Spectral deconvolution of unitary invariant modelsjgarzav/Slides_Pierre_Tarrago.pdf · 2020. 10....

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Deconvolution problem Proof of the theorem Spectral deconvolution of unitary invariant models Pierre Tarrago October 8, 2020 1 / 24

Transcript of Spectral deconvolution of unitary invariant modelsjgarzav/Slides_Pierre_Tarrago.pdf · 2020. 10....

  • Deconvolution problem Proof of the theorem

    Spectral deconvolution of unitary invariantmodels

    Pierre Tarrago

    October 8, 2020

    1 / 24

  • Deconvolution problem Proof of the theorem

    (General) spectral deconvolution problem

    Spectral measure of A :

    A = A∗ ∈ MN(C) µA =1

    N

    N∑λ eigvals of A

    δλ

    Three ingredients :

    1. A ∈ MN(C), A = A∗, the unknown ”signal” matrix,2. X = (X1, . . . ,Xr ) ∈ MN(C)r , the unknown, random ”noise”

    matrices, and

    3. W = f (A,X1,X∗1 , . . . ,Xr ,X

    ∗r ), the known ”data” matrix, with

    f ∈ C〈U,V1,V ∗1 , . . . ,Vr ,V ∗r 〉 self-adjoint polynomial.

    Main goal : recover µA from W .

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  • Deconvolution problem Proof of the theorem

    Applications

    • Wireless communication :

    W = (X1A + X2)∗(X1A + X2),

    with X1,X2 Ginibre matrices (A matrix of emission powers, X1matrix of random messages, X2 noise matrix from environment, Wsignal matrix) General linear model (Ryan, Debbah, 2007).

    • Cleaning of covariance matrix :

    W = XAX ∗

    with X Ginibre matrix (Ledoit-Péché (2011), Ledoit-Wolf (2013))

    • Generalized/structured covariance matrix :

    W = XAX ∗

    with X = BG B ≥ 0, G Ginibre matrix (Buns, Allez, Bouchaud,Potter (2016)). For example (B2)ij = exp(|i − j |/τ).

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  • Deconvolution problem Proof of the theorem

    Other approaches:

    • Covariance matrices :I El Karoui (2008) : first results on the subject.I Rao (2008) : atomic case.I Bai and al. (2010) : deconvolution using moments.I Ledoit and Wolf (2013) : ”QuEST” algorithm for estimating

    covariance matrices, numerical inversion of the Marchenko-Pasturequation.

    • Bun, Allez, Bouchaud, Potters (2016) : Rotationally InvariantEstimator of noisy matrix models, using the spectral deconvolutionas an oracle (some spectral deconvolution are achieved using theR/S-transform).

    • Hayase (2020) : use of Cauchy distribution.• Mäıda et al. (preprint, 2020) : study of the backward Fokker-Planck

    equation.

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  • Deconvolution problem Proof of the theorem

    Spectral deconvolution : main assumptions

    1. Simplest algebraic combinations: f = X + Y (additive case) orf = ZYZ∗ (multiplicative case), and we set X = Z∗Z .

    2. Control on the spectral distribution of X : E(dWass,1(µX , µ0)2) ≤ CN2for some µ0 ∈M1(R).

    3. Independence between the eigenbasis of A and X : X =law

    UXU∗ for

    U Haar unitary.

    Main phenomenon : for N large, µW ' µA � /� µX , where �,� denotesthe free additive/multiplicative convolution of measures.

    • as N goes to infinity in law : Voiculescu (1991),• as N goes to infinity, almost surely : Speicher (1993),• concentration results for fixed N : Kargin (2015), Bao, Erdösz and

    Schnelli (2017), Meckes and Meckes (2013).

    • recently, large deviation principle : Belinschi, Guionnet and Huang(2020).

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  • Deconvolution problem Proof of the theorem

    Free operations : analytic transforms

    Suppose that µ3 = µ1 � µ2. How to compute µ3 from µ1 and µ2 ?

    Cauchy transform of µ ∈M1(R) : write C± = {z ∈ C,±=z > 0},

    Gµ =

    {C+ −→ C−z 7→

    ∫R

    1z−t dµ(t)

    Stieltjes inversion formula :

    dµ(t) = − 1π

    limy→0=Gµ(t + iy).

    We also define Fµ, hµ : C+ → C+ as

    Fµ(z) =1

    Gµ(z),

    hµ(z) = Fµ(z)− z .

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  • Deconvolution problem Proof of the theorem

    Free operations : the subordination phenomenon

    (Belinschi, Bercovici, Biane, Voiculescu)Suppose that µ3 = µ1 � µ2. How to compute µ3 from µ1 and µ2 ?

