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

Institut de Physique du Globe de Paris & Université Pierre et Marie Curie (Paris VI) 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 ] = P [ A 1 ] + P [ A 2 ] − P [ A 1 ∩ A 2 ] Probability function: A measure function satisfying P [ A 0 ] = 1

Conditional probability: For a given set C ⊆ A 0 with nonzero probability, P [ C ] � = 0 , the conditional probability of any set A ⊆ A 0 , 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 .

Cells defined by constant increments of ∆ θ and ∆ ϕ : 1 n f ( θ , ϕ ) = lim (probability density) N ∆ θ ∆ ϕ ∆ θ ∆ ϕ → 0 Cells having constant surface ∆ S : 1 n f ( θ , ϕ ) = lim (volumetric probability) N ∆ S ∆ S → 0 Finite probability: � � � � P [ A ] = f ( θ , ϕ ) = dS ( θ , ϕ ) f ( θ , ϕ ) d θ d ϕ � �� � � �� � A A dS ( θ , ϕ ) = sin θ d θ d ϕ

Fisher probability density: κ f ( θ , ϕ ) = 4 π sinh κ sin θ exp ( κ sin θ ) � � P [ A ] = d θ d ϕ f ( θ , ϕ ) � �� � A Fisher (2D) volumetric probability: κ f ( θ , ϕ ) = 4 π sinh κ exp ( κ sin θ ) � � P [ A ] = dS ( θ , ϕ ) f ( θ , ϕ ) dS ( θ , ϕ ) = sin θ d θ d ϕ ; � �� � A

Volumetric probability f and probability density f : � A dV ( x 1 , x 2 . . . x n ) f ( x 2 , x 2 . . . x n ) P [ A ] = � A dx 1 dx 2 . . . dx n f ( x 2 , x 2 . . . x n ) = To pass from one to the other, express dV ( x 1 , x 2 . . . x n ) = v ( x 1 , x 2 . . . x n ) dx 1 dx 2 . . . dx n Example: for spherical coordinates in Euclidean space dV ( r , θ , ϕ ) = r 2 sin θ dr d θ d ϕ

x = r sin θ cos ϕ y = r sin θ sin ϕ z = r cos θ   ∂ x / ∂ r ∂ x / ∂θ ∂ x / ∂ϕ  = r 2 sin θ det ∂ y / ∂ r ∂ y / ∂θ ∂ y / ∂ϕ  ∂ z / ∂ r ∂ z / ∂θ ∂ z / ∂ϕ dV S ( r , θ , ϕ ) = r 2 sin θ dr d θ d ϕ dV C ( x , y , z ) = dx dy dz ⇒

Most of the positive quantities are Jeffreys quantities, like the electric resistance R . From D = | log ( R 2 / R 1 ) | it follows d ℓ = 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 � R 2 � R 2 P = d ℓ f ( R ) = dR f ( R ) . R 1 R 1

You can only forget about these “complications” if you decide to only work with Cartesian parameters , like • the logarithm of a Jeffreys parameter • the components of vectors and ordinary tensors (not pos- itive definite) • the Cartesian coordinates ( x , y , z ) in Euclidean space • the Newtonian time t (as a position on the time space, not as a period) If working with Cartesian quantities only, no difference be- tween a volumetric probability and a probability density, as d ℓ = dx . For instance, the Gaussian “distribution” � − ( x − x 0 ) 2 / ( 2 σ 2 ) � f ( x ) = f ( x ) = k exp will be used only for Cartesian parameters.

Poisson Ratio When using probability densities, the homogeneous distribu- tion is represented by the probability density 1 µ ( ν ) = , ( ν + 1 )( ν − 1/2 ) and probabilities are evaluated via � ν 2 P ( ν 1 ≤ ν < ν 2 ) = d ν f ( ν ) . ν 1 When using volumetric probabilities, the length element is given by d ν ds ( ν ) = , ( ν + 1 )( ν − 1/2 )

the homogeneous probability distribution is µ ( ν ) = const. , and probabilities are evaluated via � ν 2 P ( ν 1 ≤ ν < ν 2 ) = ds ( ν ) f ( ν ) . ν 1

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

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