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I have a large sample (a vector) $\mathbf{x}$ from a random variable $X\sim N(\mu,\sigma^2)$. The variance $\sigma^2$ is known, but the expectation $\mu$ is unknown. I would like to test the null hypothesis $H_0\colon \ \mu\leq 0$ against the alternative $H_1\colon \ \mu>0$ using a likelihood ratio (LR) test. The test statistic is $$ \text{LR}=-2\ln\frac{L(\mathbf{x}\mid\tilde\mu,\sigma^2)}{L(\mathbf{x}\mid\hat\mu,\sigma^2)}. $$ where $\tilde\mu$ is the estimate of $\mu$ under $H_0$ (thus $\tilde\mu\leq 0$) and $\hat\mu$ is the estimate of $\mu$ under $H_0 \cup H_1$ (thus $\hat\mu\in R$).

I expected the asymptotic distribution of $\text{LR}$ under $H_0$ to be $\chi^2(1)$ but I am getting only zeros in a simulation below.

Questions: Why is that? Is my simulation wrong? Or is the test statistic not supposed to have the $\chi^2(1)$ asymptotic distribution under $H_0$, and if so, why?

(Related question: "Failing to obtain $\chi^2(1)$ asymptotic distribution under $H_0$ in a likelihood ratio test: example 1")

n=1e3   # sample size
sigma=1 # standard deviation of X
m=3e3   # number of simulation runs
mu=-1   # particular instance of the null hypothesis used in the simulation

logL0s=logL1s=logLRs=rep(NA,m)
for(i in 1:m){
 set.seed(i); x=rnorm(m,mean=mu,sd=sigma)
 logL0=sum(log( dnorm(x,mean=min(0,mean(x)),sd=sigma) ))
 logL1=sum(log( dnorm(x,mean=max(0,mean(x)),sd=sigma) ))
 logLR =-2*(logL0-max(logL0,logL1)) # the -2*ln(LR) statistic from this simulation run
 logL0s[i]=logL0; logL1s[i]=logL1; logLRs[i]=logLR
}

summary(logLRs)
Richard Hardy
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  • Can you please edit your question to put the links to related questions in the body of the question? If a discussion with various upvoted comments ensues, then your initial comment may be hidden later on. – Stephan Kolassa Jan 27 '21 at 16:33
  • I won't say that I know more than you :), but I'd say that if you as the OP know about *related* questions, I think you should just put them in the body of your question - after all, they are related to it. – Stephan Kolassa Jan 27 '21 at 16:37

1 Answers1

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The LR doesn't have a $\chi^2_1$ distribution for this test.

The basic setting where there's a $\chi^2_1$ null distribution is when the null hypothesis is a subspace of the alternative hypothesis, with one less dimension. A key part of the setup is that every point in the null has points in the alternative at arbitrarily small distances away.

In a case like this one, where some points in the null are a long way from the alternative, the sampling distribution will depend on where in the null hypothesis you are.

Thomas Lumley
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  • Thank you! I was suspecting something like this. Would you have any reference that explains the intuition in a bit more detail? Or if we look at ["What are the regularity conditions for Likelihood Ratio test"](https://stats.stackexchange.com/questions/101520), which condition in particular is being violated? I looked at the list before but could not quite identify the relevant item. – Richard Hardy Jan 27 '21 at 20:55
  • It's usually thought of as part of the question, not as a regularity condition. That is, those lists are regularity conditions for testing the null hypothesis $\theta=\theta_0$ (or perhaps $(\theta,\eta)=(\theta_0,eta)$ when there are other parameters. For example, the second comment on that linked answer specifies it's a test of $\theta=\theta_0$. – Thomas Lumley Jan 28 '21 at 02:23