Suppose that $X_1,X_2,\ldots,X_n$ is a random sample from a distribution with pdf $f(x;\theta)$.

Let $\Theta$ be the parameter space. Assume that the parameter is one-dimensional.

Consider testing

\[H_0:\theta=\theta_0\quad\text{vs} H_1:\theta\ne\theta_0\]

Let $\lambda(\stackrel{\rightharpoonup}{X})=L(\theta_0)/L(\widehat{\theta})$ be the GLR for this test.

In the parameter does not define the support for the pdf, for example as in the $\text{unif}(0,\theta)$, we have:

Wilk's Theorem

Under the assumption that $H_0$ is true

\[\color{red}{ -2\ln\lambda(\stackrel{\rightharpoonup}{X})\stackrel{\text{d}}{\rightarrow}\chi^2(1) }\]

Note that

\[\begin{align} \alpha&=P(\lambda(\stackrel{\rightharpoonup}{X})\le c;\theta_0)\\ &=P(\ln\lambda(\stackrel{\rightharpoonup}{X})\le c_1;\theta_0)\\ &=P(\underbrace{-2\ln\lambda(\stackrel{\rightharpoonup}{X})}_{\color{red}{\text{approximately $\chi^2(1)$}\\\text{for large sample size $n$}}}\le c_2;\theta_0)\\ &\approx P(W\ge c_2;\theta_0)\\ &\text{where }W\sim\chi^2(1)\\ \end{align}\]

And we want to solve

\[\alpha=P(W\gt c_2;\theta_0)\\ \huge{\Downarrow}\\ c_2=\chi^2_{\alpha,1}\]

Suppose that $X_1,X_2,\ldots,X_n$ is a random sample from a distribution with pdf $f(x;\theta)$.

Suppose that $\theta$ is not involved in the support of $f$.

Suppose that $n$ is “large”.

An approximate GLRT of size $\alpha$ for testing

\[H_0: \theta=\theta_0\quad\text{vs}\quad H_1:\theta\ne\theta_0\]

is to reject $H_0$ if $\color{red}{-2\ln\lambda(\stackrel{\rightharpoonup}{X})\gt\chi^2_{\alpha,1}}$.

Example

Suppose that $X_1,X_2,\ldots,X_n$ is a random sample from the continous Pareto distribution with pdf

Fin an approximate large sample GLRT of size $\alpha=0.05$ for

\[H_0:\gamma=1.8\quad\text{vs}\quad H_1:\gamma\ne1.8\]

A likelihood is

\[L(\gamma)=\frac{\gamma^n}{\left[\prod_{i=1}^{n}(1+x)\right]^{\gamma+1}}\]

  • The MLE for $\gamma$ is

    \[\widehat{\gamma}=\frac{n}{\sum_{i=1}^{n}\ln(1+X_i)}\]

  • The restricted MLE for $\gamma$ is

    \[\widehat{\gamma}_0=1.8\]

Compute $\lambda(\stackrel{\rightharpoonup}{X})=L(\gamma_0)/L(\stackrel{\rightharpoonup}{X})$.

Compute $-2\ln\lambda(\stackrel{\rightharpoonup}{X})$.

Reject $H_0:\gamma=1.8$ if

\[-2\ln\lambda(\stackrel{\rightharpoonup}{X})\gt\chi^2_{\alpha,1} = 3.841459\]

In R:

qchisq(1-0.05,1)