    There exist ω1, ω2 : C+ → C+ such that for z ∈ C+,1. Gµ3(z) = Gµ1(ω1(z)) = Gµ2(ω2(z)) (subordination property)

    2. ω1(z) + ω2(z) = z +1

    Gµ3 (z)(free additive relation)

    Using 1. and 2., one has access to ω1(z) as a fixed pointequation/iteration limit

    ω1(z) = Kz(ω1(z)) = limn→∞

    K◦nz (w),

    for any w ∈ C+, with

    Kz(w) = hµ2(hµ1(w) + z) + z .

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  • Deconvolution problem Proof of the theorem

    Free operations : the subordination phenomenon

    (Belinschi, Bercovici, Biane, Voiculescu)Suppose that µ3 = µ1 � µ2. How to compute µ3 from µ1 and µ2 ?

    (µ1, µ2 ∈M1(R+))

    There exist ω1, ω2 : C+ → C+ such that for z ∈ C+,1. zGµ3(z) = ω1(z)Gµ1(ω1(z)) = ω2(z)Gµ2(ω2(z)) (subordination

    property for d µ̃(t) = td µ̃(t))

    2. ω1(z)ω2(z) =z

    1−z−1Fµ(z) (free multiplicative relation)

    Using 1. and 2., one has access to ω1(z) as a fixed pointequation/iteration limit

    ω1(z) = Hz(ω1(z)) = limn→∞

    H◦nz (w),

    for any w ∈ C+, with

    Hz(w) = −z

    hµ2

    (−z

    hµ1 (w)

    ) .8 / 24

  • Deconvolution problem Proof of the theorem

    Free deconvolution : the subordination method

    Suppose that µ3 = µ1 � µ2. How to compute µ2 from µ1 and µ3 ?

    Recall :

    1. Gµ3(z) = Gµ1(ω1(z)) = Gµ2(ω2(z))

    2. ω1(z) + ω2(z) = z +1

    Gµ3 (z)

    Suppose that ω2 is invertible on some domain ? ⊂ C+ :1. Gµ3(ω

    −12 (z)) = Gµ1(ω1(ω

    −12 (z))) = Gµ2(z)

    2. ω1(ω−12 (z)) + z = ω

    −12 (z) +

    1Gµ3 (ω

    −12 (z))

    = ω−12 (z) +1

    Gµ2 (z)

    Set ω3(z) = ω−12 (z), ω1′(z) = ω1(ω

    −12 (z)) :

    1. Gµ3(ω3(z)) = Gµ1(ω1′(z)) = Gµ2(z)

    2. ω1′(z) + z = ω3(z) + Fµ2(z)

    Domain of validity ?

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  • Deconvolution problem Proof of the theorem

    Free deconvolution : the subordination method

    Set Cσ = {z ∈ C,=z > σ}, and write σ21 = Var(µ1).

    Theorem (Arizmendi, Vargas, T.)There exist two analytic functions ω1, ω3 : C2√2σ1 → C

    + such that for allz ∈ C2√2σ1 ,

    1. Gµ2(z) = Gµ1(ω1(z)) = Gµ2(ω3(z)) (subordination property).

    2. ω1 + z = ω3 + Fµ1(ω1(z)) = ω3 + Fµ3(ω3(z)).

    Moreover, ω3(z) is the unique fixed point of the functionKz(w) = z − hµ1(w + Fµ3(w)− z) in C3=(z)/4 and we have

    ω3(z) = limK◦nz (w), w ∈ C3=(z)/4.

    [Interesting point : except for the first equality in 1., the latter theorem isnot related to µ2, and still holds without ”µ1 � µ2 = µ3”]

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  • Deconvolution problem Proof of the theorem

    Free deconvolution : the subordination method

    Suppose that µ3 = µ1 � µ2. How to compute µ2 from µ1 and µ3 ?

    Deconvolution method :

    1. using the latter theorem, compute ω3 and Gµ2 = Gµ3 ◦ ω3 on thehorizontal line L = {x + 2

    √2σ1i , x ∈ R}.

    2. Fact : using the Stieltjes inversion formula on L instead of R, weobtain the density f of the classical convolution µ2 ∗ C2√2σ1 , wheredCλ(t) = dCauchyλ(t) = λπ(t2+λ2) .

    3. Using classical deconvolution tools, recover µ2 by doing thedeconvolution of f by C2√2σ1 .

    Exemple of deconvolution tools (regularization needed !):

    • regularized Fourier transform,• Tychonov’s regularization,• Total Variation minimization procedure.

