统计推断(一) Hypothesis Test

1. Binary Bayesian hypothesis testing

1.0 Problem Setting

• Hypothesis
  • Hypothesis space $\mathcal{H}=\{H_0, H_1\}$H={H0​,H1​}
  • Bayesian approach: Model the valid hypothesis as an RV H
  • Prior $P_0 = p_\mathsf{H}(H_0), P_1=p_\mathsf{H}(H_1)=1-P_0$P0​=pH​(H0​),P1​=pH​(H1​)=1−P0​
• Observation
  • Observation space $\mathcal{Y}$Y
  • Observation Model $p_\mathsf{y|H}(\cdot|H_0), p_\mathsf{y|H}(\cdot|H_1)$py∣H​(⋅∣H0​),py∣H​(⋅∣H1​)
• Decision rule $f:\mathcal{Y\to H}$f:Y→H
• Cost function $C: \mathcal{H\times H} \to \mathbb{R}$C:H×H→R
  • Let $C_{ij}=C(H_j,H_i), correct hypo is H_j$Cij​=C(Hj​,Hi​),correcthypoisHj​
  • $C$C is valid if $C_{jj}<C_{ij}$Cjj​<Cij​
• Optimum decision rule $\hat{H}(\cdot) = \arg\min\limits_{f(\cdot)}\mathbb{E}[C(\mathsf{H},f(\mathsf{y}))]$H^(⋅)=argf(⋅)min​E[C(H,f(y))]

1.1 Binary Bayesian hypothesis testing

Theorem: The optimal Bayes’ decision takes the form
$L(\mathsf{y}) \triangleq \frac{p_\mathsf{y|H}(\cdot|H_1)}{p_\mathsf{y|H}(\cdot|H_0)} \overset{H_1} \gtreqless \frac{P_0}{P_1} \frac{C_{10}-C_{00}}{C_{01}-C_{11}} \triangleq \eta$L(y)≜py∣H​(⋅∣H0​)py∣H​(⋅∣H1​)​⋛H1​​P1​P0​​C01​−C11​C10​−C00​​≜η
Proof:
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Given $y^*$y∗

• if $f(y^*)=H_0$f(y∗)=H0​, $\mathbb{E}=C_{00}p_{\mathsf{H|y}}(H_0|y^*)+C_{01}p_{\mathsf{H|y}}(H_1|y^*)$E=C00​pH∣y​(H0​∣y∗)+C01​pH∣y​(H1​∣y∗)
• if $f(y^*)=H_1$f(y∗)=H1​, $\mathbb{E}=C_{10}p_{\mathsf{H|y}}(H_0|y^*)+C_{11}p_{\mathsf{H|y}}(H_1|y^*)$E=C10​pH∣y​(H0​∣y∗)+C11​pH∣y​(H1​∣y∗)

So
$\frac{p_\mathsf{H|y}(H_1|y^*)}{p_\mathsf{H|y}(H_0|y^*)} \overset{H_1} \gtreqless \frac{C_{10}-C_{00}}{C_{01}-C_{11}}$pH∣y​(H0​∣y∗)pH∣y​(H1​∣y∗)​⋛H1​​C01​−C11​C10​−C00​​
备注：证明过程中，注意贝叶斯检验为确定性检验，因此对于某个确定的 y，$f(y)=H_1$f(y)=H1​ 的概率要么为 0 要么为 1。因此对代价函数求期望时，把 H 看作是随机变量，而把 $f(y)$f(y) 看作是确定的值来分类讨论

Special cases

• Maximum a posteriori (MAP)
  • $C_{00}=C_{11}=0,C_{01}=C_{10}=1$C00​=C11​=0,C01​=C10​=1
  • $\hat{H}(y)==\arg\max\limits_{H\in\{H_0,H_1\}} p_\mathsf{H|y}(H|y)$H^(y)==argH∈{H0​,H1​}max​pH∣y​(H∣y)
• Maximum likelihood (ML)
  • $C_{00}=C_{11}=0,C_{01}=C_{10}=1, P_0=P_1=0.5$C00​=C11​=0,C01​=C10​=1,P0​=P1​=0.5
  • $\hat{H}(y)==\arg\max\limits_{H\in\{H_0,H_1\}} p_\mathsf{y|H}(y|H)$H^(y)==argH∈{H0​,H1​}max​py∣H​(y∣H)

1.2 Likelyhood Ratio Test

Generally, LRT
$L(\mathsf{y}) \triangleq \frac{p_\mathsf{y|H}(\cdot|H_1)}{p_\mathsf{y|H}(\cdot|H_0)} \overset{H_1} \gtreqless \eta$L(y)≜py∣H​(⋅∣H0​)py∣H​(⋅∣H1​)​⋛H1​​η

• Bayesian formulation gives a method of calculating $\eta$η
• $L(y)$L(y) is a sufficient statistic for the decision problem
• $L(y)$L(y) 的可逆函数也是充分统计量

