / 2exp −
(θ − μt)2
2σ 2t
⎛
⎝ ⎜
⎞
⎠ ⎟
Now all moments of f are finite
ISI distribution:
€
J(x) = −D∂P
∂x= − 1
2 σ 2 ∂P
∂xfrom
€
f (t) = −12 σ 2
2πDσ 2t
d
dxexp −
(x − μt)2
2σ 2t
⎛
⎝ ⎜
⎞
⎠ ⎟− C exp −
(x − μt − 2θ)2
2σ 2t
⎛
⎝ ⎜
⎞
⎠ ⎟
⎡
⎣ ⎢
⎤
⎦ ⎥x=θ
=θ
2πσ 2 t 3 / 2exp −
(θ − μt)2
2σ 2t
⎛
⎝ ⎜
⎞
⎠ ⎟
Now all moments of f are finite
leaky I&F neuron
€
dx
dt= −γx + I0 + δI(t) (γ = 1/τ = g/C
leaky I&F neuron
€
dx
dt= −γx + I0 + δI(t) (γ = 1/τ = g/C
€
δI(t) = 0
€
δI(t)δI( ′ t ) = σ 2δ(t − ′ t ) = 2Dδ(t − ′ t )
leaky I&F neuron
€
dx
dt= −γx + I0 + δI(t) (γ = 1/τ = g/C
€
δI(t) = 0
€
δI(t)δI( ′ t ) = σ 2δ(t − ′ t ) = 2Dδ(t − ′ t )
= Brownian motion with an added constant drift
leaky I&F neuron
€
dx
dt= −γx + I0 + δI(t) (γ = 1/τ = g/C
€
δI(t) = 0
€
δI(t)δI( ′ t ) = σ 2δ(t − ′ t ) = 2Dδ(t − ′ t )
= Brownian motion with an added constant drift
€
∂P(x, t)
∂t= −
∂J
∂x= −
∂
∂xI0 − γx( )P(x, t) − D
∂P(x, t)
∂x
⎡ ⎣ ⎢
⎤ ⎦ ⎥
leaky I&F neuron
€
dx
dt= −γx + I0 + δI(t) (γ = 1/τ = g/C
€
δI(t) = 0
€
δI(t)δI( ′ t ) = σ 2δ(t − ′ t ) = 2Dδ(t − ′ t )
= Brownian motion with an added constant drift
€
∂P(x, t)
∂t= −
∂J
∂x= −
∂
∂xI0 − γx( )P(x, t) − D
∂P(x, t)
∂x
⎡ ⎣ ⎢
⎤ ⎦ ⎥
(set γ = 1 for convenience)
Looking for stationary solution
€
∂P
∂t= 0
Looking for stationary solution
€
∂P
∂t= 0
€
dJ
dx=
d
dxI0 − x( )P(x) − D
∂P(x)
∂x
⎡ ⎣ ⎢
⎤ ⎦ ⎥= 0i.e.
Looking for stationary solution
€
∂P
∂t= 0
€
dJ
dx=
d
dxI0 − x( )P(x) − D
∂P(x)
∂x
⎡ ⎣ ⎢
⎤ ⎦ ⎥= 0
€
J(x) = I0 − x( )P(x) − D∂P(x)
∂x= const
i.e.
=>
Looking for stationary solution
€
∂P
∂t= 0
€
dJ
dx=
d
dxI0 − x( )P(x) − D
∂P(x)
∂x
⎡ ⎣ ⎢
⎤ ⎦ ⎥= 0
€
J(x) = I0 − x( )P(x) − D∂P(x)
∂x= const
Boundary conditions:
i.e.
=>
Looking for stationary solution
€
∂P
∂t= 0
€
dJ
dx=
d
dxI0 − x( )P(x) − D
∂P(x)
∂x
⎡ ⎣ ⎢
⎤ ⎦ ⎥= 0
€
J(x) = I0 − x( )P(x) − D∂P(x)
∂x= const
Boundary conditions: sink at firing threshold x
i.e.
=>
Looking for stationary solution
€
∂P
∂t= 0
€
dJ
dx=
d
dxI0 − x( )P(x) − D
∂P(x)
∂x
⎡ ⎣ ⎢
⎤ ⎦ ⎥= 0
€
J(x) = I0 − x( )P(x) − D∂P(x)
∂x= const
Boundary conditions: sink at firing threshold x source at x = 0
i.e.
