Spike detection for calcium activity

Hermine Biermé; Camille Constant; Anne Duittoz; Christine Georgelin

doi:10.1515/ijb-2020-0043

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Spike detection for calcium activity

Hermine Biermé , Camille Constant , Anne Duittoz und Christine Georgelin

Veröffentlicht/Copyright: 30. September 2021

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Aus der Zeitschrift The International Journal of Biostatistics Band 18 Heft 2

Abstract

We present in this paper a global methodology for the spike detection in a biological context of fluorescence recording of GnRH-neurons calcium activity. For this purpose we first propose a simple stochastic model that could mimic experimental time series by considering an autoregressive AR(1) process with a linear trend and specific innovations involving spiking times. Estimators of parameters with asymptotic normality are established and used to set up a statistical test on estimated innovations in order to detect spikes. We compare several procedures and illustrate on biological data the performance of our procedure.

Keywords: autoregressive process; neuronal activity; peaks detection; time series

2010 Mathematics Subject Classification Primary: 62M10; 62F12; 62F03; Secondary: 92B25

Corresponding author: Hermine Biermé, LMA UMR CNRS 7348, Université de Poitiers Bât. H3 - Site du Futuroscope, TSA 61125, 11 bd Marie et Pierre Curie, 86073 Poitiers Cedex 9, France, E-mail: hermine.bierme@math.univ-poitiers.fr

Acknowledgments

We would like to sincerely thank the reviewers for their valuable comments and challenging remarks that helped us to improve substantially this paper.

Author contribution: All the authors have accepted responsibility for the entire content of this submitted manuscript and approved submission.
Research funding: None declared.
Conflict of interest statement: The authors declare no conflicts of interest regarding this article.

Appendix A

A.1 From continuous time to discrete time

We assume to observe a process X over a time interval [0, T], T ∈ R + , satisfying

d X t = 1 τ ( μ ( t ) − X t ) d t + λ d N t + σ d W t ,

for a linear trend μ ( t ) = c + d T t , c ∈ R + , d ∈ R , τ, σ, λ ∈ (0, +∞), and W a Brownian motion independent from N a Poisson process of parameter ν ∈ (0, 1]. It follows, writing γ = 1/τ that

(18) d e γ t X t = γ e γ t c + d T t d t + σ e γ t d W t + λ e γ t d N t .

Now, we assume to have (n + 1) observations uniformly distributed on [0, T]: X ₀, X T n , …, X k T n , …, X _T. For k ∈ {0, …, n}, we will note X _n,k for X k T n . Moreover, we introduce ϕ n ≔ e − γ T n . By integrating (18) over [ k T n , ( k + 1 ) T n ] , we have:

ϕ n − ( k + 1 ) X n , k + 1 − ϕ n − k X n , k = c ϕ n − ( k + 1 ) − ϕ n − k + d n ( k + 1 ) ϕ n − ( k + 1 ) − k ϕ n − k − d γ T ϕ n − ( k + 1 ) − ϕ n − k + σ ∫ k T n ( k + 1 ) T n e γ t d W t + λ ∫ k T n ( k + 1 ) T n e γ t d N t .

Or equivalently:

X n , k + 1 = ϕ n X n , k + c 1 − ϕ n + d k n 1 − ϕ n + d n − d γ T 1 − ϕ n + σ ∫ k T n ( k + 1 ) T n e γ t − ( k + 1 ) T n d W t + λ ∑ j = N k T n N ( k + 1 ) T N e γ s j − ( k + 1 ) T n ,

where ( s j ) j are the points of the Poisson process N. Assuming n is large enough:

d n − d γ T 1 − ϕ n = d n − d γ T 1 − e − γ T n = d n − d γ T − γ T n + o 1 n = o 1 n ;

∫ k T n ( k + 1 ) T n e γ t − ( k + 1 ) T n d W t = W ( k + 1 ) T n − W k T n + o 1 n ,

with W ( k + 1 ) T n − W k T n ∼ N 0 , T n and N ( k + 1 ) T n − N k T n ∼ P ν T n . Then

P N ( k + 1 ) T n − N k T n = 1 = ν T n + o 1 n and P N ( k + 1 ) T n − N k T n = 0 = 1 − ν T n + o 1 n ,

so we can consider the following approximation

X n , k + 1 = ϕ n X n , k + c 1 − ϕ n + d k n 1 − ϕ n + σ ϵ k + 1 + λ U k + 1 ,

where for all k ∈ {0, …, n}, ϵ k ∼ N 0 , T n , U k ∼ B ν T n are independant variables. In this paper we first consider this approximation for a fix time interlaps T n = 1 .

A.2 Proof of Theorem 1

By Cramer–Wold device, it is sufficient to prove that for all u, v, w and x in R :

1 n ∑ k = 0 n − 1 Δ n , k → d n → + ∞ N 0 , l ,

where we denote Δ n , k = u + v k n + w Y k + x Z k + 1 Y k − w ρ Y ( 0 ) for all n ≥ 1 and k ∈ {0, …, n} and

l = u v w x Σ 1 u v w x t .

