Extreme Value Theorem

stat._.jun·2026년 2월 16일

Thm. (EVT)
If f:KRf: K \to \mathbb{R} is conti. on a compact KRK \subseteq \mathbb{R}, then ff attains a maximum and minumum value.

Proof.
Since KK is compact, f(K)f(K) is compact as well.
We can set α=supf(K)\alpha = \sup f(K) since f(K)f(K) is bounded and also one can observe that αf(K)\alpha \in f(K) due to the closedness of f(K)f(K).
Thus x1K\exist x_1 \in K such that α=f(x1)\alpha = f(x_1).

Thm.
Θ0=(a,b)R\Theta_0 = (a, b) \subseteq \mathbb{R} : open set
C2(Θ0)\ell \in C^2(\Theta_0) with 1) 2(θ)<0\nabla^2 \ell(\theta) <0 for any θΘ0\theta \in \Theta_0 and 2) limθΘ0(θ)=\lim_{\theta \to \partial \Theta_0} \ell(\theta) = -\infty.
Then, !θ^Θ0\exist ! \hat \theta \in \Theta_0 such that (θ^)=0\ell(\hat \theta) = 0.

Sketch of the proof.
1. EVT를 통해서 θ^\hat \thetaΘ0\Theta_0에서 \ell의 최대임을 보이자. (Compact Set을 잘 잡는게 킥...?)
2. Strictly Concave하다는 것이 조건으로 부터 자동으로 도출되기 때문에, 이를 이용해서 유일한 해가 존재함을 보일 수 있다.

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