=>
Looking for stationary solution
€
∂P
∂t= 0
€
dJ
dx=
d
dxI0 − x( )P(x) − D
∂P(x)
∂x
⎡ ⎣ ⎢
⎤ ⎦ ⎥= 0
€
J(x) = I0 − x( )P(x) − D∂P(x)
∂x= const
Boundary conditions: sink at firing threshold x source at x = 0
€
P(x) = 0, x ≥ θ
i.e.
=>
Looking for stationary solution
€
∂P
∂t= 0
€
dJ
dx=
d
dxI0 − x( )P(x) − D
∂P(x)
∂x
⎡ ⎣ ⎢
⎤ ⎦ ⎥= 0
€
J(x) = I0 − x( )P(x) − D∂P(x)
∂x= const
Boundary conditions: sink at firing threshold x source at x = 0
€
P(x) = 0, x ≥ θ
€
J(x) = 0, x > θ and x < 0
i.e.
=>
Looking for stationary solution
€
∂P
∂t= 0
€
dJ
dx=
d
dxI0 − x( )P(x) − D
∂P(x)
∂x
⎡ ⎣ ⎢
⎤ ⎦ ⎥= 0
€
J(x) = I0 − x( )P(x) − D∂P(x)
∂x= const
Boundary conditions: sink at firing threshold x source at x = 0
€
P(x) = 0, x ≥ θ
€
J(x) = 0, x > θ and x < 0
€
r = J(θ−) = −DdP
dx
⎛
⎝ ⎜
⎞
⎠ ⎟x=θ
i.e.
=>
Firing rate: current out at threshold:
Looking for stationary solution
€
∂P
∂t= 0
€
dJ
dx=
d
dxI0 − x( )P(x) − D
∂P(x)
∂x
⎡ ⎣ ⎢
⎤ ⎦ ⎥= 0
€
J(x) = I0 − x( )P(x) − D∂P(x)
∂x= const
Boundary conditions: sink at firing threshold x source at x = 0
€
P(x) = 0, x ≥ θ
€
J(x) = 0, x > θ and x < 0
€
r = J(θ−) = −DdP
dx
⎛
⎝ ⎜
⎞
⎠ ⎟x=θ
€
r = J(0+) = I0P(0) − DdP
dx
⎛
⎝ ⎜
⎞
⎠ ⎟x= 0
i.e.
=>
Firing rate: current out at threshold: = reinjection rate at reset:
Stationary solution (2)
Also need normalization:
€
dxP(x) =1−∞
∞
∫
Stationary solution (2)
Also need normalization:
€
dxP(x) =1−∞
∞
∫
Below reset level, J :
€
I0 − x( )P(x) − D∂P(x)
∂x= 0
Stationary solution (2)
Also need normalization:
€
dxP(x) =1−∞
∞
∫
Below reset level, J :
€
I0 − x( )P(x) − D∂P(x)
∂x= 0
has solution
€
P(x) = c1 exp −(x − I0)2
2D
⎡
⎣ ⎢
⎤
⎦ ⎥
Stationary solution (2)
Also need normalization:
€
dxP(x) =1−∞
∞
∫
Below reset level, J :
€
I0 − x( )P(x) − D∂P(x)
∂x= 0
has solution
€
P(x) = c1 exp −(x − I0)2
2D
⎡
⎣ ⎢
⎤
⎦ ⎥
€
P(x) = c2 exp −(x − I0)2
2D
⎡
⎣ ⎢
⎤
⎦ ⎥ dy exp
(y − I0)2
2D
⎡
⎣ ⎢
⎤
⎦ ⎥
x
θ
∫Between rest and threshold:
Stationary solution (2)
Also need normalization:
€
dxP(x) =1−∞
∞
∫
Below reset level, J :
€
I0 − x( )P(x) − D∂P(x)
∂x= 0
has solution
€
P(x) = c1 exp −(x − I0)2
2D
⎡
⎣ ⎢
⎤
⎦ ⎥
€
P(x) = c2 exp −(x − I0)2
2D
⎡
⎣ ⎢
⎤
⎦ ⎥ dy exp
(y − I0)2
2D
⎡
⎣ ⎢
⎤
⎦ ⎥
x
θ
∫Between rest and threshold:
B.C. at x :
€
r = −DdP
dx
⎛
⎝ ⎜
⎞
⎠ ⎟x=θ
= −Dc2 exp −(θ − I0)2
2D
⎡
⎣ ⎢
⎤
⎦ ⎥ −exp
(θ − I0)2
2D
⎡
⎣ ⎢
⎤
⎦ ⎥
⎛
⎝ ⎜
⎞
⎠ ⎟= Dc2
Stationary solution (2)
Also need normalization:
€
dxP(x) =1−∞
∞
∫
Below reset level, J :
€
I0 − x( )P(x) − D∂P(x)
∂x= 0
has solution
€
P(x) = c1 exp −(x − I0)2
2D
⎡
⎣ ⎢
⎤
⎦ ⎥
€
P(x) = c2 exp −(x − I0)2
2D
⎡
⎣ ⎢
⎤
⎦ ⎥ dy exp
(y − I0)2
2D
⎡
⎣ ⎢
⎤
⎦ ⎥
x
θ
∫Between rest and threshold:
B.C. at x :
€
r = −DdP
dx
⎛
⎝ ⎜
⎞
⎠ ⎟x=θ
= −Dc2 exp −(θ − I0)2
2D
⎡
⎣ ⎢
⎤
⎦ ⎥ −exp
(θ − I0)2
2D
⎡
⎣ ⎢
⎤
⎦ ⎥
⎛
⎝ ⎜
⎞
⎠ ⎟= Dc2
=>
€
c2 =r
D
Stationary solution (3)
Continuity at x = =>
€
c1 exp −I0
2
2D
⎡
⎣ ⎢
⎤
⎦ ⎥= c2 exp −
I02
2D
⎡
⎣ ⎢
⎤
⎦ ⎥ dy exp
(y − I0)2
2D
⎡
⎣ ⎢
⎤
⎦ ⎥
0
θ
∫
Stationary solution (3)
Continuity at x = =>
€
c1 exp −I0
2
2D
