Course on Inverse Problems Albert Tarantola Second Lesson: - - PowerPoint PPT Presentation

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Course on Inverse Problems Albert Tarantola Second Lesson: - - PowerPoint PPT Presentation

Princeton University Department of Geosciences Course on Inverse Problems Albert Tarantola Second Lesson: Probability (Fundamental Notions) Let be A , A 1 , A 2 A 0 . Measure function: P [ A ] 0 P [ ] = 0 P [ A 1 A 2 ]


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Princeton University

Department of Geosciences

Course on Inverse Problems

Albert Tarantola

Second Lesson: Probability (Fundamental Notions)

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Let be A , A1 , A2 · · · ⊆ A0 . Measure function: P[A] ≥ 0 P[∅] = 0 P[A1 ∪ A2] = P[A1] + P[A2] − P[A1 ∩ A2] Probability function: A measure function satisfying P[A0] = 1

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Conditional probability: For a given set C ⊆ A0 with nonzero probability, P[C] = 0 , the conditional probability of any set A ⊆ A0 , denoted P[A|C] , is defined as P[A|C] = P[A ∩ C] P[C] . One names P[A|C] the conditional probability of A given C .

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Δθ Δϕ n sinθ Δθ Δϕ n ΔS n

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Histogram tends to probability density Histogram tends to volumetric probability

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Cells defined by constant increments of ∆θ and ∆ϕ : f (θ, ϕ) = 1 N lim

∆θ ∆ϕ→0

n ∆θ ∆ϕ (probability density) Cells having constant surface ∆S : f (θ, ϕ) = 1 N lim

∆S→0

n ∆S (volumetric probability) Finite probability: P[A] =

  • A

f (θ, ϕ) = dS(θ, ϕ)

  • A

f (θ, ϕ) dS(θ, ϕ) = sin θ dθ dϕ

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Fisher probability density: f (θ, ϕ) = κ 4 π sinh κ sin θ exp(κ cos θ) P[A] =

  • A

f (θ, ϕ) Fisher (2D) volumetric probability: f (θ, ϕ) = κ 4 π sinh κ exp(κ cos θ) P[A] = dS(θ, ϕ)

  • A

f (θ, ϕ) ; dS(θ, ϕ) = sin θ dθ dϕ

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Volumetric probability f and probability density f : P[A] =

  • A dV(x1, x2 . . . xn) f (x2, x2 . . . xn)

=

  • A dx1 dx2 . . . dxn f (x2, x2 . . . xn)

To pass from one to the other, express dV(x1, x2 . . . xn) = v(x1, x2 . . . xn) dx1 dx2 . . . dxn Example: for spherical coordinates in Euclidean space dV(r, θ, ϕ) = r2 sin θ dr dθ dϕ

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x = r sin θ cos ϕ y = r sin θ sin ϕ z = r cos θ det   ∂x/∂r ∂x/∂θ ∂x/∂ϕ ∂y/∂r ∂y/∂θ ∂y/∂ϕ ∂z/∂r ∂z/∂θ ∂z/∂ϕ   = r2 sin θ dVC(x, y, z) = dx dy dz

dVS(r, θ, ϕ) = r2 sin θ dr dθ dϕ

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Most of the positive quantities are Jeffreys quantities, like the electric resistance R . From D = | log(R2/R1)| it follows dℓ = dR R Therefore, the relation between a (1D) volumetric probability f (R) and the associated probability density f (R) is f (R) = 1 R f (R) . A finite probability is computed via P =

R2

R1

dℓ f (R) =

R2

R1

dR f (R) .

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You can only forget about these “complications” if you are us- ing Cartesian parameters, like

  • the logarithm of a Jeffreys parameter
  • the components of vectors and ordinary (not positive def-

inite) tensors (and not their eigenvalues)

  • the Cartesian coordinates (x, y, z) in Euclidean space
  • the Newtonian time t (as a position on the time space,

not as a period) When working with Cartesian quantities, no difference be- tween a volumetric probability and a probability density, as dℓ = dx . For instance, the Gaussian “distribution” f (x) = f (x) = k exp − (x − x0)2/(2 σ2)

  • will be used only for Cartesian parameters.
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Poisson Ratio

When using probability densities, the homogeneous distribu- tion is represented by the probability density µ(ν) = 1

(ν + 1)(ν − 1/2)

, and probabilities are evaluated via P(ν1 ≤ ν < ν2) =

ν2

ν1

dν f (ν) . When using volumetric probabilities, the length element is given by ds(ν) = dν

(ν + 1)(ν − 1/2)

,

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the homogeneous probability distribution is µ(ν) = const. , and probabilities are evaluated via P(ν1 ≤ ν < ν2) =

ν2

ν1

ds(ν) f (ν) .

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k = +4 k = +2 k = 0 k = −2 k = −4 m = +4 m = +2 m = 0 m = −2 m = −4 Y = 0.1 Y = 1 Y = 10 Y = 100 ν = − . 9 5 ν = − . 5 ν = − . 2 5 ν = ν = . 2 5 ν = . 4 5 ν = . 4 9 9 ν = − . 9 9 9 9 κ μ 3 κ + μ Y = 3 κ − 2 μ 2 ( 3 κ + μ ) ν = k = +4 k = +2 k = −2 k = −4 k = 0

k = log(κ/κ0) m = log(μ/μ0)

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Numerical example: changing from { compressibility , bulk modulus } to { Young modulus , Poisson’s ratio }.