As usual for proving asymptotic normality in time series we consider and (m _n)-dependent approximation of (Y _k) defined through Y k ( m n ) = ∑ j = 0 m n ϕ j Z k − j and denote Δ n , k ( m n ) = u + v k n + w Y k ( m n ) + x Z k + 1 Y k ( m n ) − w ρ Y ( m n ) ( 0 ) , where ρ Y ( m n ) ( 0 ) = Var ( Y k ( m n ) ) . Let us remark that we may write

Δ n , k = Δ n , k ( m n ) + u + v k n + x Z k + 1 R m n , k + w R m n , k 2 + 2 w Y k ( m n ) R m n , k + w ρ Y ( m n ) ( 0 ) − ρ Y ( 0 )

where R m n , k = ∑ j = m n + 1 + ∞ ϕ j Z k − j = ϕ m n + 1 Y k − m n − 1 . We will show that we can find c > 0 such that

(19) ‖ 1 n ∑ k = 0 n − 1 ( Δ n , k − Δ n , k ( m n ) ) ‖ L 2 ≤ c | ϕ | m n .

Then, since |ϕ| < 1 as soon as m _n → +∞, by Slutsky theorem, it is sufficient to prove that

1 n ∑ k = 0 n − 1 Δ n , k ( m n ) → d n → + ∞ N 0 , l .

For that, since now ( Δ n , k ( m n ) ) is a sequence of m _n-dependent random variables, it is enough to check the three points of Theorem 2 from [32] concerning central limit theorem for m _n dependent centered random variables. Namely, in our setting, considering U n , k = Δ n , k ( m n ) n , it follows from

Var ∑ k = 0 n − 1 U n , k = 1 n Var ∑ k = 0 n Δ n , k ( m n ) → n → + ∞ l ;
there exists C > 0 such that ∑ k = 0 n − 1 E ( U n , k 2 ) ≤ C , for all n ≥ 1;
for all ɛ > 0, L n ( ε ) = m n 2 ∑ k = 0 n − 1 E ( U n , k 2 1 | U n , k | ≥ ε m n − 2 ) → n → + ∞ 0 .

In order to prove that and to compute explicitly the covariance matrix we need the following result.

Lemma 1

Let ( Z k ) k be a sequence of iid centered rv with E ( Z 0 4 ) < + ∞ and (Y _k) its associated AR(1) process with autoregression coefficient ϕ ∈ (−1, 1). Then, for all n ≥ 1 and k, l ∈ {0, …, n − 1} we have

( 1 ) Cov Y k , Y l 2 = E Z 0 3 1 − ϕ 3 ϕ 1 2 ( l − k ) + 3 2 | l − k | , ( 2 ) Cov Y k , Y l Z l + 1 = 0 , ( 3 ) Cov Y k Z k + 1 , Y l Z l + 1 = σ Z 2 1 − ϕ 2 E Z 0 2 1 { k = l } , ( 4 ) Cov Y k 2 , Y l Z l + 1 = 2 σ Z 4 1 − ϕ 2 ϕ 2 ( k − l ) − 1 1 { k ≥ l + 1 } , ( 5 ) Cov Y k 2 , Y l 2 = Cov Y k 2 , Y l 2 = σ Z 4 1 − ϕ 2 ϕ 2 | l − k | ( E ( Z 0 4 ) − 3 σ Z 4 ) σ Z 4 ( 1 + ϕ 2 ) + 2 ϕ 2 | l − k | 1 − ϕ 2 .

It follows that one can find a constant c > 0 such that

| Cov ( Δ n , k , Δ n , l ) | ≤ c | ϕ | | k − l | .

Proof

We first consider k and l such that 0 ≤ k ≤ l ≤ n − 1:

( 1 ) Cov Y k , Y l 2 = ∑ m , n , p = 0 + ∞ ϕ m + n + p E Z k − m Z l − n Z l − p .

Since the centered variables Z _k are iid, the terms of this sum are equal to zero except if the indices k − m, l − n and l − p are equal then if p = n = l − k + m. So we have:

Cov Y k , Y l 2 = ∑ m = 0 + ∞ ϕ 3 m + 2 ( l − k ) E Z k − l 3 = E Z 0 3 1 − ϕ 3 ϕ 2 ( l − k ) .

(2) Since Z _l+1 is independant of Y _k Y _l, we get Cov(Y _k, Y _l Z _l+1) = 0.

(3) Since Z _l+1 is independant of Y _k Y _l Z _k+1 if k < l, we have

Cov Y k Z k + 1 , Y l Z l + 1 = E Y k Z k + 1 Y l Z l + 1 1 { k = l } = E Y k 2 Z k + 1 2 1 { k = l } = E Y k 2 E Z k + 1 2 1 { k = l } = σ Z 4 1 − ϕ 2 1 { k = l } .

(4) Since Z _l+1 is independant of Y k 2 Y l , we obtain Cov Y k 2 , Y l Z l + 1 = 0 .

(5) We have

Cov Y k 2 , Y l 2 = ∑ m , n , p , q = 0 + ∞ ϕ m + n + p + q E Z k − m Z k − n Z l − p Z l − q − ρ Y ( 0 ) 2 .

