Download A Poisson point process

Review of
PROBABILITY DISTRIBUTIONS
Hideaki Shimazaki, Ph.D. http://goo.gl/viSNG
Poisson process
1　Probability distribution
Probability that a (continuous) random variable
X is in (x,x+dx).
(
P x < X < x + dx
)
0.8
Probability density function (PDF).
P ( x < X < x + dx )
f ( x ) = lim
dx→+0
dx
(
()
)
P x < X < x + dx ≈ f x dx
Cumulative distribution function (CDF)
= Life time distribution.
()
(
) ∫
F x =P X ≤x =
x
−∞
()
f s ds
Survival function
= Complementary CDF.
()
(
PDF
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gamma distrib.
0.4
0.2
0
X
0
2
4
6
CDF
1
0.8
0.6
0.4
0.2
∞
) ∫ f ( s) ds = 1− F ( x)
F x =P X >x =
0.6
x
0
0
2
4
6
Poisson process
2　The fundamental theorem of calculus
The relation between probability distribution and probability density
function.
() ∫
F x =
x
−∞
()
f s ds
d
d" x
%
F x = $ ∫ f s ds' = f x
&
dx
dx # −∞
()
Hideaki Shimazaki, Ph.D. http://goo.gl/viSNG
()
()
Poisson process
3　Probability distribution: Example
Exponential distribution
Probability density function (PDF).
f x = λe−λx
Cumulative distribution function (CDF).
F x = P X ≤ x = 1− e − λ x
Survival function.
F x = P X > x = 1− F x = e − λ x
()
()
(
()
)
(
)
()
1.0
0.8
0.6
0.4
0.2
1
Expectation E !" X #$ = λ −1
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2
3
4
5
−2
Variance var !" X #$ = λ
Poisson process
4　Samples from a distribution
PDF
0.8
0.6
gamma distrib.
0.4
If a random variable X follows a cdf FX,
then the r.v. U given by
( ) ∫
U = FX X =
X
−∞
f
X
()
fX x
0.2
0
X
0
( s) ds
follows an uniform distribution in [0,1].
2
4
6
CDF
1
U 0.8
()
FX x
0.6
0.4
0.2
0
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0
2X
4
6
Poisson process
5　Homework 1-1
Problem
( )
Confirm that U = FX X
follows an uniform distribution in [0,1].
() ∫
when X is a random variable from FX x =
x
0
()
f X s ds
Hint
Suppose a random variable (r.v.) X given by a pdf f X ( x )
Consider a r.v. Y that is transformed from X
using a monotonic function g. Y = g ( X )
The distribution of Y is given as
()
()
fY y = f X x
dx
= f X x g! x
dy
() ()
()
() ∫
Consider a particular relation given by CDF, g x = FX x =
Then we obtain fY ( y ) = f X (
Hideaki Shimazaki, Ph.D. http://goo.gl/viSNG
) ()
x g! x
−1
() ()
= fX x fX x
−1
x
−∞
−1
()
f X s ds
=1
Poisson process
6　Samples from a distribution
CDF
Inverse function method
1
Let U be an uniform r.v..
X
X that satisfiesU = ∫ −∞ f X s ds follows the density f X x .
()
The r.v. X = F
−1
(U )
()
follows the CDF given by F.
Exponential distribution:
0.8
0.6
0.4
0.2
0
0
2
4
6
f X x = λe−λx
()
( ) ∫
U = FX X =
X
λ
e − λ s ds = 1− e − λ X
−∞
A r.v. that follows an exponential distribution with rate λ is
generated using an uniform r.v. X as
( )
( )
X = FX−1 U = −λ −1 log U
Memo: The distribution that can be inverted analytically is limited, e.g., Exp., Weibull,
Pareto, etc. For other distributions, one can numerically compute CDF to obtain X.
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Poisson process
7　Introduction to
POISSON POINT PROCESS
Hideaki Shimazaki, Ph.D. http://goo.gl/viSNG
Poisson process
8　Memoryless process
A point process is memoryless if intervals X satisfy
P ( X > s + t | X > t ) = P ( X > s)
Question.
Suppose that you bought a refrigerator t
years ago. It still works well.
How long will it continue to work from now?
Event
s
0
t
t+s
X
Answer.
We wish to know the probability that the refrigerator survives another s
years given that it survived t years. This is given as P X > s + t | X > t .
(
(
)
)
From the memoryless property, it is equivalent to P X > s .