⎡
⎣ ⎢
⎤
⎦ ⎥= c2 exp −
I02
2D
⎡
⎣ ⎢
⎤
⎦ ⎥ dy exp
(y − I0)2
2D
⎡
⎣ ⎢
⎤
⎦ ⎥
0
θ
∫
i.e.,
€
c1 = c2 dy exp(y − I0)2
2D
⎡
⎣ ⎢
⎤
⎦ ⎥
0
θ
∫
Stationary solution (3)
Continuity at x = =>
€
c1 exp −I0
2
2D
⎡
⎣ ⎢
⎤
⎦ ⎥= c2 exp −
I02
2D
⎡
⎣ ⎢
⎤
⎦ ⎥ dy exp
(y − I0)2
2D
⎡
⎣ ⎢
⎤
⎦ ⎥
0
θ
∫
i.e.,
€
c1 = c2 dy exp(y − I0)2
2D
⎡
⎣ ⎢
⎤
⎦ ⎥
0
θ
∫
algebra … =>
€
r =1
π dx−I 0 / 2D
(θ −I 0 ) / 2D
∫ exp(x 2)(1+ erf x)=
1
π dx−I 0 /σ
(θ −I 0 ) /σ
∫ exp(x 2)(1+ erf x)
Stationary solution (3)
Continuity at x = =>
€
c1 exp −I0
2
2D
⎡
⎣ ⎢
⎤
⎦ ⎥= c2 exp −
I02
2D
⎡
⎣ ⎢
⎤
⎦ ⎥ dy exp
(y − I0)2
2D
⎡
⎣ ⎢
⎤
⎦ ⎥
0
θ
∫
i.e.,
€
c1 = c2 dy exp(y − I0)2
2D
⎡
⎣ ⎢
⎤
⎦ ⎥
0
θ
∫
algebra … =>
€
r =1
π dx−I 0 / 2D
(θ −I 0 ) / 2D
∫ exp(x 2)(1+ erf x)=
1
π dx−I 0 /σ
(θ −I 0 ) /σ
∫ exp(x 2)(1+ erf x)
with refractory time τr
€
r =1
τ r + π dx−I 0 /σ
(θ −I 0 ) /σ
∫ exp(x 2)(1+ erf x)
A simple model of evolution: the Bak-Sneppen model
N species,each with fitness xi, each uniformly distributed on (0,1)
A simple model of evolution: the Bak-Sneppen model
N species,each with fitness xi, each uniformly distributed on (0,1)
A simple model of evolution: the Bak-Sneppen model
N species,each with fitness xi, each uniformly distributed on (0,1)
evolutionary step:
A simple model of evolution: the Bak-Sneppen model
N species,each with fitness xi, each uniformly distributed on (0,1)
evolutionary step: eliminate the weakest species (smallest xi)
A simple model of evolution: the Bak-Sneppen model
N species,each with fitness xi, each uniformly distributed on (0,1)
evolutionary step: eliminate the weakest species (smallest xi)replace it with another species with a random xi
A simple model of evolution: the Bak-Sneppen model
N species,each with fitness xi, each uniformly distributed on (0,1)
evolutionary step: eliminate the weakest species (smallest xi)replace it with another species with a random xi
(random-neighbour version) assume another (“neighboring”)species also becomes extinct; replace it with a new one, too
A simple model of evolution: the Bak-Sneppen model
N species,each with fitness xi, each uniformly distributed on (0,1)
evolutionary step: eliminate the weakest species (smallest xi)replace it with another species with a random xi
(random-neighbour version) assume another (“neighboring”)species also becomes extinct; replace it with a new one, too
Now one or more of these new ones may get fitnesses below θreplace them and their (randomly chosen) neighbours with new ones until all fitnesses are > θ
A simple model of evolution: the Bak-Sneppen model
N species,each with fitness xi, each uniformly distributed on (0,1)
evolutionary step: eliminate the weakest species (smallest xi)replace it with another species with a random xi
(random-neighbour version) assume another (“neighboring”)species also becomes extinct; replace it with a new one, too
Now one or more of these new ones may get fitnesses below θreplace them and their (randomly chosen) neighbours with new ones until all fitnesses are > θ (avalanche)