The terms of this sum are equal to zero except if the indices k − m, k − n, l − p, l − q are all equal or equal in twos, then we deduce that

Cov Y k 2 , Y l 2 = σ Z 4 1 − ϕ 2 ϕ 2 ( l − k ) ( E ( Z 0 4 ) − 3 σ Z 4 ) σ Z 4 ( 1 + ϕ 2 ) + 2 ϕ 2 ( l − k ) 1 − ϕ 2 .

Now if we have 0 ≤ l < k ≤ n − 1, we can check the equalities (3) and (5) using symmetric property. For the other ones:

(1)
Cov Y k , Y l 2 = ∑ m , n , p = 0 + ∞ ϕ m + n + p E Z k − m Z l − n Z l − p .
The terms of this sum are equal to zero except if the indices k − m, l − n and l − p are equal then if p = n and m = n + k − l.
Cov Y k , Y l 2 = ∑ n = 0 + ∞ ϕ 3 n + k − l E Z k − l 3 = E Z 0 3 1 − ϕ 3 ϕ k − l .
(2)
Cov Y k , Y l Z l + 1 = ∑ m , n = 0 + ∞ ϕ m + n E Z k − m Z l − n Z l + 1 = 0
since the centered variables Z _k−m, Z _l−n and Z _l+1 cannot be all equal.

(3) We can prove by induction that
Cov Y k 2 , Y l Z l + 1 = 2 σ Z 4 1 − ϕ 2 ϕ 2 ( k − l ) − 1 .
Actually, for k = l + 1
Cov Y l + 1 2 , Y l Z l + 1 = Cov ϕ Y l + Z l + 1 2 , Y l Z l + 1 = ϕ 2 Cov Y l 2 , Y l Z l + 1 + Cov ( Z l + 1 2 , Y l Z l + 1 ) + 2 ϕ Var Y l Z l + 1 = 0 + 0 + 2 ϕ Var ( Y l ) Var ( Z l + 1 ) = 2 ϕ 1 − ϕ 2 σ Z 4
which proves the property for k = l + 1. Now assuming that the result holds for k > l we get
Cov Y k + 1 2 , Y l Z l + 1 = Cov ϕ Y k + Z k + 1 2 , Y l Z l + 1 = ϕ 2 Cov Y k 2 , Y l Z l + 1 + Cov ( Z k + 1 2 , Y l Z l + 1 ) + 2 ϕ Cov Y k Z k + 1 , Y l Z l + 1 .
As k > l, the two last terms are equal to zero and
Cov Y k + 1 2 , Y l Z l + 1 = ϕ 2 Cov Y k 2 , Y l Z l + 1 = 2 σ Z 4 1 − ϕ 2 ϕ 2 k + 1 − l − 1 ,
which proves using induction that (4) holds for any k ≥ l + 1.

Hence, since |ϕ| < 1, one can find a constant c > 0 such that

Cov Δ n , k , Δ n , l ≤ c | ϕ | | k − l | .

□

Since Y k m n = Y k − ϕ m n + 1 Y k − m n − 1 , we can check that there exists c > 0 such that

Cov Δ n , k − Δ n , k m n , Δ n , l − Δ n , l m n ≤ c | ϕ | 2 m n | ϕ | | k − l | and Cov Δ n , k m n , Δ n , l m n ≤ c | ϕ | | k − l | .

Hence

Var 1 n ∑ k = 0 n − 1 Δ n , k − Δ n , k m n ≤ c | ϕ | 2 m n 1 − | ϕ | ,

that proves (19) and (i) will follow from the fact that 1 n Var ∑ k = 0 n − 1 Δ n , k → n → + ∞ l . Hence, we compute

Var ∑ k = 0 n − 1 Δ n , k = Var ∑ k = 0 n − 1 u + v k n + w Y k + x Z k + 1 Y k = ∑ k , l = 0 n − 1 Cov u + v k n + w Y k + x Z k + 1 Y k , u + v l n + w Y l + x Z l + 1 Y l

= 1 n ∑ k , l = 0 n − 1 u 2 n + u v ( k + l ) + v 2 k l n Cov ( Y k , Y l ) + u w n + v w k Cov Y k , Y l 2 + ( u w n + v w l ) Cov Y k 2 , Y l + 1 n ∑ k , l = 0 n − 1 ( u x n + v k x ) Cov Y k , Y l Z l + 1 + ( u x n + v l x ) Cov Y k Z k + 1 , Y l + n x 2 Cov Y k Z k + 1 , Y l Z l + 1 + 1 n ∑ k , l = 0 n − 1 x w n Cov Y k 2 , Y l Z l + 1 + x w n Cov Y k Z k + 1 , Y l 2 + n w 2 Cov Y k 2 , Y l 2 .

Now in order to get an explicit asymptotic variance we need the following computations.

Lemma 2

For |ϕ| < 1, we have the following behaviors when n tends to +∞:

∑ i = 0 n ∑ j = 0 n ϕ | j − i | = n 1 + ϕ 1 − ϕ + o ( n ) ;
∑ i = 0 n ∑ j = 0 n i ϕ | j − i | = n 2 2 1 + ϕ 1 − ϕ + o ( n 2 ) ;
∑ i = 0 n ∑ j = 0 n i j ϕ | j − i | = n 3 3 1 + ϕ 1 − ϕ + o ( n 3 ) ;
∑ i = 0 n ∑ j = 0 n ϕ 1 2 ( j − i ) + 3 2 | j − i | = n 1 + ϕ + ϕ 2 1 − ϕ 2 + o ( n ) ;
∑ i = 0 n ∑ j = i + 1 n ϕ 2 ( j − i ) − 1 = n ϕ 1 − ϕ 2 + o ( n ) ;
∑ i = 0 n ∑ j = 0 n i ϕ 1 2 ( j − i ) + 3 2 | j − i | = n 2 2 1 + ϕ + ϕ 2 1 − ϕ 2 + o ( n 2 ) .

