Download A Course in Mathematical Statistics and Large Sample Theory by Rabi Bhattacharya, Lizhen Lin, Victor Patrangenaru PDF

By Rabi Bhattacharya, Lizhen Lin, Victor Patrangenaru

This graduate-level textbook is essentially geared toward graduate scholars of data, arithmetic, technology, and engineering who've had an undergraduate path in data, an higher department direction in research, and a few acquaintance with degree theoretic chance. It presents a rigorous presentation of the middle of mathematical statistics.
Part I of this booklet constitutes a one-semester path on easy parametric mathematical records. half II bargains with the big pattern conception of facts - parametric and nonparametric, and its contents could be coated in a single semester to boot. half III presents short bills of a couple of themes of present curiosity for practitioners and different disciplines whose paintings comprises statistical methods.

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Show that X is admissible under squared error loss: L(θ, a) = (θ − a)2 . t. loss function (θ−a) . θ Ex. 8. Show that, under squared error loss, (a) X is an admissible estimator of μ ∈ Θ1 = Rk when the sample is from N(μ, σ 2 I) with μ, σ 2 both unknown and k = 1, 2, and that (b) X is inadmissible if k ≥ 3 (Θ = Rk × (0, ∞)). ] Ex. 9. Let X be the mean of a random sample from N (μ, Σ) when μ ∈ Rk ≡ Θ1 , Σ ∈ Θ2 ≡ set of all symmetric positive definite k × k matrices. Let Θ = Θ1 × Θ2 , A = Θ1 , and let the loss function be squared error L(θ, a) = |μ − a|2 .

Proof. Suppose d is inadmissible when the parameter space is Θ = Θ1 × Θ2 . Then there exists a decision rule d1 and a point θ0 = (θ10 , θ20 ) such that R(θ, d1 ) ≤ R(θ, d) ∀ θ ∈ Θ and R(θ0 , d1 ) < R(θ0 , d). But this implies R((θ1 , θ20 ), d1 ) ≤ R((θ1 , θ20 ), d) ∀ θ1 ∈ Θ1 , R((θ10 , θ20 ), d1 ) < R((θ10 , θ20 ), d), contradicting the fact that d is admissible when the parameter space is Θ1 × {θ20 }. 6 Notes and References For Bayes estimation we refer to Ferguson (1967, Sects. 3), and Lehmann and Casella (1998, Chaps.

Xk ) have the distribution N (θ, I), θ = (θ1 , . . , θk ) ∈ Rk , I k × k identity matrix. Assume that E|g(X)|2 < ∞ and define hj (y) = E(gj (X)|Xj )Xj =y = Egj (X1 , . . , Xj−1 , y, Xj+1 , . . , Xk ). 3 (in place of g there), 1 ≤ j ≤ k. 56) E|X + g(X) − θ|2 = k + E ⎝|g(X)|2 + 2 gj (x) |x=X ⎠ . ∂xj j=1 Proof. The left side equals k E|X − θ|2 + E|g(X)|2 + 2E(X − θ) · g(X) = k + E|g(X)|2 + 2 E(Xj − θj )gj (X). j=1 Now E(Xj − θj )gj (X) = E[(Xj − θj ) · E(gj (X)|Xj )] = E(Xj − θj )hj (Xj ). 3 (with g = hj ) to get E(Xj − θj )hj (Xj ) = Ehj (Xj ) = E ∂ ∂xj gj (x) .

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