Thus, the fact that the refrigerator survived t years did not provide any
information to predict a failure of the refrigerator in future.
Hideaki Shimazaki, Ph.D. http://goo.gl/viSNG
Poisson process
9　Exponential distribution
Q. What distribution possesses the memoryless property?
Ans. The exponential distribution:
f x = λe−λx
()
The survival function can be written in a conditional form as
(
(
)
) (
P X > s+t = P X > s+t | X > t P X > t
)
Using the memoryless property, we obtain
(
(
)
) (
P X > s+t = P X > s P X > t
)
The exponential distribution satisfies the memoryless property.
(
) (
If P X > x = e − λ x , then P X > s P X > t = e
(
)
)
− λ s − λt
e
= e − λ ( s+t ) = P X > s + t
Density is given as
d
f x = − P X > x = λe−λx
dx
()
(
)
(
)
Poisson point process
If intervals of events are samples from the exponential distribution,
and the intervals are independent each other, then the process is
memoryless.
This process is called a (homogeneous) Poisson point process.
Hideaki Shimazaki, Ph.D. http://goo.gl/viSNG
Poisson process
11　Simulating a Poisson process
Question.
How can we simulate a Poisson point process?
Ans.
Use the inverse function method to generate an
inter-spike interval (ISIs) that follows an exponential
distribution.
1.0
0.8
0.6
0.4
ISI X = −λ logU
−1
0.2
1
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3
4
5
Poisson process
12　The Poisson distribution
Let’s consider a distribution of the number of spikes that occurred in T[s].
N T : the number of spikes that happens in T [s]
1 2
N T -th spike
3
T
A count distribution of a Poisson point process is given by a Poisson
distribution.
0.35
Ê Ê
0.30
(
P NT = n
)
λT )
(
=
n
n!
0.25
e − λT
0.20
Ê
0.15
‡
Ï
ÏÏ
Ï
‡
Ï
Ê
‡
‡ ÏÏÏ
Ï
Hideaki Shimazaki, Ph.D. http://goo.gl/viSNG
‡
‡
0.10
0.05
‡ ‡
Ï
Ê
Ï
Ï
Ï
Ï
‡
‡
Ï
Ï
ÏÏ
‡
Ê Ê Ê Ê Ê Ê ‡
‡ Ê
‡ Ï
‡ Ï
‡ Ï
Ê ‡
Ê ‡
Ê ‡
Ê ‡
Ê Ê
Ê
Ê
‡ Ï
Ê
‡ Ï
Ê
‡ Ï
Ê
‡ Ï
Ê
‡ Ï
Ê
‡ Ï
Ê
‡ Ï
Ê
‡ Ï
Ê
‡ Ï
Ê
‡ Ï
Ê
‡
Ê
5
10
15
20
25
30
Poisson process
13　Homework 1-2
Derive the Poisson distribution for
the first few spike counts, N T = 0,1,2.
Problem
Hint
NT = 0
If there is no spike in T, then the first spike occurred
after T. This is given by survival function of an
exponential distribution. X1
T
X2
NT = 1
T
s1
NT = 2
X3
s1
s2
T
Hideaki Shimazaki, Ph.D. http://goo.gl/viSNG
(
)
(
( )
)
P NT = 0 = P X 1 > T = F T
Let the first spike occurred at s_1, the first ISI
follows an exponential density, the second ISI
should be greater than T-s_1.
T
P N T = 1 = ∫ f s1 F T − s1 ds1
(
)
( ) (
0
)
Similarly if you have two spikes in T, T
T
(
) ∫ ∫ f (s ) f (s
P NT = 2 =
0
s1
1
2
) (
)
− s1 F T − s2 ds1 ds2
Poisson process
14　The Erlang distribution
Let’s consider a distribution of a waiting time τ until spikes occurred n times.
1 2
3
n-th spike
τ
Waiting time of multiple spikes from a Poisson process is given by an
Erlang distribution.
0.4
0.3
λ nτ n−1 − λτ
P (τ < X < τ + d τ ) =
e dτ
(n −1)!
0.2
0.1
2
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4
6
8
10
12
14
Poisson process
15　Spike count and ISI
The relation between a Poisson and Erlang distribution.
1 2
3
n-1
n
N τ spikes
τ
Sn
If Sn > τ , the number of spikes in is at most n-1.
(
(
)
P N τ < n = P Sn > τ
)
Rhs is a survival function of an Erlang distribution.
