Hypothesis testing
Definition: A hypothesis is a statement about a parameter
Goal: Decide which of 2 complementatry hypothesis is true based on data.
- $H_O$: null hypothesis
- $H_A$: alternate hypothesis
This setting originates from the experimental design.
Definition: A hypothesis test is a rule that specifies, (often) based on the value of a test statistics $W(x_1,x_n)$.
a. when to accept $H_O$ as true
b. when to reject $H_O$ and accpect $H_A$
Ex. $x_1,dots,x_n overset{text{iid}}sim N(mu,sigma^2)$ where $mu$ is unknown
a. $H_O:mu = 0$ vs $H_A:mu neq 0$ is an example of 2 complementary hypothesis.
Here $bar x$ gives information whihc hypothesis is true, so $bar x$ would be a good test statistics. Our hypothesis test is to reject $H_O$ if $bar x > c$ or $bar x < -c$, for some constant $c >0 $.
b. $H_O: mu geq 0 vs H_A: mu <0$.
If we use $bar x $ as test statistics, we would reject $H_O if bar x<-c for some c>0$.
Definition: The likelihood ratio (LR) test statistics for testing
$H_O: theta in Omega_0$ vs $ H_A: theta in Omega_1$, where $Omega=Omega_0 cup Omega_1$ is the parameter space of $theta$ and $Omega_0 cap Phi_1 = Phi$ $$lambda(x_1,dots,x_n)=frac{max limits_{theta in Omega_0}L(theta|x_1,dots,x_n)}{max limits_{theta in Omega}L(theta|x_1,dots,x_n)}$$
Note: $0leqlambdaleq 1$, $lambda approx 1$ if true value of $theta in Omega_0$(ie: $H_0$ is true); $lambda approx 0$ if $H_0$ is false (since values $theta in Omega_0$ don’t fit the data well). So accept $H_0$ if $lambda$ is close to 1, accept $H_A$ if $lambda$ if $lambda$ close to $0$.
Definition: Let $w$ be a test statitiscs, suppose the hypothesis test is the rule “reject $H_o$ if $win text{R}$”, then $text{R}$ is called the rejection region.
Ex. The rejection region for a LR test is $lbrace lambda leq c rbrace$ from some $0<c<1$.
Ex. $x_1,dots,x_n overset{text{iid}}sim N(mu,sigma^2)$ with $sigma^2$ known, $mu$ is unknown parameter.
$$H_O: mu=mu_0, H_A: muneq mu_0$$
Derive LR test statistics $lambda$.
$$begin{equation}begin{split}
L(mu|x_1,dots,x_n)&=prod limits_{i=1}^n frac{1}{sqrt{2pi}sigma}e^{-frac{(x_i-mu)^2}{2sigma^2}} \
&=frac{1}{(2pi)^{n/2}sigma^{n}}e^{-frac{sum(x_i-mu)^2}{2sigma^2}}
end{split}end{equation}$$
$$begin{equation}begin{split}
lambda(x_1,dots,x_n)&=frac{max limits_{theta in Omega_0}L(theta|x_1,dots,x_n)}{max limits_{theta in Omega}L(theta|x_1,dots,x_n)}\
&=frac{max limits_{mu=mu_0}L(theta|x_1,dots,x_n)}{max limits_{muin text{R}}L(theta|x_1,dots,x_n)} \
& = frac{frac{1}{(2pi)^{n/2}sigma^{n}}e^{-frac{sum(x_i-mu_0)^2}{2sigma^2}}}{frac{1}{(2pi)^{n/2}sigma^{n}}e^{-frac{sum(x_i-bar x)^2}{2sigma^2}}} \
& = exp(-frac{sum(x_i-mu_i)^2-sum(x_i-bar x)^2}{2sigma^2}) \
& = exp(-frac{n}{2sigma^2}(bar x-mu_0)^2)
end{split}end{equation}$$
It should notice that $$begin{equation}begin{split}sum(x_i - mu_0)^2 &= sum(x_i - bar x+bar x-mu_0)^2 \
& = sum(x_i-bar x)^2 + 2sum (x_i-bar x)(bar x - mu_0)+sum(bar x-mu_0)^2 \
&= sum(x_i - bar x)^2 + n(bar x - mu_0)^2
end{split}end{equation}$$
So the LR test rejects $H_O$ if $$begin{equation}begin{split} &e^{-frac{n}{2sigma^2}(bar x-mu_0)^2} leq C \
&Rightarrow -frac{n}{2sigma^2}(bar x-mu_0)^2 leq logC \
&Rightarrow (bar x - mu_0)^2 geq -frac{2sigma^2}{n}logC \
&Rightarrow bar x -mu_0 geq sqrt{-frac{2sigma^2}{n}logC} \
& or bar x -mu_0 leq sqrt{-frac{2sigma^2}{n}logC} \
&Rightarrow bar x geq mu_0+sqrt{-frac{2sigma^2}{n}logC} \
& or bar x leq mu_0-sqrt{-frac{2sigma^2}{n}logC}
end{split}end{equation}$$
Based on LR test, we reject $H_O: mu=mu_0$ if $bar x$ is quite different than $mu_0$ (either larger or smaller)
Definition: $w$ is test statistics, $H_O: theta in Omega_0$. A hypothesis test has level $alpha$ if $max limits_{thetainOmega_0} p(win text{R}|theta)=alpha$, where $text{R} $ the rejection region.
