证明：I类错误在边界点取到最大值_一类错误边界-优快云博客

这篇博客讨论了在一维参数空间中，对于单侧检验，最大一类错误（第一类错误）发生在参数边界的情况。错误的证明通过不正确的T_n定义和混淆了变量与常数，导致结论错误。正确的证明则展示了当参数等于边界值时，错误率达到最大，并且当参数大于边界值时，错误率趋近于0。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

问题描述：

$X1,X2,...,Xn∼i.i.dPois(λ),λ>0X_1, X_2, ..., X_n \overset{i.i.d}{\sim} Pois(\lambda), \lambda > 0$ for some unknown $λ\lambda$

$H0:λ≥λ0,H1:λ<λ0,λ0>0H_0: \lambda \ge \lambda_0, H_1: \lambda \lt \lambda_0, \lambda_0 > 0$

theorem: the maximum of the type 1 error is achieved at the boundary of $Θ1\Theta_0\ and\ \Theta_1$ for a one-sided tests, where the parameter space is 1-dimensional.

the wrong proof:

$α\alpha_\lambda = P_{\lambda \in \Theta_0}(\psi = 1) \\ where, \psi = \mathbb{I}(T_n < -q_\alpha), with\ level\ \alpha$

on this question, let $Tn=nXnˉ−λλT_n=\sqrt{n} \frac{\bar{X_n}-\lambda}{\sqrt{\lambda}}$

then, $αλ=Pλ∈Θ0(Tn<−qα)=Pλ≥λ0(nXnˉ−λλ<q−α)=Pλ≥λ0(nXnˉ−λλ<−qα)\alpha_{\lambda} = P_{\lambda \in \Theta_0}(T_n < -q_{\alpha}) = P_{\lambda \ge \lambda_0}(\sqrt{n} \frac{\bar{X_n}-\lambda}{\sqrt{\lambda}} < q_{-\alpha}) = P_{\lambda \ge \lambda_0}(\sqrt{n} \frac{\bar{X_n}-\lambda}{\sqrt{\lambda}} < -q_{\alpha})$

let $Tn,λ=nXnˉ−λλT_{n, \lambda}= \sqrt{n} \frac{\bar{X_n}-\lambda}{\sqrt{\lambda}}$ and $Tn,λ0=nXnˉ−λ0λ0T_{n, \lambda_0}= \sqrt{n} \frac{\bar{X_n}-\lambda_0}{\sqrt{\lambda_0}}$

when $λ≥λ0\lambda \ge \lambda_0$ , we have:

$Xnˉ−λ≤Xnˉ−λ0⇒Xnˉ−λλ0≤Xnˉ−λ0λ0⇒Xnˉ−λλ0≤Xnˉ−λλ≤Xnˉ−λ0λ0⇒Tn,λ≤Tn,λ0\lambda \ge \lambda_0 \\ \Rightarrow\ -\lambda \le -\lambda_0 \\ \Rightarrow\ \bar{X_n} - \lambda \le \bar{X_n} - \lambda_0 \\ \Rightarrow \frac{\bar{X_n}-\lambda}{\lambda_0} \le \frac{\bar{X_n}-\lambda_0}{\lambda_0} \\ \Rightarrow \frac{\bar{X_n}-\lambda}{\lambda_0} \le\frac{\bar{X_n}-\lambda}{\lambda} \le \frac{\bar{X_n}-\lambda_0}{\lambda_0} \\ \Rightarrow T_{n, \lambda} \le T_{n, \lambda_0}$

let event $T_{n,\lambda} \lt -q_a$ and $T_{n,\lambda_0} \lt -q_a$ ,
$\subseteq A \Leftrightarrow P(B) \le P(A) \Leftrightarrow P(T_{n,\lambda_0} \lt -q_a) \le P(T_{n,\lambda} \lt -q_a)$

which means: the minimum of the type 1 error is achieved at the boundary of $Θ0\Theta_0$

what’s the problem in this proof? --hint, the wrong $T_n$ and the confusion of variable and constant.

the correct proof:

$α\alpha_\lambda = P_{\lambda \in \Theta_0}(\psi = 1) \\ where, \psi = \mathbb{I}(T_n < -q_\alpha), with\ level\ \alpha$

on this question, let $Tn=nXnˉ−λ0λ0T_n=\sqrt{n} \frac{\bar{X_n}-\lambda_0}{\sqrt{\lambda_0}}$

then, $αλ=Pλ∈Θ0(Tn<−qα)=Pλ≥λ0(nXnˉ−λ0λ0<q−α)=Pλ≥λ0(nXnˉ−λλ<−qα)\alpha_{\lambda} = P_{\lambda \in \Theta_0}(T_n < -q_{\alpha}) = P_{\lambda \ge \lambda_0}(\sqrt{n} \frac{\bar{X_n}-\lambda_0}{\sqrt{\lambda_0}} < q_{-\alpha}) = P_{\lambda \ge \lambda_0}(\sqrt{n} \frac{\bar{X_n}-\lambda}{\sqrt{\lambda}} < -q_{\alpha})$

let $Tn,λ0=nXnˉ−λ0λ0T_{n, \lambda_0}= \sqrt{n} \frac{\bar{X_n}-\lambda_0}{\sqrt{\lambda_0}}$

when $λ=λ0\lambda = \lambda_0$ we have:

$Xnˉ⟶Pλ0\bar{X_n}\overset{\mathbb{P}}{\longrightarrow} \lambda_0$ , then

$Tn,λ0⟶dN(0,1)T_{n, \lambda_0} \overset{d}{\longrightarrow} N(0,1)$ , because of CLT

$αλ=Pλ=λ0(nXnˉ−λ0λ0<−qα)=Φ(−qα)=1−Φ(qα)\alpha_{\lambda} = P_{\lambda = \lambda_0}(\sqrt{n} \frac{\bar{X_n}-\lambda_0}{\sqrt{\lambda_0}} < -q_{\alpha}) = \Phi(-q_{\alpha}) = 1 - \Phi(q_{\alpha})$

when $λ>λ0\lambda \gt \lambda_0$ , we have:

$Xnˉ⟶Pλ\bar{X_n}\overset{\mathbb{P}}{\longrightarrow} \lambda$ , then by continuous mapping theorem

$Tn,λ0⟶Pnλ−λ0λ0⟶Pn→∞+∞T_{n, \lambda_0} \overset{\mathbb{P}}{\longrightarrow} \sqrt{n} \frac{\lambda-\lambda_0}{\sqrt{\lambda_0}} \underset{n\to\infty}{\overset{\mathbb{P}}{\longrightarrow}} +\infty$ , therefore,

$αλ=Pλ>λ0(nXnˉ−λ0λ0<−qα)⟶PPλ>λ0(nλ−λ0λ0<−qα)⟶PPλ>λ0(+∞<−qα)→0\alpha_{\lambda} = P_{\lambda > \lambda_0}(\sqrt{n} \frac{\bar{X_n}-\lambda_0}{\sqrt{\lambda_0}} < -q_{\alpha}) \\ \overset{\mathbb{P}}{\longrightarrow} P_{\lambda > \lambda_0}(\sqrt{n} \frac{\lambda-\lambda_0}{\sqrt{\lambda_0}} < -q_{\alpha}) \\ \overset{\mathbb{P}}{\longrightarrow} P_{\lambda > \lambda_0}(+\infty < -q_{\alpha}) \rightarrow 0$