P ( Sn > τ ) =
∫
∞
τ
n−1
n n−1 − λt
λ t e
dt = ∑
k=0
( λτ )
k!
k
e− λτ
P ( N τ = n ) = P ( N τ < n +1) − P ( N τ < n )
Hideaki Shimazaki, Ph.D. http://goo.gl/viSNG
( λτ )
=
n!
n
e− λτ
Poisson process
16　Instantaneous spike rate
Alternative definition of a Poisson point process
Divide T into small N bins of a small width Δ (T=NΔ).
Spike
t
Δ
t+Δ
Instantaneous spike rate P ( a spike in [t, t + Δ)) = λΔ + o ( Δ )
P (> 1 spikes in [t, t + Δ)) = o ( Δ )
P ( no spikes in [t, t + Δ)) = 1− λΔ + o ( Δ )
o (Δ)
lim
=0
o ( Δ ) is a function that approaches to zero faster than Δ : Δ→+0
Δ
Hideaki Shimazaki, Ph.D. http://goo.gl/viSNG
Poisson process
17　Relation to the Poisson distribution
The probability of a spike/no spike is an approximation of the Poisson
distribution for a small time bin.
The spike count in a small bin is approximated as
(
)
P NΔ = n =
(λΔ)
n
n!
e − λΔ =
(λΔ)
n
#
&
2
1
1−
λ
Δ
+
λ
Δ
+
%
(
n! $
2
'
( )
In particular,
#
&
2
1
P N Δ = 0 = 1%1− λΔ + λΔ +( = 1− λΔ + o Δ
2
$
'
#
&
2
1
P N Δ = 1 = λΔ %1− λΔ + λΔ +( = λΔ + o Δ
2
$
'
&
2#
2
1
P N Δ = 2 = λΔ %1− λΔ + λΔ +( = o Δ
2
$
'
(
)
(
)
(
) ( )
Hideaki Shimazaki, Ph.D. http://goo.gl/viSNG
( )
( )
( )
( )
( )
( )
Poisson process
18　Exponential / Poisson distribution revisited
The exponential distribution from an instantaneous spike rate.
The probability that no spikes happened in N-1 bins and a spike
happened in the last bin (geometric distribution). (x=NΔ)
P ( x < X < x + Δ ) = (1− λΔ )
ISI distribution: f ( x ) = lim
Δ→0
N−1
λΔ + o ( Δ )
P ( x < X < x + Δ)
= λ e− λ x
Δ
The Poisson distribution from an instantaneous spike rate.
The probability that n spikes happened in N bins (T=NΔ).
n
! N $
λ T ) − λT
N−n
n
(
P ( NT = n) = #
e as Δ → 0
& (1− λΔ ) ( λΔ ) →
n!
" n %
Hideaki Shimazaki, Ph.D. http://goo.gl/viSNG
Poisson process
19　Exponential distribution revisited
We can derive the exponential distribution from an instantaneous spike rate,
λ [spike/sec]
Divide T into small N bins of a width Δ.
Probability
N bins
Spike
Δ
a spike in a bin:
λΔ
>2 spikes in a bin:
o (Δ)
no spikes in a bin: 1− λΔ + o ( Δ )
T = NΔ
The probability that no spikes happened in N-1 bins and
a spike happened in the last bin.
P ( x < T < x + Δ ) = (1− λΔ )
ISI distribution: f ( x ) = lim
Δ→0
Hideaki Shimazaki, Ph.D. http://goo.gl/viSNG
N−1
λΔ + o ( Δ )
P ( x < T < x + Δ)
= λ e− λ x
Δ
Poisson process
20　Poisson distribution revisted
We can derive the Poisson distribution from an instantaneous spike rate,
λ [spike/sec]
Divide T into small N bins of a width Δ.
Probability
N bins
Δ
T = NΔ
Spike
a spike in a bin:
λΔ
>2 spikes in a bin:
o (Δ)
no spikes in a bin: 1− λΔ + o ( Δ )
The probability that n spikes happened in T [s].
n
! N $
λ T ) − λT
N−n
n
(
P ( NT = n) = #
e as Δ → 0
& (1− λΔ ) ( λΔ ) →
n!
n
"
%
Hideaki Shimazaki, Ph.D. http://goo.gl/viSNG
Poisson process
21　Homework 1-3
Problem
The probability that no spikes happened in N-1 bins and a
spike happened in the last bin is given as
λΔ
N−1
N
P ( x < X < x + Δ ) = (1− λΔ ) λΔ + o ( Δ ) =
(1− λΔ) + o (Δ)
1− λΔ
Show that the ISI distribution becomes an exponential
distribution.
