统计推断(一) Hypothesis Test

个人博客地址 Glooow，欢迎光临~~~

文章目录

1. Binary Bayesian hypothesis testing
- 1.0 Problem Setting
- 1.1 Binary Bayesian hypothesis testing
- - Special cases
- 1.2 Likelyhood Ratio Test
- 1.3 ROC
2. Non-Bayesian hypo test
- Neyman-Pearson criterion
3. Randomized test
- 3.1 Decision rule
- 3.2 Proposition
- 3.3 Efficient frontier
4. Minmax hypo testing
- 4.1 Decision rule

1. Binary Bayesian hypothesis testing

1.0 Problem Setting

Hypothesis
- Hypothesis space H={H0,H1}\mathcal{H}=\{H_0, H_1\}H={H0,H1}
- Bayesian approach: Model the valid hypothesis as an RV H
- Prior P0=pH(H0),P1=pH(H1)=1−P0P_0 = p_\mathsf{H}(H_0), P_1=p_\mathsf{H}(H_1)=1-P_0P0=pH(H0),P1=pH(H1)=1−P0
Observation
- Observation space Y\mathcal{Y}Y
- Observation Model py∣H(⋅∣H0),py∣H(⋅∣H1)p_\mathsf{y|H}(\cdot|H_0), p_\mathsf{y|H}(\cdot|H_1)py∣H(⋅∣H0),py∣H(⋅∣H1)
Decision rule f:Y→Hf:\mathcal{Y\to H}f:Y→H
Cost function C:H×H→RC: \mathcal{H\times H} \to \mathbb{R}C:H×H→R
- Let Cij=C(Hj,Hi),correcthypoisHjC_{ij}=C(H_j,H_i), correct hypo is H_jCij=C(Hj,Hi),correcthypoisHj
- CCC is valid if Cjj<CijC_{jj}<C_{ij}Cjj<Cij
Optimum decision rule H^(⋅)=arg⁡min⁡f(⋅)E[C(H,f(y))]\hat{H}(\cdot) = \arg\min\limits_{f(\cdot)}\mathbb{E}[C(\mathsf{H},f(\mathsf{y}))]H^(⋅)=argf(⋅)minE[C(H,f(y))]

1.1 Binary Bayesian hypothesis testing

Theorem: The optimal Bayes’ decision takes the form
L(y)≜py∣H(⋅∣H1)py∣H(⋅∣H0)⋛H1P0P1C10−C00C01−C11≜ηL(\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)⋛H1P1P0C01−C11C10−C00≜η
Proof:
KaTeX parse error: No such environment: align at position 8: \begin{̲a̲l̲i̲g̲n̲}̲ \varphi(f) &=…
Given y∗y^*y∗

if f(y∗)=H0f(y^*)=H_0f(y∗)=H0, E=C00pH∣y(H0∣y∗)+C01pH∣y(H1∣y∗)\mathbb{E}=C_{00}p_{\mathsf{H|y}}(H_0|y^*)+C_{01}p_{\mathsf{H|y}}(H_1|y^*)E=C00pH∣y(H0∣y∗)+C01pH∣y(H1∣y∗)

if f(y∗)=H1f(y^*)=H_1f(y∗)=H1, E=C10pH∣y(H0∣y∗)+C11pH∣y(H1∣y∗)\mathbb{E}=C_{10}p_{\mathsf{H|y}}(H_0|y^*)+C_{11}p_{\mathsf{H|y}}(H_1|y^*)E=C10pH∣y(H0∣y∗)+C11pH∣y(H1∣y∗)

So
pH∣y(H1∣y∗)pH∣y(H0∣y∗)⋛H1C10−C00C01−C11\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∗)⋛H1C01−C11C10−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)=H1f(y)=H_1f(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)f(y) 看作是确定的值来分类讨论

Special cases

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

1.2 Likelyhood Ratio Test

Generally, LRT
L(y)≜py∣H(⋅∣H1)py∣H(⋅∣H0)⋛H1ηL(\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)⋛H1η