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  • Deconvolution problem Proof of the theorem

    Back to the spectral deconvolution :

    Suppose that W = X + A, with µX ' µ0 and µW ' µX � µ1. How tocompute µA from µ0 and µW ?

    Deconvolution method : set σ = Var(µ0)1/2.

    1. using subordination approach, compute ω3 and G = GµW ◦ ω3 on thehorizontal line L = {x + 2

    √2σ, x ∈ R} from µ0 and µW (up to now,

    G is just a function).2. Fact : using the Stieltjes inversion formula with G on L instead of R,

    we obtain a function Ĉ close to dµA ∗ C2√2σ on R.

    3. Using classical deconvolution tools, achieve the deconvolution of Ĉby C2√2σ to obtain an estimator µ̂A of µA.

    Main question : how well does µ̂A approximate µA ?

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  • Deconvolution problem Proof of the theorem

    Simulations

    Figure: Additive case W = X + A : X ∈ MN(C) Gaussian Wigner matrix,A ∈ MN(C) diagonal with entries iid ∼ N(0, 1), N = 500.Histogram of µW , graph of Ĉ , and comparison between µ̂A and µA.

    Figure: Multiplicative case W = XAX ∗ : X ∈ MN(C) Ginibre matrix, A ∈ CWigner with Gaussian entries, N = 500.Histogram of µW , graph of Ĉ , and comparison between µ̂A and µA.

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  • Deconvolution problem Proof of the theorem

    Stability of the free deconvolution : heuristic from thesubordination property

    R R0 0

    C+C−

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  • Deconvolution problem Proof of the theorem

    Stability of the free deconvolution : heuristic from thesubordination property

    R R0 0

    Gµ2

    C+C−

    Gµ2(R)

    14 / 24

  • Deconvolution problem Proof of the theorem

    Stability of the free deconvolution : heuristic from thesubordination property

    R R0 0

    Gµ2

    Gµ3C+

    C−

    Gµ3(R)

    Gµ2(R)

    Gµ3(C+) ⊂ Gµ2(C+).

    14 / 24

  • Deconvolution problem Proof of the theorem

    Stability of the free deconvolution : heuristic from thesubordination property.

    R R0 0

    Gµ2

    Gµ3C+

    C−

    Gµ3(R)

    Gµ2(R)

    R + i2√

    C2√2σ

    14 / 24

  • Deconvolution problem Proof of the theorem

    Stability of the free deconvolution : heuristic from thesubordination property

    R R0 0

    Gµ3 ◦ ω3

    Gµ2

    Gµ3C+

    C−

    Gµ3(R)

    Gµ2(R)

    R + i2√

    C2√2σ

    Subordination of the deconvolution.

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  • Deconvolution problem Proof of the theorem

    Stability of the free deconvolution : heuristic from thesubordination property

    R R0 0

    Gµ3 ◦ ω3

    Gµ2

    Gµ3C+

    C−

    Gµ3(R)

    Gµ2(R)

    R + i2√

    Theoretical limit

    D

    Maximum domain of recovery of Gµ2 .

    14 / 24

  • Deconvolution problem Proof of the theorem

    Stability of the free deconvolution : heuristic from thesubordination property

    R R0 0

    Gµ3C+

    C−

    Gµ3(R)

    Gµ2(R)

    R + i2√

    C2√2σ/D

    Extension of Gµ2 from C2√

    2Var(µ0)/Dmax to C+.

    In the case C2√

    2Var(µ0), this equivalent to a classical deconvolution.

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  • Deconvolution problem Proof of the theorem

    What about the classical case ?

    Suppose

    • µ1, µ2 ∈M1(R),• we know µ1 and µ3 with ‖dµ3, d(µ1 ∗ µ2)‖L2 ≤ �.

    How well can we recover µ2 ? Theoretically :

    Φµ3 ' Φµ1Φµ2 ,

    with Φµ the Fourier transform of µ. BUT, in most cases,

    Φµ1(t) −−−→t→∞ 0,

    which prevent the naive method µ̂2 ' Φ−1(Φµ3/Φµ1).1. Without regularization, nothing can be done : need for hypothesis

    on µ2.

    2. Smoother µ1, rougher µ2 ⇒ Faster exploding of Φµ3/Φµ1 at ∞ ⇒Harder regularization.

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  • Deconvolution problem Proof of the theorem

    The classical case :

    (Butucea, Fan, Lacour and al.)

    µ̂2 ' Φ−1(Φµ3/Φµ1)

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  • Deconvolution problem Proof of the theorem

    The classical case :

    (Butucea, Fan, Lacour and al.)