A simple model of evolution: the Bak-Sneppen model
N species,each with fitness xi, each uniformly distributed on (0,1)
evolutionary step: eliminate the weakest species (smallest xi)replace it with another species with a random xi
(random-neighbour version) assume another (“neighboring”)species also becomes extinct; replace it with a new one, too
Now one or more of these new ones may get fitnesses below θreplace them and their (randomly chosen) neighbours with new ones until all fitnesses are > θ (avalanche)
start over
A simple model of evolution: the Bak-Sneppen model
N species,each with fitness xi, each uniformly distributed on (0,1)
evolutionary step: eliminate the weakest species (smallest xi)replace it with another species with a random xi
(random-neighbour version) assume another (“neighboring”)species also becomes extinct; replace it with a new one, too
Now one or more of these new ones may get fitnesses below θreplace them and their (randomly chosen) neighbours with new ones until all fitnesses are > θ (avalanche)
start over
Want to know the avalanche length distribution
getting a master equationP(n,t) = prob that n species have fitness values < θ at time t
getting a master equationP(n,t) = prob that n species have fitness values < θ at time tAt each step 2 species are reassigned random new fitnesses
getting a master equationP(n,t) = prob that n species have fitness values < θ at time tAt each step 2 species are reassigned random new fitnessestransition matrix: Tmn:
getting a master equationP(n,t) = prob that n species have fitness values < θ at time tAt each step 2 species are reassigned random new fitnessestransition matrix: Tmn:
€
Tn−1,n = (1−θ)2
Tnn = 2θ(1−θ)
Tn +1,n = θ 2
getting a master equationP(n,t) = prob that n species have fitness values < θ at time tAt each step 2 species are reassigned random new fitnessestransition matrix: Tmn:
€
Tn−1,n = (1−θ)2
Tnn = 2θ(1−θ)
Tn +1,n = θ 2
simple random walk in n with first step n=0 -> n=1, thereafter
getting a master equationP(n,t) = prob that n species have fitness values < θ at time tAt each step 2 species are reassigned random new fitnessestransition matrix: Tmn:
€
Tn−1,n = (1−θ)2
Tnn = 2θ(1−θ)
Tn +1,n = θ 2
net drift per step:
mean square change:
€
Tn +1,n − Tn−1,n = θ 2 − (1−θ)2 = 2 θ − 12( )
Tn +1,n + Tn−1,n =1− 2θ(1−θ)
simple random walk in n with first step n=0 -> n=1, thereafter
getting a master equationP(n,t) = prob that n species have fitness values < θ at time tAt each step 2 species are reassigned random new fitnessestransition matrix: Tmn:
€
Tn−1,n = (1−θ)2
Tnn = 2θ(1−θ)
Tn +1,n = θ 2
net drift per step:
mean square change:
€
Tn +1,n − Tn−1,n = θ 2 − (1−θ)2 = 2 θ − 12( )
Tn +1,n + Tn−1,n =1− 2θ(1−θ)
simple random walk in n with first step n=0 -> n=1, thereafter
Walk (avalanche) ends when n=0 again for the first time.
