generalized radon transforms and mathematical economics filethegeneralizedhouthakker–johansenmodel...

Generalized Radon transforms and mathematicaleconomics

Alexey [email protected]

last update: May 12, 2017

Alexey Agaltsov On the generalized Radon transform

The generalized Houthakker–Johansen model

The first microfounded model of production goes back to(Houthakker, 1955), (Johansen, 1972)It was fruitfully applied to study of macroeconomic processes(Petrov, Pospelov, Shananin); e.g. impact of technologicalinnovation on production (Johansen, 1972)Globalization led to qualitative changes in productionprocesses (increase in substitutability of factors), which are outof scope of the classical modelA natural generalization was proposed in (Sato, 1975),(Shananin, 1997), which allows to overcome this problem


The generalized Houthakker–Johansen model

Technologies are parametrized by vectors x ∈ Rn≥0

Each technology x has a capacity f (x) ≥ 0 (number ofproduction units using this technology)

Technologies are described by the unit cost of productionqp(x) = q(p1x1, . . . , pnxn) (here p are the prices of resources)

The maximal possible profit for the industry is given by theprofit function (Πqf )(p0, p) (p0 the price of the final product):

(Πqf )(p0, p) =

∫Rn≥0

max{0, p0 − qp(x)}f (x) dx


Production technologies

Neoclassical n-input, 1-output technologies Q(n): smooth,1-homogeneous q : Rn

>0 → R>0 with bounded level setsAction of Rn

>0 on Q(n), (p, q) 7→ qp:

qp(x1, . . . , xn) = q(p1x1, . . . , pnxn)

Partial composition ◦i : Q(m)×Q(n)→ Q(m + n − 1):

(f ◦i g)(x1, . . . , xm+n−1)

= f (x1, . . . , xi−1, g(xi , . . . , xi+m−1), xi+m, . . . )


Main operators

Let q ∈ Q(n), h : R≥0 → R≥0.The generalized Radon transform:

(Rqf )(p) =

∫q−1p (1)

f (x)dSx

|∇qp(x)|

Radon-type integral operators:

(Rhq f )(p) =

∫Rn≥0

h(qp(x)) f (x) dx

The profit function:

(Πqf )(p0, p) =

∫Rn≥0

max{0, p0 − qp(x)}f (x) dx


Main questions

Characterization. Determine the scope of the model. When agiven profit function Π of an industry can be constructed using ourframework, i.e. Π = Πqf for some q, f ?

Uniqueness. If Π = Πqf , when such a representation is unique?

Inversion. If Π = Πqf for unique q and f , how to find q and f ?

Identification. Given trade statistics, how to identify compatible qand f ?


Main spaces

Weighted spaces Lrc(Rn≥0) with finite norms

‖f ‖r ,c =

(∫Rn≥0

|f (x)|rx rc−I dx)1/r

, 1 ≤ r <∞,

‖f ‖∞,c = inf{K ≥ 0 : |f (x)xc | ≤ K for a.e. x ∈ Rn≥0},

where c ∈ Rn>0, I = (1, . . . , 1)

Rq, Rhq , Πq are continious from LrI−c(Rn

≥0) to Lrc(Rn≥0)

The Mellin transform

(Mf )(z) =

∫Rn≥0

xz−I f (x) dx , z ∈ Cn

is isometric from L2c(Rn

≥0) to L2(<z = c)


Injectivity

Question U. If Π = Πqf , when such a representation is unique?

Let S , H ⊆ Cn. We say thato S is 1-meagre in H iff S ∩ H is nowhere dense in H;o S is 2-meagre in H iff S ∩ H has measure zero in H;o S is ∞-meagre in H iff S ∩ H = ∅.

Theorem (A, Inverse Problems 2016)

Let q ∈ Q(n), c ∈ Rn>0, r ∈ {1, 2,∞}. The following statements

are equivalent:Πq is injective in LrI−c(Rn

≥0).Rq is injective in LrI−c(Rn

≥0).The nullset of Me−q is r -meagre in the plane <z = c .