i.e. $alpha=probability$ of rejecting $H_O$ when $H_O$ is true. $alpha$ is also called type-I error rate.
Continue our example above:
Let $mathbf{C}=-sqrt{-frac{2sigma^2}{n}logC}$. What value of $mathbf{C}$ achieves a level $alpha$ test?
Ans: reject $H_O$ if $bar xleq mu_0+mathbf{C}$ or $bar x leq mu_0-mathbf{C}$
If $H_O$ is true, the $mu=mu_0$ and so $bar xsim N(mu, frac{sigma^2}{n})Rightarrow frac{bar x-mu_0}{sigma/sqrt{n}}sim N(0,1)$ under $H_O$
$$begin{equation}begin{split}
alpha &=max limits_{mu = mu_0}p(bar xgeq mu_0+mathbf{C} or bar xleq mu_0-mathbf{C}) \
&=p(frac{bar x-mu_0}{sigma/sqrt{n}} geq frac{mathbf{C}}{sigma/sqrt{n}})+p(frac{bar x-mu_0}{sigma/sqrt{n}} leq frac{-mathbf{C}}{sigma/sqrt{n}})
end{split}end{equation}$$where $bar x sim N(mu,frac{sigma^2}{n})$
$Rightarrow frac{mathbf{C}}{sigma/sqrt{n}}$ is the $1-frac{alpha}{2}$ quantile pf $N(0,1)$.
So the final rule to achieve $alpha$ level test is reject $H_O$ if $frac{bar x-mu_0}{sigma/sqrt{n}}geq Z_{1-frac{alpha}{2}}$ or $frac{bar x-mu_0}{sigma/sqrt{n}}leq -Z_{1-frac{alpha}{2}}$
$Rightarrow |frac{bar x-mu_0}{sigma/sqrt{n}}|geq Z_{1-frac{alpha}{2}}$,this is the one-sample z-test for a mean.
If $x_1,dots,x_n overset{text{iid}}sim N(mu,sigma^2)$ with both $mu, sigma^2$ unknown, the LR test for $H_O:mu=mu_0$ vs $H_A:muneq mu_0$ yields one-sample test. HINT: in numerator of LR, we need $max limits_{mu=mu_0,sigma^2>0}L(mu,sigma^2|x_1,dots,x_n)$
i.e. substitute $mu=mu_0$, maximize w.r.t $sigma^2$ alone.
Often, we would not know the exact distribution of LR test statistics $lambda_0$. The, we can use the large sample approximation.
Theory: $H_O: theta in Omega_0, H_A:theta in Omega_1, Omega = Omega_0 cup Omega_1$. $lambda=$ LR test statistics based on $x_1,dots,x_n$. When n is large: if $H_O$ is true, then $$-2loglambdaoverset{text{approx}}sim chi_p^2$$ for any $theta in Omega_0$.
$p = dim(Omega)-dim(Omega_0)$, i.e. difference in # free parameters.
LR test rejects $H_O$ if $lambda leq C$, then $-2loglambda geq underbrace{-2logC}_{mathbf{C}}$
and $alpha = p(-2loglambda geq mathbf{C})$ gives the approx level $alpha$ test.
So the approx level $alpha$ of LR test is reject $H_O$ if $-2loglambdageq (1-alpha)$ quantile of $chi_p^2$.
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