LRT(Likelihood Ratio Test)

deejayosamu·2025년 8월 11일

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LRT of size α\alpha reject H0H_0 if
L(θ^Ω0;x)L(θ^MLE;x)k\frac{L(\hat{\theta}^{\Omega_0};\underline{x})}{L(\hat{\theta}^{MLE};\underline{x})} \leq k
with k satisfying Pθ0(L(θ^Ω0;x)L(θ^MLE;x)k)=αk \text{ satisfying } P_{\theta_0}(\frac{L(\hat{\theta}^{\Omega_0};\underline{x})}{L(\hat{\theta}^{MLE};\underline{x})} \leq k) = \alpha

  • log form
    LRT of size α\alpha reject H0H_0 if
    2[l(θ^Ω0;x)l(θ^Ω;x)]c2[l(\hat{\theta}^{\Omega_0};\underline{x})-l(\hat{\theta}^{\Omega};\underline{x})] \leq c
    with maxθΩ0P(l(θ^Ω0;x)l(θ^Ω;x))=α\underset{\theta \in \Omega_0}{max} P(l(\hat{\theta}^{\Omega_0};\underline{x})-l(\hat{\theta}^{\Omega};\underline{x}))=\alpha

ex1) X1,...,XniidN(μ,σ2),X_1,...,X_n \overset{iid}{\sim} N(\mu,\sigma^2), σ2\sigma^2 is known & μΩ=(,)\mu \in \Omega=(-\infty,\infty)
H0:μ=μ0H_0: \mu=\mu_0 vs H1:μμ0H_1:\mu \neq \mu_0
lrt-ex1

ex2) T-test
X1,...,XniidN(θ1,θ2),X_1,...,X_n \overset{iid}{\sim} N(\theta_1,\theta_2), θ2\theta_2 is unknown, <θ1<,-\infty<\theta_1<\infty, θ2>0\theta_2>0
H0:θ1=θ0H_0: \theta_1=\theta_0 vs H1:θ1θ0H_1:\theta_1 \neq \theta_0
lrt-ex2

ex3) Chi-square test
X1,...,XniidN(μ,θ),X_1,...,X_n \overset{iid}{\sim} N(\mu,\theta), <μ<,-\infty<\mu<\infty, θ>0\theta>0
H0:θ=θ0H_0: \theta=\theta_0 vs H1:θθ0H_1:\theta \neq \theta_0
lrt-ex3

ex4) F-test
X1,...,XniidN(μ1,θ1), Y1,...,YmiidN(μ2,θ2)X_1,...,X_n \overset{iid}{\sim} N(\mu_1,\theta_1), \space Y_1,...,Y_m \overset{iid}{\sim} N(\mu_2,\theta_2)
H0:θ1=θ2H_0: \theta_1=\theta_2 vs H1:θ1θ2H_1: \theta_1 \neq \theta_2
lrt-ex4

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