Injectivity: idea

Recall that (Rqf )(p) =∫qp(x)=1 f (x) dSx

|∇qp(x)|

If f (x) = f1(qp1(x)) + · · ·+ fN(qpN (x)), then Πq is injective:

∫Rn≥0

|f (x)|2dx =N∑

k=1

∫ ∞0

fk(t)t−1(Rqf )(pk

t

)dt.

Generalizes to infinite sums. When functions of the formϕ(qp(x)) with varying ϕ and p span L1

I−c(Rn≥0), L2

I−c(Rn≥0)?


Injectivity: idea

Question. When functions of the form ϕ(qp(·)) with varying ϕ, pspan L1

I−c(Rn≥0), L2

I−c(Rn≥0)?

⇓ Wiener Tauberian theorems: translates φ(· − a) of φ ∈ L1(Rn)span L1(Rn) iff Fϕ(ξ) 6= 0 ∀ξ. Similar in L2(Rn).

⇓ Define Ec : CRn≥0 → CRn

by (Ec f )(y) = ecy f (ey ). Then:

Ec is an isometry from Lrc(Rn≥0) to Lr (Rn),

Ec maps the Mellin transform to the Fourier transformEc maps the action (p, q)→ qp to the additive translation

Conclusion. Functions of the form ϕ(qp(·)) with varying ϕ, p spanL1I−c(Rn

≥0) iff Me−q(z) 6= 0 for <z = c . Similar in L2I−c(Rn

≥0).


Injectivity: example

A CES function is a function of the form

q(x) = C (a1xα1 + · · ·+ anx

αn )

1α ,

where C > 0, aj ≥ 0, a1 + · · ·+ an = 1 (and here α ∈ (0, 1])(Arrow, Solow et al, 1961)A nested CES function is obtained from CES functions usingfinite compositions (Sato, 1967). It allows to take into accountsuch effects as capital-skill complementarity (Griliches, 1969)

Theorem (A, http://arxiv.org/abs/1508.02014)

If q is a nested CES , then Πq injective in LrI−c , r ∈ {1, 2,∞}.


Injectivity: example: idea

Question. How the injectivity behaves with respect to composition◦i of technologies? Define f (z) = Me−f (z)/Γ(Σ(z)).

⇓ Πf is injective in L2I−c(Rn

≥0) iff f (z) 6= 0 for <z = c a.e.

⇓ One can show that f ◦i g(z ◦i w) = f (z ◦i Σ(w))g(w)

⇓ One can show that q(z) = a−z/αΓ(Σ(z))B(z/α)

αn−1CΣ(z) for CES q

Conclusion. Injectivity propagates.


Main questions

Question E. Determine the scope of the model. When a givenprofit function Π of an industry can be constructed using ourframework, i.e. Π = Πqf for some q, f ?

Question U. If Π = Πqf , when such a representation is unique?⇒ f is uniquely determined by q iff Me−q has no zeros

[A, http://arxiv.org/abs/1508.02014]Question I. If Π = Πqf for unique q and f , how to find q and f ?


Characterisation

Question. How to decide whether Π is in image of Πq?

⇓ We know that Πq = Rh′′q , h′′(t) = max{0, 1− t}.

Question’. How to decide whether F is in image of Rh′′q ?

⇓ We know the answer for the Laplace transform Rh′q′ , where

h′(t) = e−t , q′(x) = x1 + · · ·+ xn

Theorem (S. Bernstein, V. Hilbert, S. Bochner)

Let F ∈ RRn≥0 . Then F = Rh′

q′µ for some finite Borel µ ≥ 0 iff F iscompletely monotone, i.e. F is smooth with non-negative evenderivatives and non-positive odd derivatives.

Question”. How to relate the image of Rh′′q to the image of Rh′

q′ ?


Characterization

Question”. How to relate the image of Rh′′q to the image of Rh′

q′ ?

A remarkable property

M(Rh′′q F ) = Γ−1 ·MF ·Me−q ·Mh′′,

M(Rh′q′F ) = Γ−1 ·MF ·Me−q

′ ·Mh′

It implies

Rh′q′F = M−1Me−q

′ ·Mh′

Me−q ·Mh′′M(Rh

qF )

Explicit formulas:

Me−q′(z) = Γ(z1) · · · Γ(zn),

Mh′(s) = Γ(s),

Mh′′(s) = 1s(s+1) .


