7. Statistical estimation

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1 7. Statistical estimation Convex Optimization Boyd & Vandenberghe maximum likelihood estimation optimal detector design experiment design 7 1 Parametric distribution estimation distribution estimation problem: estimate probability density p(y) of a random variable from observed values parametric distribution estimation: choose from a family of densities p x (y), indexed by a parameter x maximum likelihood estimation maximize (over x) log p x (y) y is observed value l(x) = log p x (y) is called log-likelihood function can add constraints x C explicitly, or define p x (y) = 0 for x C a convex optimization problem if log p x (y) is concave in x for fixed y Statistical estimation 7 2

2 Linear measurements with IID noise linear measurement model y i = a T i x + v i, i = 1,..., m x R n is vector of unknown parameters v i is IID measurement noise, with density p(z) y i is measurement: y R m has density p x (y) = m p(y i a T i x) maximum likelihood estimate: any solution x of maximize l(x) = m log p(y i a T i x) (y is observed value) Statistical estimation 7 3 examples Gaussian noise N (0, σ 2 ): p(z) = (2πσ 2 /2 e z2 /(2σ 2), ML estimate is LS solution l(x) = m 2 log(2πσ2 ) 1 2σ 2 Laplacian noise: p(z) = (1/(2a))e z /a, m (a T i x y i ) 2 l(x) = m log(2a) 1 a ML estimate is l 1 -norm solution uniform noise on [ a, a]: m a T i x y i { m log(2a) a T l(x) = i x y i a, i = 1,..., m otherwise ML estimate is any x with a T i x y i a Statistical estimation 7 4

3 Logistic regression random variable y {0, 1} with distribution p = prob(y = 1) = exp(at u + b) 1 + exp(a T u + b) a, b are parameters; u R n are (observable) explanatory variables estimation problem: estimate a, b from m observations (u i, y i ) log-likelihood function (for y 1 = = y k = 1, y k+1 = = y m = 0): k l(a, b) = log exp(a T u i + b) m exp(a T u i + b) 1 + exp(a T u i + b) = concave in a, b k (a T u i + b) i=k+1 m log(1 + exp(a T u i + b)) Statistical estimation 7 5 PSfrag example replacements (n = 1, m = 50 measurements) prob(y = 1) u circles show 50 points (u i, y i ) solid curve is ML estimate of p = exp(au + b)/(1 + exp(au + b)) Statistical estimation 7 6

4 (Binary) hypothesis testing detection (hypothesis testing) problem given observation of a random variable X {1,..., n}, choose between: hypothesis 1: X was generated by distribution p = (p 1,..., p n ) hypothesis 2: X was generated by distribution q = (q 1,..., q n ) randomized detector a nonnegative matrix T R 2 n, with 1 T T = 1 if we observe X = k, we choose hypothesis 1 with probability t 1k, hypothesis 2 with probability t 2k if all elements of T are 0 or 1, it is called a deterministic detector Statistical estimation 7 7 detection probability matrix: D = [ T p T q ] = [ ] 1 Pfp P fn P fp 1 P fn P fp is probability of selecting hypothesis 2 if X is generated by distribution 1 (false positive) P fn is probability of selecting hypothesis 1 if X is generated by distribution 2 (false negative) multicriterion formulation of detector design minimize (w.r.t. R 2 +) (P fp, P fn ) = ((T p) 2, (T q) 1 ) subject to t 1k + t 2k = 1, k = 1,..., n t ik 0, i = 1, 2, k = 1,..., n variable T R 2 n Statistical estimation 7 8

5 scalarization (with weight λ > 0) minimize (T p) 2 + λ(t q) 1 subject to t 1k + t 2k = 1, t ik 0, i = 1, 2, k = 1,..., n an LP with a simple analytical solution { (1, 0) pk λq (t 1k, t 2k ) = k (0, 1) p k < λq k a deterministic detector, given by a likelihood ratio test if p k = λq k for some k, any value 0 t 1k 1, t 1k = 1 t 2k is optimal (i.e., Pareto-optimal detectors include non-deterministic detectors) minimax detector minimize max{p fp, P fn } = max{(t p) 2, (T q) 1 } subject to t 1k + t 2k = 1, t ik 0, i = 1, 2, k = 1,..., n an LP; solution is usually not deterministic Statistical estimation 7 9 PSfrag replacements example P = Pfn P fp solutions 1, 2, 3 (and endpoints) are deterministic; 4 is minimax detector Statistical estimation 7 10

6 Experiment design m linear measurements y i = a T i x + w i, i = 1,..., m of unknown x R n measurement errors w i are IID N (0, 1) ML (least-squares) estimate is ( m ˆx = a i a T i m y i a i error e = ˆx x has zero mean and covariance ( m E = E ee T = a i a T i confidence ellipsoids are given by {x (x ˆx) T E 1 (x ˆx) β} experiment design: choose a i {v 1,..., v p } (a set of possible test vectors) to make E small Statistical estimation 7 11 vector optimization formulation minimize (w.r.t. S n +) E = ( p k=1 m kv k vk T subject to m k 0, m m p = m m k Z variables are m k (# vectors a i equal to v k ) difficult in general, due to integer constraint relaxed experiment design assume m p, use λ k = m k /m as (continuous) real variable minimize (w.r.t. S n +) E = (1/m) ( p k=1 λ kv k vk T subject to λ 0, 1 T λ = 1 common scalarizations: minimize log det E, tr E, λ max (E),... can add other convex constraints, e.g., bound experiment cost c T λ B Statistical estimation 7 12

7 D-optimal design minimize log det ( p k=1 λ kv k vk T subject to λ 0, 1 T λ = 1 interpretation: minimizes volume of confidence ellipsoids dual problem maximize log det W + n log n subject to vk T W v k 1, k = 1,..., p interpretation: {x x T W x 1} is minimum volume ellipsoid centered at origin, that includes all test vectors v k complementary slackness: for λ, W primal and dual optimal λ k (1 v T k W v k ) = 0, k = 1,..., p optimal experiment uses vectors v k on boundary of ellipsoid defined by W Statistical estimation 7 13 example (p = 20) λ 1 = 0.5 PSfrag replacements λ 2 = 0.5 design uses two vectors, on boundary of ellipse defined by optimal W Statistical estimation 7 14

8 derivation of dual of page 7 13 first reformulate primal problem with new variable X: minimize subject to L(X, λ, Z, z, ν) = log det X 1 +tr log det X 1 X = p k=1 λ kv k vk T, λ 0, 1T λ = 1 ( Z ( X )) p λ k v k vk T z T λ+ν(1 T λ 1) k=1 minimize over X by setting gradient to zero: X 1 + Z = 0 minimum over λ k is unless v T k Zv k z k + ν = 0 dual problem maximize n + log det Z ν subject to vk T Zv k ν, k = 1,..., p change variable W = Z/ν, and optimize over ν to get dual of page 7 13 Statistical estimation 7 15

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