Lecture 9: Population genetics, first-passage problems

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


2 alleles, N haploid organismschoose 2 individual at random: 1 dies, the other reproduces


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



n1 + 1 with probability x(1 – x) x = n1/Nn1 – 1 with probability x(1 – x)



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)




So

€

n1(t + δt) = n1(t)

n1(t + δt) − n1(t + δt)( )2

= 2n1(t)n2(t) = 2n1(t) 1− n1(t)( )




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)




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)


€

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)


€

δt = 1N

⇒∂P(x, t)

∂t= 1

2 σ 2 ∂ 2

∂x 2x(1− x)P(x, t)[ ], σ 2 =

2

N


boundary conditions: P(0,t) = P(1,t) = 0


€

δt = 1N

⇒∂P(x, t)

∂t= 1

2 σ 2 ∂ 2

∂x 2x(1− x)P(x, t)[ ], σ 2 =

2

N


boundary conditions: P(0,t) = P(1,t) = 0(once an allele dies out, it can not come back)


€

δt = 1N

⇒∂P(x, t)

∂t= 1

2 σ 2 ∂ 2

∂x 2x(1− x)P(x, t)[ ], σ 2 =

2

N



stochastic differential equation:

€

dx = σ x(1− x)dW (t)


€

δt = 1N

⇒∂P(x, t)

∂t= 1

2 σ 2 ∂ 2

∂x 2x(1− x)P(x, t)[ ], σ 2 =

2

N



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)( )


€

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)( )


€

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)( )


€

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)( )


€

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⇒[ ]


€

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)


€

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( )


€

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:


€

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:


€

d x 2 = 2 μ12 − μ12 + μ21( )x[ ]xdt + σ 2x(1− x)dt


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


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

( )


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

( )


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

( )


€

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

( )


€

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

( )


€

H →2μ12μ21

μ12 + μ21( )2 = x 1− x ; x 2 − x

2∝σ 2 → 0


€

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

( )


€

H →2μ12μ21

μ12 + μ21( )2 = x 1− x ; x 2 − x

2∝σ 2 → 0


€

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)


€

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


€

p1 =w1x

w1x + w2(1− x), p2 =

w2(1− x)

w1x + w2(1− x)


n1 + 1 with probability p1(1 – x)


€

p1 =w1x

w1x + w2(1− x), p2 =

w2(1− x)

w1x + w2(1− x)


n1 + 1 with probability p1(1 – x)n1 – 1 with probability p2x


€

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):




€

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):



€

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)



€

dx

dt= sx(1− x)

logx

1− x⋅1− x0

x0

⎛

⎝ ⎜

⎞

⎠ ⎟= stsolution:



€

dx

dt= sx(1− x)

logx

1− x⋅1− x0

x0

⎛

⎝ ⎜

⎞

⎠ ⎟= st

x =1

1+1− x0

x0

exp(−st)

solution:



€

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


€

CdV

dt= I(t) V measured from resting potential

Neurons


€

CdV

dt= I(t)

CdV

dt= −gV + I(t)

V measured from resting potential

with leak g = membrane conductance

Neurons


€

CdV

dt= I(t)

CdV

dt= −gV + I(t)



(experimental fact:) input current is noisy, very small τc compared tomembrane time constant τ = C/g

Neurons


€

CdV

dt= I(t)

CdV

dt= −gV + I(t)



(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.


“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?



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


with no leak:

Assume at t = 0, x = 0boundary condition at θ: P(θ) = 0.

€

CV ≡ x

dx


with no leak:


€

CV ≡ x

dx


We have solved this problem when there is no threshold:

with no leak:


€

CV ≡ x

dx



€

P(x, t) =1

2πσ 2texp −

x 2

2σ 2t

⎛

⎝ ⎜

⎞

⎠ ⎟

with no leak:


€

CV ≡ x

dx



€

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

⎛

⎝ ⎜

⎞

⎠ ⎟

⎡

⎣ ⎢

⎤

⎦ ⎥



€

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

∞

∫



€

P(x, t) =1

2πσ 2texp −

x 2

2σ 2t

⎛

⎝ ⎜

⎞

⎠ ⎟− exp −

(x − 2θ)2

2σ 2t

⎛

⎝ ⎜

⎞

⎠ ⎟

⎡

⎣ ⎢

⎤

⎦ ⎥


€

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:



€

P(x, t) =1

2πσ 2texp −

x 2

2σ 2t

⎛

⎝ ⎜

⎞

⎠ ⎟− exp −

(x − 2θ)2

2σ 2t

⎛

⎝ ⎜

⎞

⎠ ⎟

⎡

⎣ ⎢

⎤

⎦ ⎥


€

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.