Characterization

Set Tq = M−1ρqM, where ρq(z) = Γ(z1+···+zn+2)Γ(z1)···Γ(zn)Me−q(z) .

Denote Qregc (n) = {q ∈ Q(n) : ρq ∈ L2 ∪ L∞(<z = c)}

Theorem (A, Func. Ann. App. 2015)

Let q ∈ Qregc (n), c ∈ Rn

>0. Then Π ∈ RRn≥0 is of the form Π = Πqµ

with Borel µ ≥ 0 such that∫x−cdµ <∞ iff

Π ∈ L2c(Rn

≥0), TqΠ ∈ L1c(Rn

≥0), TqΠ is completely monotone


Main questions

Question E. Determine the scope of the model. When a givenprofit function Π of an industry can be constructed using ourframework, i.e. Π = Πqf for some q, f ?⇒ Π = Πqf for some f iff TqΠ is completely monotone, where

Tq is a Mellin multiplier given by an explicit formula[A, Funk. Anal. Appl. 2015]


[A, http://arxiv.org/abs/1508.02014]Question I. If Π = Πqf for some q and f , how to find q and f ?


Inversion

Question. If Π = Πqf and q is known, how to find f ?A remarkable property:

M(Πqf ) = Γ−1 ·Me−q ·Mf

N-smooth functions CN,σc (Rn

>0), N, σ > n, with finite norm

‖f ‖CN,σc

= sup|α|≤N,y∈Rn

(1 + |y |)σn

∣∣∣∣∂|α|u∂yα

∣∣∣∣, u(y) = ecy f (ey )

Theorem (A, Proc. of MIPT 2014)

Let q ∈ Q(n), Me−q(z) 6= 0 for <z = c a.e., f ∈ CN,σI−c (Rn

>0),Π = Πqf . Set s = z1 + · · ·+ zn. Then f = fR + f err

R , where

fR(x) = (2π)−n∫c+iBR

xz−IΓ(s + 2)

(Me−q)(z)· (MΠ)(z) dz ,

‖f errR ‖CN,σ

I−c≤ C (n,N, σ)‖f ‖

CN,σI−c

Rn−N .


Main questions

Question E. Determine the scope of the model. When a givenprofit function Π of an industry can be constructed using ourframework, i.e. Π = Πqf for some q, f ?⇒ Π = Πqf for some f iff TqΠ is completely monotone, where

Tq is a Mellin multiplier given by an explicit formula[A, Funk. Anal. Appl. 2015]


[A, http://arxiv.org/abs/1508.02014]Question I. If Π = Πqf for some q and f , how to find q and f ?⇒ explicit inversion formula

[A, Proceedings of MIPT 2014]


Further comments

The class QCES(n) of CES technologies

q(x) = C (a1xα1 + · · ·+ anx

αn )

1α ,

where C > 0, aj ≥ 0, a1 + · · ·+ an = 1 (and α ∈ (0, 1]).

⇒ If Πq1f1 = Πq2f2, q1 6= q2 are QCES(n), and f1, f2 ≥ 0 decayfast, then f1 = f2 = 0. Otherwise, there are counter-examples[A, Proceedings of MIPT 2014]

⇒ There is a more simple characterization[A, Proceedings of MIPT 2013]


References

[1] A. AgaltsovCharacterization and inversion theorems for a generalized Radon transformProceedings of MIPT 5 (4), 2013, 48–61

[2] A. AgaltsovInversion and uniqueness theorems for Radon-type integral operatorsProceedings of MIPT 6 (2), 2014, 3–14

[3] A. AgaltsovA characterization theorem for a generalized Radon transform arising in a modelof mathematical economicsFunk. Anal. Appl. 49 (3), 2015, 201–204

[4] A. AgaltsovOn the injectivity of a generalized Radon transform arising in a model ofmathematical economicsInverse Problems 32 (11), 2016, 115022-1-17

[5] A. Agaltsov, E. Molchanov, A. ShananinInverse problems in models of resource distributionJournal of Geometric Analysis, 2017, doi:10.1007/s12220-017-9840-1


generalized radon transforms and mathematical economics filethegeneralizedhouthakker–johansenmodel...

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