Why is the test statistic of a likelihood ratio test distributed chi-squared?
$2(\ln \text{ L}_{\rm alt\ model} – \ln \text{ L}_{\rm null\ model} ) \sim \chi^{2}_{df_{\rm alt}-df_{\rm null}}$
Likelihood-Ratio Test – Why It Is Distributed Chi-Squared
chi-squared-testdistributionslikelihood-ratio
Related Solutions
You can do this using simulation.
Write a function that does your test and accepts the lambdas and sample size(s) as arguments (you have a good start above).
Now for a given set of lambdas and sample size(s) run the function a bunch of times (the replicate function in R is great for that). Then the power is just the proportion of times that you reject the null hypothesis, you can use the mean function to compute the proportion and prop.test to give a confidence interval on the power.
Here is some example code:
tmpfunc1 <- function(l1, l2=l1, n1=10, n2=n1) {
x1 <- rpois(n1, l1)
x2 <- rpois(n2, l2)
m1 <- mean(x1)
m2 <- mean(x2)
m <- mean( c(x1,x2) )
ll <- sum( dpois(x1, m1, log=TRUE) ) + sum( dpois(x2, m2, log=TRUE) ) -
sum( dpois(x1, m, log=TRUE) ) - sum( dpois(x2, m, log=TRUE) )
pchisq(2*ll, 1, lower=FALSE)
}
# verify under null n=10
out1 <- replicate(10000, tmpfunc1(3))
mean(out1 <= 0.05)
hist(out1)
prop.test( sum(out1<=0.05), 10000 )$conf.int
# power for l1=3, l2=3.5, n1=n2=10
out2 <- replicate(10000, tmpfunc1(3,3.5))
mean(out2 <= 0.05)
hist(out2)
# power for l1=3, l2=3.5, n1=n2=50
out3 <- replicate(10000, tmpfunc1(3,3.5,n1=50))
mean(out3 <= 0.05)
hist(out3)
My results (your will differ with a different seed, but should be similar) showed a type I error rate (alpha) of 0.0496 (95% CI 0.0455-0.0541) which is close to 0.05, more precision can be obtained by increasing the 10000 in the replicate command. The powers I computed were: 9.86% and 28.6%. The histograms are not strictly necessary, but I like seeing the patterns.
- In general, a higher value of the likelihood (or log-likelihood) is preferred. The relationship to the value of 0 is irrelevant. However, there are a number of caveats: if you add more parameters (predictors) to the model, the likelihood can only go up, so selecting the maximum value would always give the model with the most variables. There are multiple ways around it via various information criteria, that penalize the model by its number of parameters. Look up the Akaike Information Criterion (AIC) or the Bayesian Information Criterion (BIC).
- I don't know exactly what the "likelihood ratio chi-squared" is, but it could be the value of the likelihood-ratio test statistic compared to the null model (not the other competing models). It is unlikely to be the Wald or the score test statistic. However, if the competing models are nested (one is a special case of the other) you can use the log-likelihood values to compute the likelihood-ratio test statistics of one model versus the other. $$ \mathrm{LRT} = -2 *( \log L(M_1) - \log L(M_2)) \sim \chi^2_{p_2-p_1}$$ under $H_0$, where $p_1$ and $p_2$ are the number of parameters
- There are no "typical" values for the value of the log-likelihood, just as there are no typical values for the sum of all the
yvalues in your data. They are only meaningful when comparing different fits to the same set of values for the outcome.
Best Answer
As mentioned by @Nick this is a consequence of Wilks' theorem. But note that the test statistic is asymptotically $\chi^2$-distributed, not $\chi^2$-distributed.
I am very impressed by this theorem because it holds in a very wide context. Consider a statistical model with likelihood $l(\theta \mid y)$ where $y$ is the vector observations of $n$ independent replicated observations from a distribution with parameter $\theta$ belonging to a submanifold $B_1$ of $\mathbb{R}^d$ with dimension $\dim(B_1)=s$. Let $B_0 \subset B_1$ be a submanifold with dimension $\dim(B_0)=m$. Imagine you are interested in testing $H_0\colon\{\theta \in B_0\}$.
The likelihood ratio is $$lr(y) = \frac{\sup_{\theta \in B_1}l(\theta \mid y)}{\sup_{\theta \in B_0}l(\theta \mid y)}. $$ Define the deviance $d(y)=2 \log \big(lr(y)\big)$. Then Wilks' theorem says that, under usual regularity assumptions, $d(y)$ is asymptotically $\chi^2$-distributed with $s-m$ degrees of freedom when $H_0$ holds true.
It is proven in Wilk's original paper mentioned by @Nick. I think this paper is not easy to read. Wilks published a book later, perhaps with an easiest presentation of his theorem. A short heuristic proof is given in Williams' excellent book.