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Дата изменения: Wed Apr 1 23:53:38 2015
Дата индексирования: Sun Apr 10 11:57:39 2016
Кодировка:
Astro 193 : 2015 Apr 1
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Follow-up
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Spearman's and Kendall's Class on Apr 10 Gibbs sampler acceptance rates: third time's the charm Adaptive MCMC

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Bias-Variance Tradeoff / Kernel Density Estimation Signal Processing: Fourier Transforms


Adaptive MCMC
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Run a chain for a short number of iterations in the kth block with modified step size (Hyungsuk Tak) compute acceptance rate if >0.4/npar, inflate step size by f+=e+min{0.07,1/k} if <0.2/npar, deflate step size by f­=e­min{0.07,1/k} f± 1 as k throw out iterations from before f± converges

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Bias-variance tradeoff
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Given sample x = {x1, x2, ..., xn} ~ f(x), obtain estimate f(x) Least disruptive solution: histogram · f(x) = (1/h) {Ii: |xi-x|h}
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f(x) = mean({Yi: |xi-x|h}) with regression Y = f(x) +

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Affected by noise fluctuations. Also, how to choose h? Minimize Risk, mean squared error R(f(x),f(x)) = E[(f(x)-f(x))І] = E[fІ] - 2E[ff] + E[fІ] = fІ - 2fE[f] + E[fІ] + E[f]І - E[f]І = E[f]І - 2fE[f] + fІ + E[fІ] -E[f]І = E[f]І - 2fE[f] + fІ + E[fІ] -E[f]І = (E[f]-f)І + var[f] = bias2 + variance

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When h is small, bias is small and variance is large; when h is large, bias is large and variance is small. Optimize h to minimize Risk. As h0, R ~ o(h4 + 1/nh), h
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optimal

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~ o(1/n1/5)
­1/5

in Gaussian case, hoptimal = 1.06 f n


Kernel Density Estimation
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A "kernel" K(x) is a non-negative real-valued function such that xR dx K(x) = 1 ; E[x] = x
R

dx x K(x) = 0 ; E[x2] = xR dx x K(x) <

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The Kernel density estimator for a bandwidth h, fh(x) = [1/nh]
i=1..n

K([x-xi]/h)

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Types of Kernels
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Boxcar, Gaussian, Triangle, Tophat RMF
i=1..n li(x)Yi

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For linear smoothers, fh(x) = li = K([x-xi]/h] /
i=1..n

K([x-xi]/h), centered on xi, weighting all of the sample

Can write fh = LY in matrix notation Effective degrees of freedom, = trace(L) =
·
i=1..n

Lii

Choosing the bandwidth via cross-validation (Jackknife/leave-one-out) 2 CV|h = (1/n) i=1..n (Yi - fh(-i)(xi)) Generalized CV, GCV(h) (1/n) [1-/n]
-2 i=1..n

(Yi-fh(xi))

2