Calculating the $p$-value with Wilks' Theorem I am stuck at understanding the following: According to Wilks' Theorem, the test-statistic $$q_0 = -2\cdot \ln\frac{\mathcal L\left( S = 0\right)}{\mathcal L\left( \hat{S}\right)}$$ follows a $\chi^{2}$-distribution with $1$ d. o. f. (I am aware that there is also a more general form of this theorem, but let's stick to the simplified version for now).
In a lecture, it was said that we can thus compute the $p$-values from the Gaussian quantiles as follows:
$$p_{0} = 1 - \Phi\left( \sqrt{q_0}\right)$$
Could anybody elaborate on this conclusion, please, particularly why Wilks' Theorem plays a role here?
 A: In general, if we have a test statistic $x$ with a known CDF $C(x_{\rm thresh})=P(x\leq x_{\rm thresh})$ under the null hypothesis, then the (one-sided) $p$-value for measuring $x=x_0$ or larger assuming the null hypothesis is $p=P(x\geq x_0)=1-P(x\leq x_0)=1-C(x_0)$.
We can relate this to the PDF $f(x)$ (assuming the null hypothesis, in your case $S=0$) as follows. The CDF is
\begin{equation}
C(x_0) = \int_{-\infty}^{x_0} f(x) dx
\end{equation}
Assuming $f(x)$ is normalized so $\int_{-\infty}^{\infty} dx f(x) = 1$, then the CDF is related to the $p$-value by
\begin{equation}
p = \int_{x_0}^{\infty} f(x) dx = 1 - \int_{-\infty}^{x_0} f(x) dx = 1-C(x_0)
\end{equation}
as stated above.
Since $q$ is drawn from a $\chi^2$ distribution with 1 dof, then $\sqrt{q}$ is drawn from a Gaussian distribution. Assuming $\Phi(x)$ is the CDF for a zero mean, unit variance Gaussian random variable, then $1-\Phi(\sqrt{q_0})$ would give you the $p$-value for measuring $q$ to be as large or larger than $q_0$.
I think Wilk's theorem is only used to say that the log likelihood ratio approaches the $\chi^2$ distribution asymptotically.