Then we obtain that 1 n Var ∑ k = 0 n − 1 Δ n , k → n → + ∞ l , where

l = u 2 σ Z 2 1 − ϕ 2 1 − ϕ 1 + ϕ + u v σ Z 2 1 − ϕ 2 1 + ϕ 1 − ϕ + v 2 σ Z 2 1 − ϕ 2 1 + ϕ 3 ( 1 − ϕ ) + 2 u w E Z 0 3 σ Z 2 ( 1 − ϕ 3 ) 1 + ϕ + ϕ 2 1 − ϕ 2 + v w E Z 0 3 1 − ϕ 3 1 + ϕ + ϕ 2 2 ( 1 − ϕ 2 ) + w 2 σ Z 4 1 − ϕ 2 1 + ϕ 2 1 − ϕ 2 E Z 0 4 − 3 σ Z 4 σ Z 4 ( 1 + ϕ 2 ) + 2 1 − ϕ 2 + 4 w x σ Z 4 1 − ϕ 2 ϕ 1 − ϕ 2 + x 2 σ Z 4 1 − ϕ 2 = σ Z 2 ( 1 − ϕ ) 2 u 2 + u v + 2 u w E ( Z 0 3 ) σ Z 2 ( 1 + ϕ ) + v 2 3 + E Z 0 3 σ Z 2 ( 1 + ϕ ) v w + σ Z 2 ( 1 − ϕ ) 2 w 2 ( E Z 0 4 − 3 σ Z 4 ) σ Z 2 ( 1 + ϕ ) 2 + 2 σ Z 2 1 + ϕ 2 ( 1 − ϕ ) ( 1 + ϕ ) 3 + 4 w x σ Z 2 ϕ ( 1 + ϕ ) 2 + x 2 σ Z 2 1 − ϕ 1 + ϕ = u v w x Σ 1 u v w x t ≠ 0 .

For (ii) note that for k < n

| U n , k | = 1 n Δ n , k m n ≤ 1 n V k ,

where V k = | u | + | v | W k + | w | ( W k 2 + ρ Y ( 0 ) ) + | x ‖ Z k + 1 | W k with W k = ∑ j = 0 + ∞ | ϕ | j | Z k − j | . Since E ( Z 0 4 ) < + ∞ , we have for all α ∈ (0, 4]

‖ W k ‖ L α ≤ ‖ Z 0 ‖ L α 1 1 − | ϕ | .

Hence, by Cauchy–Schwarz inequality,

‖ V k ‖ L 2 ≤ | u | + | v | ‖ W k ‖ L 2 + | w | ‖ W k ‖ L 4 2 + ρ Y ( 0 ) + | x | ‖ Z k + 1 ‖ L 4 ‖ W k ‖ L 4 ≤ C ,

where

C = | u | + | v | ‖ Z 0 ‖ L 2 1 1 − | ϕ | + | w | ‖ Z 0 ‖ L 4 2 1 ( 1 − | ϕ | ) 2 + ρ Y ( 0 ) + | x | ‖ Z 0 ‖ L 4 2 1 1 − | ϕ | .

and (ii) holds since ∑ k = 0 n − 1 E ( U n , k 2 ) ≤ 1 n ∑ k = 0 n − 1 ‖ V k ‖ L 2 2 ≤ C 2 .

For (iii) we remark that since E ( V 0 2 ) < + ∞ one has

E ( V k 2 1 V k > n 1 / 8 ) = E ( V 0 2 1 V 0 > n 1 / 8 ) → n → + ∞ 0 ,

and we can choose m n = min E ( V 0 2 1 V 0 > n 1 / 8 ) − 1 / 4 , n 1 / 8 → n → + ∞ + ∞ , in such a way that for ɛ > 0,

L n ( ε ) = m n 2 ∑ k = 0 n − 1 E U n , k 2 1 m n 2 | U n , k | > ε ≤ m n 2 ∑ k = 0 n − 1 1 n E V k 2 1 m n 2 n | V k | > ε ≤ m n 2 E ( V 0 2 1 V 0 > ε n 1 / 4 ) ,

where we have used the fact that m n 2 n ≤ n − 1 / 4 in the last inequality. Since ɛn ^1/4 > n ^1/8, for n sufficiently large, we obtain

L n ( ε ) ≤ m n − 2 ,

such that (iii) holds.

A.3 Proof of Theorem 2

Let us write

H n ( X ) = A n ( X ) t A n ( X ) = n ∑ k = 0 n − 1 k n ∑ k = 0 n − 1 X k ∑ k = 0 n − 1 k n ∑ k = 0 n − 1 k n 2 ∑ k = 0 n − 1 k n X k ∑ k = 0 n − 1 X k ∑ k = 0 n − 1 k n X k ∑ k = 0 n − 1 X k 2 .