充分统计量

1.3 ROC

• Detection probability $P_D = P(\hat{H}=H_1 | \mathsf{H}=H_1)$PD​=P(H^=H1​∣H=H1​)
• False-alarm probability $P_F = P(\hat{H}=H_1 | \mathsf{H}=H_0)$PF​=P(H^=H1​∣H=H0​)

性质（重要！）

• LRT 的 ROC 曲线是单调不减的

2. Non-Bayesian hypo test

• Non-Bayesian 不需要先验概率或者代价函数

Neyman-Pearson criterion

$\max_{\hat{H}(\cdot)}P_D \ \ \ s.t. P_F\le \alpha$H^(⋅)max​PD​   s.t.PF​≤α

Theorem(Neyman-Pearson Lemma)：NP 准则的最优解由 LRT 得到，其中 $\eta$η 由以下公式得到
$P_F=P(L(y)\ge\eta | \mathsf{H}=H_0) = \alpha$PF​=P(L(y)≥η∣H=H0​)=α
Proof：

物理直观：同一个 $P_F$PF​ 时 LRT 的 $P_D$PD​ 最大。物理直观来看，LRT 中判决为 H1 的区域中 $\frac{p(y|H_1)}{p(y|H_0)}$p(y∣H0​)p(y∣H1​)​ 都尽可能大，因此 $P_F$PF​ 相同时 $P_D$PD​ 可最大化

备注：NP 准则最优解为 LRT，原因是

• 同一个 $P_F$PF​ 时， LRT 的 $P_D$PD​ 最大
• LRT 取不同的 $\eta$η 时，$P_F$PF​ 越大，则 $P_D$PD​ 也越大，即 ROC 曲线单调不减

3. Randomized test

3.1 Decision rule

• Two deterministic decision rules $\hat{H'}(\cdot),\hat{H''}(\cdot)$H′^(⋅),H′′^(⋅)

• Randomized decision rule $\hat{H}(\cdot)$H^(⋅) by time-sharing
  $\hat{\mathrm{H}}(\cdot)=\left\{\begin{array}{ll}{\hat{H}^{\prime}(\cdot),} & {\text { with probability } p} \\ {\hat{H}^{\prime \prime}(\cdot),} & {\text { with probability } 1-p}\end{array}\right.$H^(⋅)={H^′(⋅),H^′′(⋅),​ with probability p with probability 1−p​

  • Detection prob $P_D=pP_D'+(1-p)P_D''$PD​=pPD′​+(1−p)PD′′​
  • False-alarm prob $P_F=pP_F'+(1-P)P_F''$PF​=pPF′​+(1−P)PF′′​

• A randomized decision rule is fully described by $p_{\mathsf{\hat{H}|y}}(H_m|y)$pH^∣y​(Hm​∣y) for m=0,1

3.2 Proposition

1. Bayesian case: cannot achieve a lower Bayes’ risk than the optimum LRT

   Proof: Risk for each y is linear in $p_{\mathrm{H} | \mathbf{y}}\left(H_{0} | \mathbf{y}\right)$pH∣y​(H0​∣y), so the minima is achieved at 0 or 1, which degenerate to deterministic decision
   KaTeX parse error: No such environment: align at position 8: \begin{̲a̲l̲i̲g̲n̲}̲ \varphi(\mathb…

2. Neyman-Pearson case:

   1. continuous-valued: For a given $P_F$PF​ constraint, randomized test cannot achieve a larger $P_D$PD​ than optimum LRT
   2. discrete-valued: For a given $P_F$PF​ constraint, randomized test can achieve a larger $P_D$PD​ than optimum LRT. Furthermore, the optimum rand test corresponds to simple time-sharing between the two LRTs nearby

3.3 Efficient frontier

Boundary of region of achievable $(P_D,P_F)$(PD​,PF​) operation points

• continuous-valued: ROC of LRT
• discrete-valued: LRT points and the straight line segments

Facts

• $P_D \ge P_F$PD​≥PF​
• efficient frontier is concave function
• $\frac{dP_D}{dP_F}=\eta$dPF​dPD​​=η

4. Minmax hypo testing

prior: unknown, cost fun: known

4.1 Decision rule

• minmax approach
  $\hat H(\cdot)=\arg\min_{f(\cdot)}\max_{p\in[0,1]} \varphi(f,p)$H^(⋅)=argf(⋅)min​p∈[0,1]max​φ(f,p)

• optimal decision rule
  $\hat H(\cdot)=\hat{H}_{p_*}(\cdot) \\ p_* = \arg\max_{p\in[0,1]} \varphi(\hat H_p, p)$H^(⋅)=H^p∗​​(⋅)p∗​=argp∈[0,1]max​φ(H^p​,p)