    µ̂2 ' Φ−1(1[−K ,K ]Φµ3/Φµ1)

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  • Deconvolution problem Proof of the theorem

    The classical case :

    (Butucea, Fan, Lacour and al.)

    µ̂2 ' Φ−1(1[−K ,K ]Φµ3/Φµ1)

    Then, suppose there exist bi , γi , si > 0 with

    • a ≤ |Φµ1(t)|(1 + t2)γ1/2 exp(b1|t|s1) ≤ A, a,A > 0•∫R |Φµ2(t)|

    2(1 + t2)γ2 exp(2b2|t|s2) < L with L 0

    s2 = 0 �2γ2

    2γ1+2γ2+1 (− log �)−2s2/s1

    s2 > 0 (− log �)2γ1+1

    b2 � complicated, if s1 = s2 : �b2

    b2+b1 (− log(�))ξ

    Optimal bounds !

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  • Deconvolution problem Proof of the theorem

    The classical case :

    (Butucea, Fan, Lacour and al.)

    µ̂2 ' Φ−1(1[−K ,K ]Φµ3/Φµ1)

    Then, suppose there exist bi , γi , si > 0 with

    • a ≤ |Φµ1(t)|(1 + t2)γ1/2 exp(b1|t|s1) ≤ A, a,A > 0•∫R |Φµ2(t)|

    2(1 + t2)γ2 exp(2b2|t|s2) < L with L 0

    s2 = 0 �2γ2

    2γ1+2γ2+1 (− log �)−2s2/s1

    s2 > 0 (− log �)2γ1+1

    b2 � complicated, if s1 = s2 : �b2

    b2+b1 (− log(�))ξ

    Optimal bounds !

    16 / 24

  • Deconvolution problem Proof of the theorem

    The classical case :

    (Butucea, Fan, Lacour and al.)

    µ̂2 ' Φ−1(1[−K ,K ]Φµ3/Φµ1)

    Then, suppose there exist bi , γi , si > 0 with

    • a ≤ |Φµ1(t)|(1 + t2)γ1/2 exp(b1|t|s1) ≤ A, a,A > 0•∫R |Φµ2(t)|

    2(1 + t2)γ2 exp(2b2|t|s2) < L with L 0

    s2 = 0 �2γ2

    2γ1+2γ2+1 (− log �)−2s2/s1

    s2 > 0 (− log �)2γ1+1

    b2 � complicated, if s1 = s2 : �b2

    b2+b1 (− log(�))ξ

    Optimal bounds !

    16 / 24

  • Deconvolution problem Proof of the theorem

    The classical case :

    (Butucea, Fan, Lacour and al.)

    µ̂2 ' Φ−1(1[−K ,K ]Φµ3/Φµ1)

    Then, suppose there exist bi , γi , si > 0 with

    • a ≤ |Φµ1(t)|(1 + t2)γ1/2 exp(b1|t|s1) ≤ A, a,A > 0•∫R |Φµ2(t)|

    2(1 + t2)γ2 exp(2b2|t|s2) < L with L 0

    s2 = 0 �2γ2

    2γ1+2γ2+1 (− log �)−2s2/s1

    s2 > 0 (− log �)2γ1+1

    b2 � complicated, if s1 = s2 : �b2

    b2+b1 (− log(�))ξ

    Optimal bounds !

    16 / 24

  • Deconvolution problem Proof of the theorem

    The classical/free Cauchy deconvolution :

    (Butucea, Fan, Lacour and al.)

    If µ1 = Cλ, Φµ1(t) = exp(−λ|t|).Hence,

    • a1 = 1,γ1 = 0, b1 = λ, s1 = 1.•∫R |Φµ2(t)|

    2(1 + t2)γ2 exp(2b2|t|s2) < L with L 0

    s2 = 0 �2γ2

    2γ1+2γ2+1 (− log �)−2s2/s1

    s2 > 0 (− log �)2γ1+1

    b2 � complicated, if s1 = s2 : �b2

    b2+b1 (− log(�))ξ

    17 / 24

  • Deconvolution problem Proof of the theorem

    The classical/free Cauchy deconvolution :

    (Butucea, Fan, Lacour and al.)