In view of (3), if (X _n,k) satisfies (2) for some θ = ( m , b , ϕ ) ∈ R 2 × − 1,1 , we can write X n , k = c n + d k n + Y k with Y a stationary centered solution of the AR(1) Eq. (4), namely

Y k + 1 = ϕ Y k + Z k + 1 ,

and c _n, d are given by (5) i.e. c n = c − b n ( 1 − ϕ ) 2 , d = b 1 − ϕ and c = m 1 − ϕ .

Hence, according to Corollary 1 and Lemma 1 above, the following convergences hold a.s and in L ²:

1 n ∑ k = 0 n − 1 X n , k = c n + d 1 n ∑ k = 0 n − 1 k n + 1 n ∑ k = 0 n − 1 Y k → n → + ∞ c + d / 2 , 1 n ∑ k = 0 n − 1 k n X n , k = c n 1 n ∑ k = 0 n − 1 k n + d 1 n ∑ k = 0 n − 1 k n 2 + 1 n ∑ k = 0 n − 1 k n Y k → n → + ∞ c / 2 + d / 3 , 1 n ∑ k = 0 n − 1 X n , k 2 = c n 2 + 2 c n d 1 n ∑ k = 0 n − 1 k n + d 2 1 n ∑ k = 0 n − 1 k n 2 + 1 n ∑ k = 0 n − 1 Y k 2 + + 2 c n 1 n ∑ k = 0 n − 1 Y k + 2 d 1 n ∑ k = 0 n − 1 k n Y k → n → + ∞ c 2 + d 2 / 3 + c d + ρ Y ( 0 ) ,

with ρ Y ( 0 ) = σ Z 2 / ( 1 − ϕ 2 ) as the stationary solution of the AR(1) equation. Therefore

1 n H n ( X ) → n → + ∞ H ≔ 1 1 / 2 c + d / 2 1 / 2 1 / 3 c / 2 + d / 3 c + d / 2 c / 2 + d / 3 c 2 + d 2 / 3 + c d + ρ Y ( 0 ) a.s. and in L 2 .

Note that for all ( u , v , w ) ∈ R 3 ,

u v w H u v w t = ( u + c w ) 2 + 1 3 ( v + d w ) 2 + w 2 ρ Y ( 0 ) + ( u + c w ) ( v + d w ) = ( u + c w ) + 1 2 ( v + d w ) 2 + 1 12 ( v + d w ) 2 + w 2 ρ Y ( 0 ) ,

and therefore the matrix H is positive definite.

Now let us consider for θ ̃ = ( m ̃ , b ̃ , ϕ ̃ ) ∈ R 3 , the contrast function

M n ( θ ̃ ) = ( X n − A n ( X ) θ ̃ ) t ( X n − A n ( X ) θ ̃ ) = ∑ k = 0 n − 1 X n , k + 1 − ϕ ̃ X n , k − m ̃ − b ̃ k n 2 ,

where X n ≔ ( X n , k + 1 ) 0 ≤ k ≤ n − 1 . Let us write θ = (m, b, ϕ) the true parameter such that

X n , k + 1 − ϕ X n , k − m − b k n = Z k + 1 ,

meaning that X _n − A _n(X)θ = Z for Z ≔ ( Z k + 1 ) 0 ≤ k ≤ n − 1 . Then θ ̂ n = arg min θ ̃ ∈ R 3 M n ( θ ̃ ) satisfies J n ( θ ̂ n ) = 0 for J _n = ∇M _n. But J n ( θ ̃ ) = − 2 t A n ( X ) ( X n − A n ( X ) θ ̃ ) and J n ( θ ) = J n ( θ ) − J n ( θ ̂ n ) = − 2 H n ( X ) ( θ − θ ̂ n ) . On the other hand, since X _n − A _n(X)θ = Z, we get

J n ( θ ) = − 2 ∑ k = 0 n − 1 Z k + 1 ∑ k = 0 n − 1 k n Z k + 1 ∑ k = 0 n − 1 X n , k Z k + 1 .

We will prove that − 1 2 n J n ( θ ) → n → + ∞ 0 a.s. and in L ² with − 1 2 n J n ( θ ) → d n → + ∞ N ( 0 , Σ ) , with Σ = σ Z 2 H . Since (X _n,k) satisfies (3) for θ the true parameter, we can write

1 n ∑ k = 0 n − 1 X n , k Z k + 1 = c n × 1 n ∑ k = 0 n − 1 Z k + 1 + d × 1 n ∑ k = 0 n − 1 k n Z k + 1 + 1 n ∑ k = 0 n − 1 Y k Z k + 1 .

Each empirical mean will tend to 0 a.s. and in L ² according to Corollary 1 (choosing ϕ = 0 for (i)). This allows to conclude for the first point using the fact that (c _n) is a bounded sequence. Using the fact that ( θ − θ ̂ n ) = 1 n H n − 1 − 1 2 n J n ( θ ) we can conclude for the a.s. convergence. Now to prove the convergence in distribution, we use again the Cramer–Wold device (see Proposition 6.3.1 of [10] for instance) and consider for ( u , v , w ) ∈ R 3 \ { 0 } ,

1 n ∑ k = 0 n − 1 u + v k n + w X n , k Z k + 1 = 1 n ∑ k = 0 n − 1 ( u + c n w ) + ( v + d w ) k n + w Y k Z k + 1 .