Bayesian formulation gives a method of calculating η\etaη
L(y)L(y)L(y) is a sufficient statistic for the decision problem
L(y)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 PD=P(H^=H1∣H=H1)P_D = P(\hat{H}=H_1 | \mathsf{H}=H_1)PD=P(H^=H1∣H=H1)
False-alarm probability PF=P(H^=H1∣H=H0)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⁡H^(⋅)PDs.t.PF≤α\max_{\hat{H}(\cdot)}P_D \ \ \ s.t. P_F\le \alpha H^(⋅)maxPD s.t.PF≤α

Theorem(Neyman-Pearson Lemma)：NP 准则的最优解由 LRT 得到，其中 η\etaη 由以下公式得到
PF=P(L(y)≥η∣H=H0)=αP_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 PFP_FPF 时 LRT 的 PDP_DPD 最大。物理直观来看，LRT 中判决为 H1 的区域中 p(y∣H1)p(y∣H0)\frac{p(y|H_1)}{p(y|H_0)}p(y∣H0)p(y∣H1) 都尽可能大，因此 PFP_FPF 相同时 PDP_DPD 可最大化

备注：NP 准则最优解为 LRT，原因是

同一个 PFP_FPF 时， LRT 的 PDP_DPD 最大

LRT 取不同的 η\etaη 时，PFP_FPF 越大，则 PDP_DPD 也越大，即 ROC 曲线单调不减

3. Randomized test

3.1 Decision rule

Two deterministic decision rules H′^(⋅),H′′^(⋅)\hat{H'}(\cdot),\hat{H''}(\cdot)H′^(⋅),H′′^(⋅)
Randomized decision rule H^(⋅)\hat{H}(\cdot)H^(⋅) by time-sharing
H^(⋅)={H^′(⋅),with probability pH^′′(⋅),with probability 1−p\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 PD=pPD′+(1−p)PD′′P_D=pP_D'+(1-p)P_D''PD=pPD′+(1−p)PD′′
- False-alarm prob PF=pPF′+(1−P)PF′′P_F=pP_F'+(1-P)P_F''PF=pPF′+(1−P)PF′′
A randomized decision rule is fully described by pH^∣y(Hm∣y)p_{\mathsf{\hat{H}|y}}(H_m|y)pH^∣y(Hm∣y) for m=0,1

3.2 Proposition

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

Proof: Risk for each y is linear in pH∣y(H0∣y)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…
Neyman-Pearson case:
1. continuous-valued: For a given PFP_FPF constraint, randomized test cannot achieve a larger PDP_DPD than optimum LRT
2. discrete-valued: For a given PFP_FPF constraint, randomized test can achieve a larger PDP_DPD 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 (PD,PF)(P_D,P_F)(PD,PF) operation points

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

Facts

PD≥PFP_D \ge P_FPD≥PF
efficient frontier is concave function
dPDdPF=η\frac{dP_D}{dP_F}=\etadPFdPD=η