    If µ1 = Cλ, Φµ1(t) = exp(−λ|t|).Hence,

    • a1 = 1,γ1 = 0, b1 = λ, s1 = 1.•∫R |Φµ2(t)|

    2(1 + t2)γ2 exp(2b2|t|s2) < L with L 0

    s2 = 0 �2γ2

    2γ1+2γ2+1 (− log �)−2s2/s1

    s2 > 0 (− log �)2γ1+1

    b2 � complicated, if s1 = s2 : �b2

    b2+b1 (− log(�))ξ

    17 / 24

  • Deconvolution problem Proof of the theorem

    The classical classical/free Cauchy deconvolution :

    For us : µ1 = C2√2σ1 , Φµ1(t) = exp(−λ|t|), µ2 = µA and dµ3 = Ĉ .Hence, if

    ‖Ĉ − d(C2√2σ ∗ µA)‖L2 ≤ �

    an appropriate K = K (�) yields that ‖d µ̂A − dµA‖2L2 is bounded by

    dµA is Ck C (k)(− log �)−2

    dµA can be analytically �b2

    b2+2√

    2σ (− log(�))ξextended to R×]− b2, b2[

    There is a version for atomic measures : the condition to ensures wellrecovery is a minimum separation distance t(λ) between atoms of µ2.

    Main goal : to estimate ‖Ĉ − d(C2√2σ ∗ µA)‖L2 .

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  • Deconvolution problem Proof of the theorem

    Convergence result on the subordination methodRecall : W = UXU∗ + A, U Haar unitary, A,X ∈ MN(C) self-adjoint.

    TheoremSuppose that N ≥ K1.Then,

    E(‖Ĉ − d(C2√2σ1 ∗ µA)‖

    2L2

    )≤ K2

    N2+

    K3N3

    +K4N4

    ,

    with K1,K2,K3,K4 depending explicitly on the first six moments of Aand X .

    CorollarySuppose that dLevy (µA, µ) ≤ 1N , with dµ analytically extendable toR + i ]− τ, τ [, τ ∈]0,∞], then for N ≥ K1,

    P(‖µ̂A − µA‖L2 > δ) ≤C

    δ2

    (K2N2

    +K3N3

    +K4N4

    ) ττ+2√

    2σ1

    ,

    for some C depending on the regularity of µ.

    18 / 24

  • Deconvolution problem Proof of the theorem

    Convergence result on the subordination method :simulations

    Figure: Simulation of ‖Ĉ − d(C2√2σ1 ∗ µA)‖L2 in the additive case for N from50 to 2000 (with a sampling of size 100 for each size), theoretical bound inTheorem, and ratio of the theoretical bound on the simulated error.

    19 / 24

  • Deconvolution problem Proof of the theorem

    Convergence result on the subordination method :simulations

    Figure: Simulation of ‖Ĉ − d(Cκ ∗ µA)‖L2 in the multiplicative case for N from100 to 2000 (with a sampling of size 100 for each size) , theoretical bound inTheorem, and ratio of the theoretical bound on the simulated error.

    20 / 24

  • Deconvolution problem Proof of the theorem

    Matricial subordination in the additive case

    (Pastur and Vasilchuk, Kargin)

    W = UXU∗ + A

    [For R ∈ MN(C),R = R∗, GR(z) = (z − R)−1, GµR (z) = tr((R − z)−1)]Set fA(z) = − tr(AGW (z)) and fX (z) = − tr(XGW (z)). Then, set

    ωA(z) = z +E(fX (z))E(GµW (z))

    , ωX (z) = z +E(fA(z))E(GµW (z))

    . (1)

    The functions ωA, ωB are called matricial subordination functions.µ3 = µA � µX W = UXU∗ + A

    ω̃A(z) + ω̃X (z) = z +1

    Gµ3 (z)ωA(z) + ωX (z) = z +

    1EGµW (z)

    Gµ3(z) = GµA(ω̃A(z)) = GµX (ω̃X (z))EGµW (z) =GµA(ω̃A(z)) + RA(z)

    =GµX (ω̃X (z)) + RX (z)

    1. Multiplicative case ? (XUAU∗X − z)−1 X−1(UAU∗− zX 2)−1X−1

    2. Estimates on RA(z),RX (z) ? Estimates on |GµW − EGµW | ?21 / 24

  • Deconvolution problem Proof of the theorem

    Matricial subordination in the additive case

    RA(z) =E [(GµW − EGµW )(φ− Eφ)]− E [(fX − EfX )(ψ − Eψ))]

    EGµW,

    with φ = tr(GA(ωA(z))UXU∗G̃H) and ψ = tr(GA(ωA(z))GH).

    • Using Nevanlinna theory and Weingarten formulas gives an upperbound on 1EGµW

    in terms of first moments of A and X ,

    • Using Poincaré inequality on UN , matrix Hölder inequalities andWeingarten formulas give an upper bound on Var(GµW ), Var(fX ),Var(φ), Var(ψ) in terms of the first moments of X and A.[Reminder of Poincaré inequalities : if f : UN → R has mean zero,then

    ∫UN|f |2 ≤ 1N

    ∫UN‖∇f ‖2]

    Finally,

    |RA(z)| ≤K (=z)N2

    ,

    where K depends on the first moments of A and X .