Hint
Approximate the terms by the Taylor expansions.
1 2 1 3
log
1−
x
=
−x
−
x − x 
1
( )
= 1+ x + x 2 + x 3 
2
3
1− x
N
(1− λΔ) = exp #$ N log (1− λΔ)%&
λΔ
2
# '
*%
= λΔ 1+ λΔ + ( λΔ ) +
1
2
=
exp
N
−
λ
Δ
−
λ
Δ
−
(
+.
(
)
1− λΔ
,&
2
$ )
= λΔ + o ( Δ )
/ 1
2
= e− λ x 11− λ 2 xΔ +4
0 2
3
Hence we obtain
P ( x < X < x + Δ)
f
x
=
lim
= λ e− λ x
P ( x < X < x + Δ ) = λΔ exp [−λ x ] + o ( Δ ) , and ( ) Δ→0
Δ
{
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}
Poisson process
22　Homework 1-4
Problem Probability that n spikes occur in N bins (T [s]) is given as
! N $
n
N−n
P ( NT = n) = #
& ( λΔ ) (1− λΔ )
" n %
Show that the probability becomes the Poisson distribution for large N and
small Δ with a relation given by a constant T=NΔ.
Hint
Using an approximation given by the Taylor expansions (See previous slides).
n
# λΔ &
N!
N!
N
n
P ( NT = n) =
λΔ ) exp [−λΔN ]
%
( (1− λΔ ) ≈
(
( N − n)!n! $ 1− λΔ '
( N − n)!n!
Use the relation, T=NΔ, to obtain
N!
1
n
P ( NT = n) =
λ
T
exp [−λT ]
(
)
n
( N − n)!N n!
Use the Stirling’s formula ln N! ~ N ln N − N
N
" n% " n%
N!
−n
and prove that ln
=
$1− ' ln $1− ' − n = 1⋅ ln e − n → 0
n
( N − n)!N # N & # N &
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Poisson process
23　Likelihood function
t1
Likelihood function
t2
tn
0
T
p (t1, t2 ,…, tn ∩ N T = n ) = λ n exp [−λT ]
Proof
p (t1, t2 ,…, tn ∩ N T = n ) Δ n
n
= f (t1 ) Δ∏ f (ti − ti−1 ) Δ ⋅ P (tn+1 > T | tn )
i=2
n
= λ exp (−λ t1 ) Δ∏ λ exp &'−λ (ti − ti−1 )() Δ ⋅ exp &'−λ (T − tn )()
i=2
= λ n Δ exp [−λT ]
Here the ISI distribution is f ( x ) = λ exp [−λ x ]
Hideaki Shimazaki, Ph.D. http://goo.gl/viSNG
Poisson process
24　INHOMOGENEOUS POISSON
PROCESS
Hideaki Shimazaki, Ph.D. http://goo.gl/viSNG
Poisson process
25　Inhomogeneous Poisson process
Instantaneous rate λ (t )
Point process
t t+Δ
P ( a spike in [t, t + Δ)) = λ (t ) Δ + o ( Δ )
Time-dependent rate
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Poisson process
26　Inhomogeneous Poisson process
Instantaneous rate λ (t )
Point process
ti−1
ti
t
#
f (t | ti−1 ) = λ (t ) exp %− ∫ λ (u) du&(
$ ti−1
'
for
t > ti−1
Note that the above formula extends the exponential ISI of
homogeneous Poisson process.
Hideaki Shimazaki, Ph.D. http://goo.gl/viSNG
Poisson process
27　Homework 1-5
Problem Derive the ISI density of the inhomogeneous Poisson process.