4. Minmax hypo testing

prior: unknown, cost fun: known

4.1 Decision rule

minmax approach
H^(⋅)=arg⁡min⁡f(⋅)max⁡p∈[0,1]φ(f,p)\hat H(\cdot)=\arg\min_{f(\cdot)}\max_{p\in[0,1]} \varphi(f,p) H^(⋅)=argf(⋅)minp∈[0,1]maxφ(f,p)
optimal decision rule
H^(⋅)=H^p∗(⋅)p∗=arg⁡max⁡p∈[0,1]φ(H^p,p)\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]maxφ(H^p,p)
要想证明上面的最优决策，首先引入 mismatch Bayes decision
H^q(y)={H1,L(y)≥1−qqC10−C00C01−C11H0,otherwise\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−qC01−C11C10−C00otherwise
代价函数如下，可得到 φ(H^q,p)\varphi(\hat H_q,p)φ(H^q,p) 与概率 ppp 成线性关系
φ(H^q,p)=(1−p)[C00(1−PF(q))+C10PF(q)]+p[C01(1−PD(q))+C11PD(q)]\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))+C10PF(q)]+p[C01(1−PD(q))+C11PD(q)]
Lemma: Max-min inequality
max⁡xmin⁡yg(x,y)≤min⁡ymax⁡xg(x,y)\max_x\min_y g(x,y) \le \min_y\max_x g(x,y) xmaxyming(x,y)≤yminxmaxg(x,y)
Theorem:
min⁡f(⋅)max⁡p∈[0,1]φ(f,p)=max⁡p∈[0,1]min⁡f(⋅)φ(f,p)\min_{f(\cdot)}\max_{p\in[0,1]}\varphi(f,p)=\max_{p\in[0,1]}\min_{f(\cdot)}\varphi(f,p) f(⋅)minp∈[0,1]maxφ(f,p)=p∈[0,1]maxf(⋅)minφ(f,p)
Proof of Lemma: Let h(x)=min⁡yg(x,y)h(x)=\min_y g(x,y)h(x)=minyg(x,y)
g(x)≤f(x,y),∀x∀y⟹max⁡xg(x)≤max⁡xf(x,y),∀y⟹max⁡xg(x)≤min⁡ymax⁡xf(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)⟹xmaxg(x)⟹xmaxg(x)≤f(x,y),∀x∀y≤xmaxf(x,y),∀y≤yminxmaxf(x,y)
Proof of Thm: 先取 ∀p1,p2∈[0,1]\forall p_1,p_2 \in [0,1]∀p1,p2∈[0,1]，可得到
φ(H^p1,p1)=min⁡fφ(f,p1)≤max⁡pmin⁡fφ(f,p)≤min⁡fmax⁡pφ(f,p)≤max⁡pφ(H^p2,p)\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)=fminφ(f,p1)≤pmaxfminφ(f,p)≤fminpmaxφ(f,p)≤pmaxφ(H^p2,p)
由于 p1,p2p_1,p_2p1,p2 任取时上式都成立，因此可以取 p1=p2=p∗=arg⁡max⁡pφ(H^p,p)p_1=p_2=p_*=\arg\max_p \varphi(\hat H_p, p)p1=p2=p∗=argmaxpφ(H^p,p)

要想证明定理则只需证明 φ(H^p∗,p∗)=max⁡pφ(H^p∗,p)\varphi(\hat H_{p_*},p_*)=\max_p \varphi(\hat H_{p_*}, p)φ(H^p∗,p∗)=maxpφ(H^p∗,p)

由前面可知 φ(H^q,p)\varphi(\hat H_q,p)φ(H^q,p) 与 ppp 成线性关系，因此要证明上式
- 若 p∗∈(0,1)p_* \in (0,1)p∗∈(0,1)，只需 ∂φ(H^q∗,p)∂p∣for any p=0\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∗=1p_* = 1p∗=1，只需 ∂φ(H^q∗,p)∂p∣for any p>0\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=1p=1p=1；q∗=0q_*=0q∗=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⁡pφ(H^p,p)p_*=\arg\max_p \varphi(\hat H_p, p)p∗=argmaxpφ(H^p,p)，其中 p∗p_*p∗ 满足
0=∂φ(H^p∗,p)∂p=(C01−C00)−(C01−C11)PD(p∗)−(C10−C00)PF(p∗)\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:
dφ(H^p,p)dp∣p=q=∂φ(H^q,p)∂p∣p=q=∂φ(H^q,p)∂p∣for any p\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

其他内容请看：
统计推断(一) Hypothesis Test
统计推断(二) Estimation Problem
统计推断(三) Exponential Family
统计推断(四) Information Geometry
统计推断(五) EM algorithm
统计推断(六) Modeling
统计推断(七) Typical Sequence
统计推断(八) Model Selection
统计推断(九) Graphical models
统计推断(十) Elimination algorithm
统计推断(十一) Sum-product algorithm