    22 / 24

  • Deconvolution problem Proof of the theorem

    Back to the deconvolutionWe suppose here for simplicity µX = µ0. We compare then two systems :

    1. the deconvolution system

    ω1(z) + z = ω3(z) +

    1

    GµX (ω1(z)), (1)

    ω1(z) + z = ω3(z) +1

    GW (ω3(z)). (2)

    2. the matricial subordination system of W = UXU∗ + A at ω3(z) :�X := RX (ω3(z), �A := RA(ω3(z)).ωX (ω3) + ωA(ω3) = ω3 +

    1

    GµA(ωA(ω3)) + �A, (a)

    ωX (ω3) + ωA(ω3) = ω3 +1

    GµX (ωX (ω3)) + �X= ω3 +

    1

    EGW (ω3). (b)

    Successively deduce :

    1. ωX (ω3) ' ω1 : (1), (2), (b) and the invertibility domain of 1/Gµ,2. ωA(ω3) ' z : ωX ' ω1, (2) and (b),3. GµA(z) ' GµA(ωA(ω3)) ' EGW (ω3) ' GW (ω3) : ωA(ω3) ' z , then

    (a) and (b), and finally Poincaré inequalities.

    23 / 24

  • Deconvolution problem Proof of the theorem

    Back to the deconvolutionWe suppose here for simplicity µX = µ0. We compare then two systems :

    1. the deconvolution system

    ω1(z) + z = ω3(z) +

    1

    GµX (ω1(z)), (1)

    ω1(z) + z = ω3(z) +1

    GW (ω3(z)). (2)

    2. the matricial subordination system of W = UXU∗ + A at ω3(z) :�X := RX (ω3(z), �A := RA(ω3(z)).ωX (ω3) + ωA(ω3) = ω3 +

    1

    GµA(ωA(ω3)) + �A, (a)

    ωX (ω3) + ωA(ω3) = ω3 +1

    GµX (ωX (ω3)) + �X= ω3 +

    1

    EGW (ω3). (b)

    Successively deduce :

    1. ωX (ω3) ' ω1 : (1), (2), (b) and the invertibility domain of 1/Gµ,2. ωA(ω3) ' z : ωX ' ω1, (2) and (b),3. GµA(z) ' GµA(ωA(ω3)) ' EGW (ω3) ' GW (ω3) : ωA(ω3) ' z , then

    (a) and (b), and finally Poincaré inequalities.

    23 / 24

  • Deconvolution problem Proof of the theorem

    Back to the deconvolutionWe suppose here for simplicity µX = µ0. We compare then two systems :

    1. the deconvolution system

    ω1(z) + z = ω3(z) +

    1

    GµX (ω1(z)), (1)

    ω1(z) + z = ω3(z) +1

    GW (ω3(z)). (2)

    2. the matricial subordination system of W = UXU∗ + A at ω3(z) :�X := RX (ω3(z), �A := RA(ω3(z)).ωX (ω3) + ωA(ω3) = ω3 +

    1

    GµA(ωA(ω3)) + �A, (a)

    ωX (ω3) + ωA(ω3) = ω3 +1

    GµX (ωX (ω3)) + �X= ω3 +

    1

    EGW (ω3). (b)

    Successively deduce :

    1. ωX (ω3) ' ω1 : (1), (2), (b) and the invertibility domain of 1/Gµ,2. ωA(ω3) ' z : ωX ' ω1, (2) and (b),3. GµA(z) ' GµA(ωA(ω3)) ' EGW (ω3) ' GW (ω3) : ωA(ω3) ' z , then

    (a) and (b), and finally Poincaré inequalities.

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  • Deconvolution problem Proof of the theorem

    Back to the deconvolutionWe suppose here for simplicity µX = µ0. We compare then two systems :

    1. the deconvolution system

    ω1(z) + z = ω3(z) +

    1

    GµX (ω1(z)), (1)

    ω1(z) + z = ω3(z) +1

    GW (ω3(z)). (2)

    2. the matricial subordination system of W = UXU∗ + A at ω3(z) :�X := RX (ω3(z), �A := RA(ω3(z)).ωX (ω3) + ωA(ω3) = ω3 +

    1

    GµA(ωA(ω3)) + �A, (a)

    ωX (ω3) + ωA(ω3) = ω3 +1

    GµX (ωX (ω3)) + �X= ω3 +

    1

    EGW (ω3). (b)

    Successively deduce :

    1. ωX (ω3) ' ω1 : (1), (2), (b) and the invertibility domain of 1/Gµ,2. ωA(ω3) ' z : ωX ' ω1, (2) and (b),3. GµA(z) ' GµA(ωA(ω3)) ' EGW (ω3) ' GW (ω3) : ωA(ω3) ' z , then

    (a) and (b), and finally Poincaré inequalities.