The convergence will follow from a Lindeberg condition for triangular array of martingales [33]. Actually, let us write Δ n , k + 1 = ( u + c n w ) + ( v + d w ) k n + w Y k Z k + 1 and S n , l = ∑ k = 0 l Δ n , k + 1 . We may consider the filtration F n , l = σ ( Z k , k ≤ l ) = F l . It follows that

E ( Δ n , k + 1 | F n , k ) = E ( Δ n , k + 1 | F k ) = ( u + c n w ) + ( v + d w ) k n + w Y k E ( Z k + 1 ) = 0 a.s.

since Z _k+1 is centered and independent from F k and (S _n,l) is a martingale triangular array. Then let us write S n s n ≔ S n , n − 1 s n , where s n 2 = Var ( S n , n − 1 ) . Hence according to Theorem 2 of [33] or [34] if

(20) 1 s n 2 ∑ k = 1 n − 1 E ( Δ n , k + 1 2 | F k ) → P n → + ∞ 1 ,

(21) and 1 s n 2 ∑ k = 1 n − 1 E ( Δ n , k + 1 2 1 | Δ n , k + 1 | > ε s n | F k ) → P n → + ∞ 0 , for all ε > 0 ,

we have S n s n → d n → + ∞ N ( 0,1 ) .

So let us first compute the asymptotic variance. We write ⟨ S ⟩ n = ∑ k = 1 n − 1 E ( Δ n , k + 1 2 | F k ) and note that s n 2 = E ( ⟨ S ⟩ n ) . In our setting, it is clear that

E ( Δ n , k + 1 2 | F k ) = σ Z 2 ( u + c n w ) + ( v + d w ) k n + w Y k 2 .

By Corollary 1 we have 1 n ∑ k = 0 n − 1 Y k → 0 , 1 n ∑ k = 0 n − 1 k n Y k → 0 and 1 n ∑ k = 0 n − 1 Y k 2 → ρ Y ( 0 ) a.s. and therefore,

1 n ∑ k = 0 n − 1 E ( Δ n , k + 1 2 | F k ) → n → + ∞ s 2 ≔ σ Z 2 ( u + c w ) 2 + 1 3 ( v + d w ) 2 + w 2 ρ Y ( 0 ) + ( u + c w ) ( v + d w ) a.s.

Note that we may deduce from these lines that the asymptotic covariance matrix is given by

Σ = σ Z 2 1 1 2 c + d 2 1 2 1 3 c 2 + d 3 c + d 2 c 2 + d 3 c 2 + d 2 3 + ρ Y ( 0 ) + c d = σ Z 2 H .

Moreover, since (u, v, w) ≠ 0 and H is positive definite we check that s ² > 0. Then,

s n 2 n = E 1 n ∑ k = 0 n − 1 E ( Δ n , k + 1 2 | F k ) → n → + ∞ s 2 ,

and (20) follows by writing

1 s n 2 ∑ k = 0 n − 1 E ( Δ n , k + 1 2 | F k ) = n s n 2 × 1 n ∑ k = 0 n − 1 E ( Δ n , k + 1 2 | F k ) .

Now it remains to prove the Lindenberg condition (21). So let us fix N ∈ N * large enough such that for all n ≥ 2N we have s n > s 2 N , choose C > 0 such that |Δ_n,k+1| ≤ C(1 + |Y _k|)|Z _k+1| and remark that for n ≥ 2N, we have

1 n ∑ k = 0 n − 1 E ( Δ n , k + 1 2 1 | Δ n , k + 1 | > ε s n | F k ) ≤ 1 n ∑ k = 0 n − 1 E ( Δ n , k + 1 2 1 | Δ n , k + 1 | > ε s N | F k ) ≤ C 2 n ∑ k = 0 n − 1 E ( ( 1 + | Y k | ) 2 Z k + 1 2 1 | C ( 1 + | Y k | ) | Z k + 1 | > ε s N | F k ) .

Hence, by stationarity, we get

E 1 n ∑ k = 0 n − 1 E ( Δ n , k + 1 2 1 | Δ n , k + 1 | > ε s n | F k ) = C 2 E ( ( 1 + | Y 0 | ) 2 Z 1 2 1 | ( 1 + | Y 0 | ) | Z 1 | > ε s N / C ) → n → + ∞ 0 ,

since E ( ( 1 + | Y 0 | ) 2 Z 1 2 ) < + ∞ , and allows to get (21). We have therefore S n s n → d n → + ∞ N ( 0,1 ) and consequently, by Slutsky’s theorem, − 1 2 n J n ( θ ) = s n n S n s n → d n → + ∞ N ( 0 , s 2 ) . But − 1 2 n J n ( θ ) = 1 n H n ( X ) n ( θ − θ ̂ n ) , and we can write

n ( θ − θ ̂ n ) = 1 n H n ( X ) − 1 − 1 2 n J n ( θ ) .

