GEO7600 Inverse Theory 09 Sep 2008

Inverse Theory: Goals are to (1) Solve for parameters from observational data; (2) Know something about the range of models that can fit the data within uncertainties

Recall from last time that model parameter uncertainty can be estimated from the model covariance matrix Cm:

**********

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Cm =σ 2 GT

G ⎛

⎝ ⎜

⎞

⎠ ⎟

−1

Parameter variance for each model parameter is then:

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σmi2 = σ 2 G

TG

⎛

⎝ ⎜

⎞

⎠ ⎟

−1 ⎡

⎣ ⎢ ⎢

⎤

⎦ ⎥ ⎥ii

And:

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mi = ˜ m i ±σ mi

What should we expect Emin = eTe to be?

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Emin = eT

e = d −G ˜ m ( )T

d −G ˜ m ( )

Can substitute ; after lots of algebra and using the identity:

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d = Gmt +ε

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xT

Ax = Tr Ax xT ⎡

⎣ ⎢ ⎤ ⎦ ⎥

We get:

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Emin = Tr IN −G(GT

G)−1GT ⎡

⎣ ⎢ ⎤ ⎦ ⎥εε

T

And:

€

Emin = Tr IN −G(GT

G)−1GT ⎡

⎣ ⎢ ⎤ ⎦ ⎥ εε

T

If we assume uncorrelated errors:

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εεT

= Cε =σ 2 IN

Then:

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Emin =σ 2Tr IN ⎡ ⎣ ⎢

⎤ ⎦ ⎥−σ 2Tr G(G

TG)−1G

T ⎡ ⎣ ⎢

⎤ ⎦ ⎥= N − M( )σ

2

Useful take-home points:

One can always fit the data exactly if N = M.

If your measurement errors are unknown and N–M is “large”,

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˜ σ 2 =Emin

N − M

The latter is useful because often-times when σ2 is estimated independently, we find that !This generally indicates either (1) unanticipated noisein the measurements, (2) correlated errors or (3) (& very likely) the model is under-parameterized.Hence we define a chi-squared parameter€

˜ σ 2 >σ 2

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χ 2 =

ei2

i=1

N

∑σ 2 N − M( )

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χmin2 =1

The χ2 is commonly used to evaluate data fit and optimize the choice of number of parameters: 1) If , can safely add more model parameters 2) If , too many parameters (model is fitting noise).

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χmin2 >1

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χmin2 <1

Solution appraisal:Assume: zero-mean, jointly normal, uncorrelated errors

Estimate: Confidence intervals expressed as %: 100(1–)%

Case 1: Data error variance is known (= σ2)

-z +z1- /2/2€

mi = ˜ m i ± zσ mi

Desired confidence interval is ±z of the normal (z) distribution function

Can get this from standard statistical tables or codes

Suppose we want the 95% confidence interval:Typically use the F distribution = 1–/2

95% conf 1– = 0.95

1–/2 = 0.975

Looking up on a standard table, find F(z) = 0.975 when z = 1.96 (i.e. not quite 2σ).

Case 2: Use estimated error variance from

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˜ σ 2 =Emin

N − M

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mi = ˜ m i ± t1−α

2

N−M σ mi

(Look up the t-distribution as you would z-distribution in math probability tables, or use corresponding routines in Matlab or other stat codes).

For a multi-parameter model, in reality have confidence regions: hyperellipsoids in model parameter space

m1

m2

Emin

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˜ m 1€

˜ m 2 QuickTime™ and aTIFF (Uncompressed) decompressor

are needed to see this picture.

E.g. Mars…