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  • Deconvolution problem Proof of the theorem

    Back to the deconvolutionWe suppose here for simplicity µX = µ0. We compare then two systems :

    1. the deconvolution system

    ω1(z) + z = ω3(z) +

    1

    GµX (ω1(z)), (1)

    ω1(z) + z = ω3(z) +1

    GW (ω3(z)). (2)

    2. the matricial subordination system of W = UXU∗ + A at ω3(z) :�X := RX (ω3(z), �A := RA(ω3(z)).ωX (ω3) + ωA(ω3) = ω3 +

    1

    GµA(ωA(ω3)) + �A, (a)

    ωX (ω3) + ωA(ω3) = ω3 +1

    GµX (ωX (ω3)) + �X= ω3 +

    1

    EGW (ω3). (b)

    Successively deduce :

    1. ωX (ω3) ' ω1 : (1), (2), (b) and the invertibility domain of 1/Gµ,2. ωA(ω3) ' z : ωX ' ω1, (2) and (b),3. GµA(z) ' GµA(ωA(ω3)) ' EGW (ω3) ' GW (ω3) : ωA(ω3) ' z , then

    (a) and (b), and finally Poincaré inequalities.

    23 / 24

  • Deconvolution problem Proof of the theorem

    Back to the deconvolutionWe suppose here for simplicity µX = µ0. We compare then two systems :

    1. the deconvolution system

    ω1(z) + z = ω3(z) +

    1

    GµX (ω1(z)), (1)

    ω1(z) + z = ω3(z) +1

    GW (ω3(z)). (2)

    2. the matricial subordination system of W = UXU∗ + A at ω3(z) :�X := RX (ω3(z), �A := RA(ω3(z)).ωX (ω3) + ωA(ω3) = ω3 +

    1

    GµA(ωA(ω3)) + �A, (a)

    ωX (ω3) + ωA(ω3) = ω3 +1

    GµX (ωX (ω3)) + �X= ω3 +

    1

    EGW (ω3). (b)

    Successively deduce :

    1. ωX ' ω1 : (1), (2), (b) and the invertibility domain of 1/Gµ,2. ωA(ω3) ' z : ωX ' ω1, (2) and (b),3. GµA(z) ' GµA(ωA(ω3)) ' EGW (ω3) ' GW (ω3) : ωA(ω3) ' z , then

    (a) and (b), and finally Poincaré inequalities.

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  • Deconvolution problem Proof of the theorem

    Back to the deconvolutionWe suppose here for simplicity µX = µ0. We compare then two systems :

    1. the deconvolution system

    ω1(z) + z = ω3(z) +

    1

    GµX (ω1(z)), (1)

    ω1(z) + z = ω3(z) +1

    GW (ω3(z)). (2)

    2. the matricial subordination system of W = UXU∗ + A at ω3(z) :�X := RX (ω3(z), �A := RA(ω3(z)).ωX (ω3) + ωA(ω3) = ω3 +

    1

    GµA(ωA(ω3)) + �A, (a)

    ωX (ω3) + ωA(ω3) = ω3 +1

    GµX (ωX (ω3)) + �X= ω3 +

    1

    EGW (ω3). (b)

    Successively deduce :

    1. ωX ' ω1 : (1), (2), (b) and the invertibility domain of 1/Gµ,2. ωA(ω3) ' z : ωX ' ω1, (2) and (b),3. GµA(z) ' GµA(ωA(ω3)) ' EGW (ω3) ' GW (ω3) : ωA(ω3) ' z , then

    (a) and (b), and finally Poincaré inequalities.