  要想证明上面的最优决策，首先引入 mismatch Bayes decision
  $\hat{\mathrm{H}}_q(y)=\left\{ \begin{array}{ll}{H_1,} & {L(y) \ge \frac{1-q}{q}\frac{C_{10}-C_{00}}{C_{01}-C_{11}}} \\ {H_0,} & {otherwise}\end{array}\right.$H^q​(y)={H1​,H0​,​L(y)≥q1−q​C01​−C11​C10​−C00​​otherwise​
  代价函数如下，可得到 $\varphi(\hat H_q,p)$φ(H^q​,p) 与概率 $p$p 成线性关系
  $\varphi(\hat H_q,p)=(1-p)[C_{00}(1-P_F(q))+C_{10}P_F(q)] + p[C_{01}(1-P_D(q))+C_{11}P_D(q)]$φ(H^q​,p)=(1−p)[C00​(1−PF​(q))+C10​PF​(q)]+p[C01​(1−PD​(q))+C11​PD​(q)]
  Lemma: Max-min inequality
  $\max_x\min_y g(x,y) \le \min_y\max_x g(x,y)$xmax​ymin​g(x,y)≤ymin​xmax​g(x,y)
  Theorem:
  $\min_{f(\cdot)}\max_{p\in[0,1]}\varphi(f,p)=\max_{p\in[0,1]}\min_{f(\cdot)}\varphi(f,p)$f(⋅)min​p∈[0,1]max​φ(f,p)=p∈[0,1]max​f(⋅)min​φ(f,p)
  Proof of Lemma: Let $h(x)=\min_y g(x,y)$h(x)=miny​g(x,y)
  $\begin{aligned} g(x) &\leq f(x, y), \forall x \forall y \\ \Longrightarrow \max _{x} g(x) & \leq \max _{x} f(x, y), \forall y \\ \Longrightarrow \max _{x} g(x) & \leq \min _{y} \max _{x} f(x, y) \end{aligned}$g(x)⟹xmax​g(x)⟹xmax​g(x)​≤f(x,y),∀x∀y≤xmax​f(x,y),∀y≤ymin​xmax​f(x,y)​
  Proof of Thm: 先取 $\forall p_1,p_2 \in [0,1]$∀p1​,p2​∈[0,1]，可得到
  $\varphi(\hat H_{p_1},p_1)=\min_f \varphi(f,p_1) \le \max_p \min_f \varphi(f,p) \le \min_f \max_p \varphi(f, p) \le \max_p \varphi(\hat H_{p_2}, p)$φ(H^p1​​,p1​)=fmin​φ(f,p1​)≤pmax​fmin​φ(f,p)≤fmin​pmax​φ(f,p)≤pmax​φ(H^p2​​,p)
  由于 $p_1,p_2$p1​,p2​ 任取时上式都成立，因此可以取 $p_1=p_2=p_*=\arg\max_p \varphi(\hat H_p, p)$p1​=p2​=p∗​=argmaxp​φ(H^p​,p)

  要想证明定理则只需证明 $\varphi(\hat H_{p_*},p_*)=\max_p \varphi(\hat H_{p_*}, p)$φ(H^p∗​​,p∗​)=maxp​φ(H^p∗​​,p)

  由前面可知 $\varphi(\hat H_q,p)$φ(H^q​,p) 与 $p$p 成线性关系，因此要证明上式

  • 若 $p_* \in (0,1)$p∗​∈(0,1)，只需 $\left.\frac{\partial \varphi\left(\hat{H}_{q^{*}}, p\right)}{\partial p}\right|_{\text {for any } p}=0$∂p∂φ(H^q∗​,p)​∣∣∣∣​for any p​=0，等式自然成立
  • 若 $p_* = 1$p∗​=1，只需 $\left.\frac{\partial \varphi\left(\hat{H}_{q^{*}}, p\right)}{\partial p}\right|_{\text {for any } p} > 0$∂p∂φ(H^q∗​,p)​∣∣∣∣​for any p​>0，最优解就是 $p=1$p=1；$q_*=0$q∗​=0 同理

  根据下面的引理，可以得到最优决策就是 Bayes 决策 $p_*=\arg\max_p \varphi(\hat H_p, p)$p∗​=argmaxp​φ(H^p​,p)，其中 $p_*$p∗​ 满足
  $\begin{aligned} 0 &=\frac{\partial \varphi\left(\hat{H}_{p_{*}}, p\right)}{\partial p} \\ &=\left(C_{01}-C_{00}\right)-\left(C_{01}-C_{11}\right) P_{\mathrm{D}}\left(p_{*}\right)-\left(C_{10}-C_{00}\right) P_{\mathrm{F}}\left(p_{*}\right) \end{aligned}$0​=∂p∂φ(H^p∗​​,p)​=(C01​−C00​)−(C01​−C11​)PD​(p∗​)−(C10​−C00​)PF​(p∗​)​
  Lemma:
  $\left.\frac{\mathrm{d} \varphi\left(\hat{H}_{p}, p\right)}{\mathrm{d} p}\right|_{p=q}=\left.\frac{\partial \varphi\left(\hat{H}_{q}, p\right)}{\partial p}\right|_{p=q}=\left.\frac{\partial \varphi\left(\hat{H}_{q}, p\right)}{\partial p}\right|_{\text {for any } p}$dpdφ(H^p​,p)​∣∣∣∣∣∣​p=q​=∂p∂φ(H^q​,p)​∣∣∣∣∣∣​p=q​=∂p∂φ(H^q​,p)​∣∣∣∣∣∣​for any p​