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Material Type: Notes; Class: Prb&Rand Proc; Subject: Electrical Engineering And Computer Science; University: University of Michigan - Ann Arbor; Term: Fall 2001;
Typology: Study notes
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EECS 501 ESTIMATION: MLE, MAP, LS Fall 2001
Model: A known model of system or process with unknown parameter a. Data: An observation R of a random variable r whose pdf depends on a. Model→ fr|a(R|A): If knew a = A, would know pdf of observation r. Goal: Estimate a from R and conditional pdf fr|a(R|A): Compute ˆa(R). Example: Flip coin 10 times. Data: #heads in 10 independent flips. Model: Binomial pmf for r. Unknown parameter: a=Pr[heads].
BLUE: Best (minimum variance) Linear Unbiased Estimator of constant x from y = Hx + v, E[v] = 0 is ˆx(Y ) = (H′H)−^1 H′Y. Proof: p. 290.
2a. MEP: Min Error Prob: c(e) =
0 if |e| < ≤; 1 if |e| > ≤.
“close only counts in horseshoes” “a miss is as good as a mile” E[c(e)] = 1 −
−∞ dR^
∫ (^) aˆ(R)+≤ a ˆ(R)−≤ dA fr,a(R, A) = 1^ −^2 ≤^
−∞ fr,a(R,^ ˆa(R))dR. This is minimized when fr,a(R, ˆa(R)) maximized for each R.
MAP: Max A Posteriori: ˆaM AP (R) = argmaxA [fr|a(R|A)fa(A)] (compare MLE).
Compute: (^) ∂A∂ [log fr|a(R|A) + log fa(A)] = 0. MEP criterion→MAP solution.
2b. LSE: Least Squares Estimation criterion: c(e) = e^2. Penalize big errors. LSE: ˆaLS (R) = E[a|r = R] =
∫ Afr|a(R|A)fa(A)dA fr|a(R|A′)fa(A′)dA′
Denominator just fr(R) : no effect on argmax of A Proof: Page 298. Moment of inertia minimized around center of mass.
Bias: Let a be an unknown constant Aact so that fa(A) = δ(A − Aact). DEF: ˆa(R) is unbiased if E[ˆa(r)] = Aact ↔ E[e] = 0. How to compute: E[ˆa(r)] =
ˆa(R)fr|a(R|A)δ(A − Aact)dR dA =
ˆa(R)fr|a(R|Aact)dR. MSE: ˆa(R) unbiased→ E[(ˆa(r) − Aact)^2 ] = σ^2 ˆa(r) →MSE =variance of ˆa(R).
EECS 501 ESTIMATION EXAMPLES Fall 2001 Given: Flip coin with Pr[heads]=a. Data: #heads in 10 independent flips. Model: pmf pr|a(R|A) =
R
Goal: Estimate a=Pr[heads] from r=#heads in 10 flips and a priori fa(A).
MLE: (^) ∂A∂ [log
R
Bias: E[ˆaM LE (r)] = E[ 10 r ] = 10 A 10 act = Aact → ˆaM LE (r) unbiased. MSE: E[(ˆaM LE (r) − Aact)^2 ] = σ^210 r (since unbiased) = 10 Aact 100 (1− Aact).
EX2: Now suppose have fa(A) = 1 for 0 ≤ A ≤ 1 (Bayesian problem). MAP: log fa(A) = 0 →same algebra→ ˆaM AP (R) = ˆaM LE (R) = 10 R. Have: Uniform a priori pdf a ∼ N (0, σ^2 → ∞) → ˆaM AP (R) = ˆaM LE (R).
EX3: Now suppose have fa(A) = 2A for 0 ≤ A ≤ 1 (Bayesian problem). MAP: (^) ∂A∂ [log
R
EX4: Now suppose have fa(A) = 1 for 0 ≤ A ≤ 1 (Bayesian problem).
LSE: ˆaLS (R) = E[a|r = R] =
(^10) R )AR (^) (1−A) 10 −R (^) dA ∫ (^1) 0 (^
(^10) R )AR(1−A) 10 −RdA =^
R+
Ref: Schaum’s Outline Math. Handbook, (15.24) on p. 95. ˆaLS (5) = 12. Note: Even with a uniform a priori distribution for a, ˆaLS still slanted!
LLSE: min E[(a − aˆ(r))^2 ] such that ˆa(R) = cR + b for some constants b, c. Soln: (^) ∂c∂ E[(a − cr − b)^2 ] = 0 → aˆLLSE (R) = E[a] + λ σar (^2) r (R − E[r]). & (^) ∂b∂ E[(a − cr − b)^2 ] = 0. This is Linear Least Squares Estimator.
LSE: r, a jointly Gaussian→
r a
E[r] E[a]
σ^2 r λra λra σ^2 a
→ aˆLS (R) = E[a|r = R] = E[a] + λ σar (^2) r (R − E[r]) = ˆaLLSE (R)! Fact: Two very different problems have the same solution!
Norm: Normalized form: (ˆa(R) − E[a])/σa = ρar (R − E[r])/σr.
MSE: E[(a − ˆa(r))^2 ] = σ^2 a − λ
(^2) ar σ r^2 →^ E
ˆa(r)−a σa
= 1 − ρ^2 ar.