    23 / 24

  • Deconvolution problem Proof of the theorem

    Back to the deconvolutionWe suppose here for simplicity µX = µ0. We compare then two systems :

    1. the deconvolution system

    ω1(z) + z = ω3(z) +

    1

    GµX (ω1(z)), (1)

    ω1(z) + z = ω3(z) +1

    GW (ω3(z)). (2)

    2. the matricial subordination system of W = UXU∗ + A at ω3(z) :�X := RX (ω3(z), �A := RA(ω3(z)).ωX (ω3) + ωA(ω3) = ω3 +

    1

    GµA(ωA(ω3)) + �A, (a)

    ωX (ω3) + ωA(ω3) = ω3 +1

    GµX (ωX (ω3)) + �X= ω3 +

    1

    EGW (ω3). (b)

    Successively deduce :

    1. ωX ' ω1 : (1), (2), (b) and the invertibility domain of 1/Gµ,2. ωA(ω3) ' z : ωX ' ω1, (2) and (b),3. GµA(z) ' GµA(ωA(ω3)) ' EGW (ω3) ' GW (ω3) : ωA(ω3) ' z , then

    (a) and (b), and finally Poincaré inequalities.

    23 / 24

  • Deconvolution problem Proof of the theorem

    Back to the deconvolutionWe suppose here for simplicity µX = µ0. We compare then two systems :

    1. the deconvolution system

    ω1(z) + z = ω3(z) +

    1

    GµX (ω1(z)), (1)

    ω1(z) + z = ω3(z) +1

    GW (ω3(z)). (2)

    2. the matricial subordination system of W = UXU∗ + A at ω3(z) :�X := RX (ω3(z), �A := RA(ω3(z)).ωX (ω3) + ωA(ω3) = ω3 +

    1

    GµA(ωA(ω3)) + �A, (a)

    ωX (ω3) + ωA(ω3) = ω3 +1

    GµX (ωX (ω3)) + �X= ω3 +

    1

    EGW (ω3). (b)

    Successively deduce :

    1. ωX ' ω1 : (1), (2), (b) and the invertibility domain of 1/Gµ,2. ωA(ω3) ' z : ωX ' ω1, (2) and (b),3. GµA(z) ' GµA(ωA(ω3))'EGW (ω3) ' GW (ω3) : ωA(ω3) ' z , then

    (a) and (b), and finally Poincaré inequalities.

    23 / 24

  • Deconvolution problem Proof of the theorem

    Back to the deconvolutionWe suppose here for simplicity µX = µ0. We compare then two systems :

    1. the deconvolution system

    ω1(z) + z = ω3(z) +

    1

    GµX (ω1(z)), (1)

    ω1(z) + z = ω3(z) +1

    GW (ω3(z)). (2)

    2. the matricial subordination system of W = UXU∗ + A at ω3(z) :�X := RX (ω3(z), �A := RA(ω3(z)).ωX (ω3) + ωA(ω3) = ω3 +

    1

    GµA(ωA(ω3)) + �A, (a)

    ωX (ω3) + ωA(ω3) = ω3 +1

    GµX (ωX (ω3)) + �X= ω3 +

    1

    EGW (ω3). (b)

    Successively deduce :

    1. ωX ' ω1 : (1), (2), (b) and the invertibility domain of 1/Gµ,2. ωA(ω3) ' z : ωX ' ω1, (2) and (b),3. GµA(z) ' GµA(ωA(ω3)) ' EGW (ω3) ' GW (ω3) : ωA(ω3) ' z , then

    (a) and (b), and finally Poincaré inequalities.

    23 / 24

  • Deconvolution problem Proof of the theorem

    L2-estimates

    We finally get for =(z) ≥ 2√

    2Var(µ0)

    |GµA(z)− GW (ω3)| ≤C1(=z)|z |N2

    +C2(=z)|GW (ω3)− EGW (ω3)|

    |z |.

    Recall

    • E(|GW (ω3(z))− EGW (ω3(z))|2) ≤ C(=z)N2 ,• t 7→ 1|t+iη| ∈ L

    2(R).Thus

    E(∫

    R

    ∣∣∣=GW (ω3(t + 2√2Var(µ0)i))−=GµA(t + 2√2Var(µ0)i)∣∣∣2dt)≤K2N2

    +K3N3

    +K4N4

    .

    24 / 24

  • Deconvolution problem Proof of the theorem

    L2-estimates

    We finally get for =(z) ≥ 2√

    2Var(µ0)

    |GµA(z)− GW (ω3)| ≤C1(=z)|z |N2

    +C2(=z)|GW (ω3)− EGW (ω3)|

    |z |.

    Recall

    • E(|GW (ω3(z))− EGW (ω3(z))|2) ≤ C(=z)N2 ,• t 7→ 1|t+iη| ∈ L

    2(R).Thus,

    E(‖Ĉ − d(µA ∗ C2√2Var(µ0)‖

    2L2

    )≤K2N2

    +K3N3

    +K4N4

    .

    24 / 24

  • Deconvolution problem Proof of the theorem

    THANK YOU !

    24 / 24

    Deconvolution problemThe free deconvolution

    Proof of the theorem