getting a master equationP(n,t) = prob that n species have fitness values < θ at time tAt each step 2 species are reassigned random new fitnessestransition matrix: Tmn:
€
Tn−1,n = (1−θ)2
Tnn = 2θ(1−θ)
Tn +1,n = θ 2
net drift per step:
mean square change:
€
Tn +1,n − Tn−1,n = θ 2 − (1−θ)2 = 2 θ − 12( )
Tn +1,n + Tn−1,n =1− 2θ(1−θ)
simple random walk in n with first step n=0 -> n=1, thereafter
Walk (avalanche) ends when n=0 again for the first time.
critical case (no drift): θ =½
TrafficNagel-Paczuski model:cars can move with speed v=+1 step/time unit or 0.
TrafficNagel-Paczuski model:cars can move with speed v=+1 step/time unit or 0.jam/kø/queue/file/stau = n cars in a row not moving
TrafficNagel-Paczuski model:cars can move with speed v=+1 step/time unit or 0.jam/kø/queue/file/stau = n cars in a row not movingfirst car in stau can change speed from 0 to +1 with prob p/step
TrafficNagel-Paczuski model:cars can move with speed v=+1 step/time unit or 0.jam/kø/queue/file/stau = n cars in a row not movingfirst car in stau can change speed from 0 to +1 with prob p/stepnew cars enter the stau with prob q
TrafficNagel-Paczuski model:cars can move with speed v=+1 step/time unit or 0.jam/kø/queue/file/stau = n cars in a row not movingfirst car in stau can change speed from 0 to +1 with prob p/stepnew cars enter the stau with prob q transition matrix for stau length:
TrafficNagel-Paczuski model:cars can move with speed v=+1 step/time unit or 0.jam/kø/queue/file/stau = n cars in a row not movingfirst car in stau can change speed from 0 to +1 with prob p/stepnew cars enter the stau with prob q transition matrix for stau length:
€
Tn−1,n = q(1− p)
Tnn = (1− p)(1− q) + pq
Tn +1,n = p(1− q)
TrafficNagel-Paczuski model:cars can move with speed v=+1 step/time unit or 0.jam/kø/queue/file/stau = n cars in a row not movingfirst car in stau can change speed from 0 to +1 with prob p/stepnew cars enter the stau with prob q transition matrix for stau length:
€
Tn−1,n = q(1− p)
Tnn = (1− p)(1− q) + pq
Tn +1,n = p(1− q)
€
Tn +1,n − Tn−1,n = p(1− q) − q(1− p) = p − q
Tn +1,n + Tn−1,n = p(1− q) + q(1− p) = p + q − 2pq
net drift per step:
mean square change:
TrafficNagel-Paczuski model:cars can move with speed v=+1 step/time unit or 0.jam/kø/queue/file/stau = n cars in a row not movingfirst car in stau can change speed from 0 to +1 with prob p/stepnew cars enter the stau with prob q transition matrix for stau length:
€
Tn−1,n = q(1− p)
Tnn = (1− p)(1− q) + pq
Tn +1,n = p(1− q)
€
Tn +1,n − Tn−1,n = p(1− q) − q(1− p) = p − q
Tn +1,n + Tn−1,n = p(1− q) + q(1− p) = p + q − 2pq
net drift per step:
mean square change:
biased random walk again
TrafficNagel-Paczuski model:cars can move with speed v=+1 step/time unit or 0.jam/kø/queue/file/stau = n cars in a row not movingfirst car in stau can change speed from 0 to +1 with prob p/stepnew cars enter the stau with prob q transition matrix for stau length:
€
Tn−1,n = q(1− p)
Tnn = (1− p)(1− q) + pq
Tn +1,n = p(1− q)
€
Tn +1,n − Tn−1,n = p(1− q) − q(1− p) = p − q
Tn +1,n + Tn−1,n = p(1− q) + q(1− p) = p + q − 2pq
net drift per step:
mean square change:
biased random walk again, critical (long-tail distribution of staulengths, lifetimes) fpr p = q.