Hint Let the last spike occurred at time 0. The probability of a spike occurrence in [t, t+Δ) is given as N−1
P ( t < X < t + Δ ) = ∏ #$1− λ ( kΔ ) Δ%& ⋅ #$λ ( NΔ ) Δ%& + o ( Δ )
k=1
N
λ ( NΔ ) Δ
=
⋅ ∏ #$1− λ ( kΔ ) Δ%& + o ( Δ )
1− λ ( NΔ ) Δ k=1
Using the following approximation, #N
%
#N
%
#
%
∏$1− λ (kΔ) Δ& = exp)∑ log {1− λ (kΔ) Δ}* = exp)∑{−λ (kΔ) Δ + o (Δ)}*
$ k=1
&
$ k=1
&
k=1
λ ( NΔ ) Δ
= λ ( NΔ ) Δ + o ( Δ )
1− λ ( NΔ ) Δ
t
P (t < X < t + Δ )
%
(
f
t
=
lim
=
λ
t
exp
−
λ
u
du
(
)
(
)
(
)
The ISI density is given as
'& ∫ 0
*)
Δ→0
Δ
N
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Poisson process
28　Likelihood function
λ (t )
Likelihood function
0
T
t1
tn
t2
n
T
%
p (t1, t2 ,…, tn ∩ N T = n ) = ∏ λ (ti ) exp '− ∫ λ (u) du(*
& 0
)
i=1
n
Proof
p (t1, t2 ,…, tn ∩ N T = n ) Δ = f (t1 ) Δ∏ f (ti | ti−1 ) Δ ⋅ P (tn+1 > T | tn )
n
i=2
n
T
'
= ∏ λ (ti ) Δ exp )− ∫ λ (u) du*,
( 0
+
i=1
ti
#
&
Here the ISI distribution is given as f (ti | ti−1 ) = λ (ti ) exp %− ∫ λ (u) du(
$ ti−1
'
Hideaki Shimazaki, Ph.D. http://goo.gl/viSNG
Poisson process
29　What we learned
1
2
3
4
5
6
•  Memoryless property of a process.
•  Exponential distribution = Poisson process.
•  Poisson count distribution, Waiting time distribution (Erlang).
•  Definition of a Poisson process using instantaneous firing rate.
•  Exponential and a Poisson count distribution revisited.
•  Inhomogeneous Poisson process: ISI distribution and likelihood
Hideaki Shimazaki, Ph.D. http://goo.gl/viSNG
Poisson process
30　Tomorrow, we will learn
1
2
3
4
5
•  Renewal process: Conditional intensity function and ISI density.
•  non-Poisson process (Point process written by the conditional intensity
function)
•  Time-rescaling theorem
•  How to simulate a point process via the time-rescaling theorem.
•  How to assess a point process model via the time-rescaling theorem.
(Q-Q plot and K-S test)
Hideaki Shimazaki, Ph.D. http://goo.gl/viSNG
Poisson process
31　Follow-up from the lecture feedback
Stimulus signal
x (t )
Tomorrow’s topic.
Today’s topic.
Instantaneous spike-rate λ (t )
Point process
t t+Δ
P ( a spike in [t, t + Δ)) = λ (t ) Δ + o ( Δ )
Hideaki Shimazaki, Ph.D. http://goo.gl/viSNG
Poisson process
32　Likelihood function
Density function
An observed sample
f ( x ) = λ e− λ x
X
Likelihood function
L ( λ ) = f ( X ) = λ e− λ X
Multiple observation
X1, X 2 ,, X n
n
Likelihood function
L ( λ ) = ∏ p ( Xi | λ )
i=1
Maximum likelihood estimation (MLE)
n
λMLE = argmax ∏ p ( Xi | λ )
λ
Hideaki Shimazaki, Ph.D. http://goo.gl/viSNG
i=1
Poisson process
33　Likelihood of a Poisson process
Likelihood function of a homogeneous Poisson process
Inter-spike Intervals (ISIs)
p (t1, t2 ,…, tn ∩ N T = n | λ ) = p (t1, t2 − t1,…, tn − tn−1 ∩ N T = n )
n
f (ti − ti−1 ) = λ e
− λ (ti −ti−1 )
= ∏ f (ti − ti−1; λ ) ⋅ F (T − tn ; λ )
i=1
= λ n exp [−λT ]
Likelihood function of a inhomogeneous Poisson process
p (t1, t2 ,…, tn ∩ N T = n | λ0:T ) = p (t1, t2 − t1,…, tn − tn−1 ∩ N T = n )
n
ti
#
f (ti − ti−1 ) = λ (ti ) exp %− ∫ λ (u) du&(
$ ti−1
'
= ∏ f (ti − ti−1; λti :ti−1 ) ⋅ F (T − ti ; λtn :T )
i=1
n
T
&
= ∏ λ (ti ) exp (− ∫ λ (u) du)+
' 0
*
i=1
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Poisson process
34