Again, by Slutsky’s theorem, we may deduce that

n ( θ − θ ̂ n ) → d n → + ∞ H − 1 N 0 , Σ = N 0 , H − 1 Σ t H − 1 = N 0 , σ Z 2 H − 1 .

Note that det ( H ) = ρ Y ( 0 ) 12 , so

Σ 2 = σ Z 2 H − 1 = ( 1 − ϕ 2 ) c 2 + 4 ρ Y ( 0 ) c d − 6 ρ Y ( 0 ) − c c d − 6 ρ Y ( 0 ) d 2 + 12 ρ Y ( 0 ) − d − c − d 1 ,

with c = m 1 − ϕ , d = b 1 − ϕ and ρ Y ( 0 ) = 1 1 − ϕ 2 σ Z 2 .

A.4 Numerical results

We first present some histograms obtained with 1000 simulations, ν = 0.3, λ = 1, σ = 0.2, and n = 1000. The red lines correspond to the theoretical Gaussian distribution with variance computed according to Theorem 2.

$Figure 21: Histograms for a = 5, b = −5. First row: ( m ̂ n ( 1 ) − m ) $({\hat{m}}_{n}^{(1)}-m)$ . Second row: ( b ̂ n ( 1 ) − b ) $({\hat{b}}_{n}^{(1)}-b)$ . Third row: ( ϕ ̂ n ( 1 ) − ϕ ) $({\hat{\phi }}_{n}^{(1)}-\phi )$ .$

Figure 21:

Histograms for a = 5, b = −5. First row: ( m ̂ n ( 1 ) − m ) . Second row: ( b ̂ n ( 1 ) − b ) . Third row: ( ϕ ̂ n ( 1 ) − ϕ ) .

$Figure 22: Histograms for c = 5, d = −5, n = 1000. First row: ( c ̂ n ( 1 ) − c ) $({\hat{c}}_{n}^{(1)}-c)$ . Second row: ( d ̂ n ( 1 ) − d ) $({\hat{d}}_{n}^{(1)}-d)$ .$

Figure 22:

Histograms for c = 5, d = −5, n = 1000. First row: ( c ̂ n ( 1 ) − c ) . Second row: ( d ̂ n ( 1 ) − d ) .

Figure 23:

Test over one simulation for c = 15, d = −10, b = d(1 − ϕ), a = c(1 − ϕ) − λν, X ₀ = c _n, with c _n = c − d/(n(1 − ϕ)). First line: σ = 0.1, second line: σ = 0.3, third line: σ = 0.5.

Figure 24:

Test over one simulation for a = 5, b = −5, d = b/(1 − ϕ), c _n = (a + λν)/(1 − ϕ) − b/(n(1 − ϕ)²), X ₀ = c _n. First line: σ = 0.1, second line: σ = 0.3, third line: σ = 0.5.

$Figure 25: Means and standard deviation of the estimators of ϕ, c n and d, according Theorem 2 (first column) or linear regression (second colum), over 100 simulations of ( X n , k ) 0 ≤ k ≤ n ${({X}_{n,k})}_{0\le k\le n}$ , for n ∈ {100, 200, …, 5000} with ϕ = 0.9, ν = 0.3, λ = 1, c = 15, d = −10, X 0 = c n , with c n = c − d/(n(1 − ϕ)).$

Figure 25:

Means and standard deviation of the estimators of ϕ, c _n and d, according Theorem 2 (first column) or linear regression (second colum), over 100 simulations of ( X n , k ) 0 ≤ k ≤ n , for n ∈ {100, 200, …, 5000} with ϕ = 0.9, ν = 0.3, λ = 1, c = 15, d = −10, X ₀ = c _n, with c _n = c − d/(n(1 − ϕ)).

Table 2:

TPR for tests over one simulation for ν = 0.3, λ = 1, c = 15, d = −10, b = d(1 − ϕ), a = c(1 − ϕ) − λν, X ₀ = c _n, with c _n = c − d/(n(1 − ϕ)) and n = 100, as shown in Figure 23 with tolerance α = β = 0.01.

Φ	σ = 0.1			σ = 0.3			σ = 0.5
	0.1	0.5	0.9	0.1	0.5	0.9	0.1	0.5	0.9
T ⁽¹⁾	1	1	1	0.9032	0.9677	0.8387	0.8710	0.5484	0.3226
T ⁽²⁾	1	1	1	0.9032	0.9677	0.8387	0.8710	0.5484	0.2903
T ̃ ( 1 )	1	1	1	0.7419	0.9355	0.8065	0.6452	0.2258	0.0645
T ̃ ( 2 )	1	1	1	0.7097	0.9355	0.8065	0.6452	0.2258	0.0645
T c ̃ ( 1 )	1	1	1	0.6219	0.8065	0.7419	0.6129	0.2258	0.1613
T c ̃ ( 2 )	1	1	1	0.6219	0.7742	0.7419	0.6129	0.2903	0.0645
T c ( 1 )	1	1	1	0.5484	0.7419	0.6774	0.6129	0.1613	0.0645
T c ( 2 )	1	1	1	0.5484	0.7419	0.6774	0.6129	0.1613	0.0645

The number of jumps is 31.