€

J(x) = −D∂P

∂x= − 1

2 σ 2 ∂P

∂x


€

f (t) =d

dx

1

2πσ 2texp −

x 2

2σ 2t

⎛

⎝ ⎜

⎞

⎠ ⎟−

1

2πσ 2texp −

(x − 2θ)2

2σ 2t

⎛

⎝ ⎜

⎞

⎠ ⎟

⎡

⎣ ⎢

⎤

⎦ ⎥x=θ



€

J(x) = −D∂P

∂x= − 1

2 σ 2 ∂P

∂x


€

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:


€

t = tf (t)dt0

∞

∫ ~tdt

t 3 / 2∫ = ∞


€

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 )


€

dx


solution with no boundary:

€

P(x, t) =1

2πσ 2texp −

(x − μt)2

2σ 2t

⎛

⎝ ⎜

⎞

⎠ ⎟


€

dx



€

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

⎛

⎝ ⎜

⎞

⎠ ⎟


€

dx



€

P(x, t) =1

2πσ 2texp −

(x − μt)2

2σ 2t

⎛

⎝ ⎜

⎞

⎠ ⎟


€

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

⎛

⎝ ⎜

⎞

⎠ ⎟


€

dx



€

P(x, t) =1

2πσ 2texp −

(x − μt)2

2σ 2t

⎛

⎝ ⎜

⎞

⎠ ⎟


€

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

⎛

⎝ ⎜

⎞

⎠ ⎟


€

dx



€

P(x, t) =1

2πσ 2texp −

(x − μt)2

2σ 2t

⎛

⎝ ⎜

⎞

⎠ ⎟


€

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 )


€

∂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 )


€

∂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


€

∂P

∂t= 0

€

dJ

dx=

d

dxI0 − x( )P(x) − D

∂P(x)

∂x

⎡ ⎣ ⎢

⎤ ⎦ ⎥= 0i.e.


€

∂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.

=>


€

∂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.

=>


€

∂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.

=>


€

∂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.

=>


€

∂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


€

P(x) = 0, x ≥ θ

i.e.

=>


€

∂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


€

P(x) = 0, x ≥ θ

€

J(x) = 0, x > θ and x < 0

i.e.

=>


€

∂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


€

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:


€

∂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


€

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−∞

∞

∫



€

dxP(x) =1−∞

∞

∫

Below reset level, J :

€

I0 − x( )P(x) − D∂P(x)

∂x= 0



€

dxP(x) =1−∞

∞

∫


€

I0 − x( )P(x) − D∂P(x)

∂x= 0

has solution

€

P(x) = c1 exp −(x − I0)2

2D

⎡

⎣ ⎢

⎤

⎦ ⎥



€

dxP(x) =1−∞

∞

∫


€

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:



€

dxP(x) =1−∞

∞

∫


€

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


B.C. at x :

€

r = −DdP

dx

⎛

⎝ ⎜

⎞

⎠ ⎟x=θ

= −Dc2 exp −(θ − I0)2

2D

⎡

⎣ ⎢

⎤

⎦ ⎥ −exp

(θ − I0)2

2D

⎡

⎣ ⎢

⎤

⎦ ⎥

⎛

⎝ ⎜

⎞

⎠ ⎟= Dc2



€

dxP(x) =1−∞

∞

∫


€

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


B.C. at x :

€

r = −DdP

dx

⎛

⎝ ⎜

⎞

⎠ ⎟x=θ

= −Dc2 exp −(θ − I0)2

2D

⎡

⎣ ⎢

⎤

⎦ ⎥ −exp

(θ − I0)2

2D

⎡

⎣ ⎢

⎤

⎦ ⎥

⎛

⎝ ⎜

⎞

⎠ ⎟= Dc2

=>

€

c2 =r

D


Continuity at x = =>

€

c1 exp −I0

2

2D

⎡

⎣ ⎢

⎤

⎦ ⎥= c2 exp −

I02

2D

⎡

⎣ ⎢

⎤

⎦ ⎥ dy exp

(y − I0)2

2D

⎡

⎣ ⎢

⎤

⎦ ⎥

0

θ

∫



€

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

∫



€

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)



€

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


with refractory time τr

€

r =1

τ r + π dx−I 0 /σ

(θ −I 0 ) /σ


A simple model of evolution: the Bak-Sneppen model

N species,each with fitness xi, each uniformly distributed on (0,1)





evolutionary step:



evolutionary step: eliminate the weakest species (smallest xi)



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





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






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:


€

Tn−1,n = (1−θ)2

Tnn = 2θ(1−θ)

Tn +1,n = θ 2


€

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


€

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)



€

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)


Walk (avalanche) ends when n=0 again for the first time.


€

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)


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:


€

Tn−1,n = q(1− p)

Tnn = (1− p)(1− q) + pq

Tn +1,n = p(1− q)


€

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:


€

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


€

Tn−1,n = q(1− p)

Tnn = (1− p)(1− q) + pq

Tn +1,n = p(1− q)

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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.

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Lecture 9: Population genetics, first-passage problems