Table 3:

FPR for tests over one simulation for ν = 0.3, λ = 1, c = 15, d = −10, b = d(1 − ϕ), a = c(1 − ϕ) − λν, X ₀ = c _n, with c _n = c − d/(n(1 − ϕ)) and n = 100, as shown in Figure 23 with tolerance α = β = 0.01.

Φ	σ = 0.1			σ = 0.3			σ = 0.5
	0.1	0.5	0.9	0.1	0.5	0.9	0.1	0.5	0.9
T ⁽¹⁾	0.0145	0.0145	0	0.0145	0.0290	0.0290	0.1159	0.0145	0
T ⁽²⁾	0.0145	0	0	0.0145	0.0290	0.0290	0.1159	0.0145	0
T ̃ ( 1 )	0	0	0	0.0145	0	0.0145	0.0435	0	0
T ̃ ( 2 )	0	0	0	0.0145	0	0.0145	0.0435	0	0
T c ̃ ( 1 )	0	0	0	0.0145	0	0	0.0145	0	0
T c ̃ ( 2 )	0	0	0	0.0145	0	0	0.0145	0	0
T c ( 1 )	0	0	0	0	0	0	0.0145	0	0
T c ( 2 )	0	0	0	0	0	0	0.0145	0	0

The number of jumps is 31.

Table 4:

TPR for tests over one simulation for ν = 0.3, λ = 1, a = 5, b = −5, b = d(1 − ϕ), a = c(1 − ϕ) − λν, X ₀ = c _n, with c _n = c − d/(n(1 − ϕ)) and n = 100, as shown in Figure 24 with tolerance α = β = 0.01.

Φ	σ = 0.1			σ = 0.3			σ = 0.5
	0.1	0.5	0.9	0.1	0.5	0.9	0.1	0.5	0.9
T ⁽¹⁾	1	1	1	0.9	0.9	0.7333	0.5	0.6	0.3
T ⁽²⁾	1	1	1	0.9	0.9	0.7333	0.5333	0.6	0.2667
T ̃ ( 1 )	1	1	1	0.7333	0.8	0.6333	0.2	0.4667	0
T ̃ ( 2 )	1	1	1	0.7333	0.8	0.6333	0.2	0.4667	0
T c ̃ ( 1 )	1	1	0.9667	0.6667	0.6333	0.6333	0.2333	0.4	0
T c ̃ ( 2 )	1	1	0.9667	0.6667	0.6667	0.6333	0.2333	0.4333	0
T c ( 1 )	1	1	0.9667	0.5333	0.6333	0.5667	0.2	0.2333	0
T c ( 2 )	1	1	0.9667	0.5333	0.6333	0.5667	0.2	0.6667	0

The number of jumps is 30.

Table 5:

FPR for tests over one simulation for ν = 0.3, λ = 1, a = 5, b = −5, b = d(1 − ϕ), a = c(1 − ϕ) − λν, X ₀ = c _n, with c _n = c − d/(n(1 − ϕ)) and n = 100, as shown in Figure 24 with tolerance α = β = 0.01.

Φ	σ = 0.1			σ = 0.3			σ = 0.5
	0.1	0.5	0.9	0.1	0.5	0.9	0.1	0.5	0.9
T ⁽¹⁾	0	0.0286	0	0	0	0.0286	0.0143	0.0571	0
T ⁽²⁾	0	0.0286	0	0	0	0.0286	0.0143	0.0571	0
T ̃ ( 1 )	0	0.0143	0	0	0	0.0143	0	0.0143	0
T ̃ ( 2 )	0	0.0143	0	0	0	0.0143	0	0.0143	0
T c ̃ ( 1 )	0	0	0	0	0	0.0143	0	0.0143	0
T c ̃ ( 2 )	0	0	0	0	0	0.0143	0	0.0143	0
T c ( 1 )	0	0	0	0	0	0.0143	0	0	0
T c ( 2 )	0	0	0	0	0	0	0.0143	0	0

The number of jumps is 30.

Now we present typical realizations of trajectories for several choices of parameters but with the same jumps (in red star) in order to illustrate the parameters effect on shapes and tests. We set ν = 0.3, λ = 1, n = 100 and consider tests: in blue cross detected jumps with T ⁽¹⁾ for level α = 0.01, in green cross detected jumps with the multiple test correction T ̃ ( 1 ) for level β = 0.01, in black plus detected jumps with the multiple test correction T ̃ c ( 1 ) for level β = 0.01 and magenta plus indicate the detected jumps for T c ( 1 ) for level α = 0.01. We add circle for the corresponding detected jumps using T ⁽²⁾ in blue, T ̃ ( 2 ) in green, T ̃ c ( 2 ) in black and T c ( 2 ) in magenta. The blue line, respectively green line, is the estimated line with the first, respectively the second, estimators. The dot red line corresponds to the straight line with parameters (c _n, d) induced by our choices of values. The corresponding true and false positive rates are given in tables. We first present results for c = 15, d = −10 in Figure 23 and Tables 2 and 3.

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Received: 2020-04-03

Revised: 2021-03-17

Accepted: 2021-07-31

Published Online: 2021-09-30

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Artikel in diesem Heft

https://doi.org/10.1515/ijb-2020-0043

Schlagwörter für diesen Artikel

autoregressive process; neuronal activity; peaks detection; time series