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

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 $$H_0\colon \ \mu=0$$ against the alternative $$H_1\colon \ \mu>0$$ using a likelihood ratio (LR) test. The test statistic is $$\text{LR}=-2\ln\frac{L(\mathbf{x}\mid 0,\sigma^2)}{L(\mathbf{x}\mid\hat\mu,\sigma^2)}.$$ where $$\hat\mu$$ is the estimate of $$\mu$$ under $$H_0 \cup H_1$$ (thus $$\hat\mu\geq0$$).

I expected the asymptotic distribution of $$\text{LR}$$ under $$H_0$$ to be $$\chi^2(1)$$ but I am getting something else 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?

n=3e3   # sample size
sigma=1 # standard deviation of X
m=3e3   # number of simulation runs

logL0s=logL1s=logLRs=rep(NA,m)
for(i in 1:m){
set.seed(i); x=rnorm(n,mean=0,sd=sigma)
logL0=sum(log( dnorm(x,mean=0,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
}

# Critical values: asymptotic nominal vs. empirical
crit.val=qchisq(p=0.95,df=1)
empirical.crit.val=quantile(x=logLRs,probs=0.95)
print(paste0("Asymptotic critical value = ",round(crit.val,2),", simulated critical value = ",round(empirical.crit.val,2)))

# Test size: asymptotic nominal vs. empirical
empirical.size=length(which( logLRs > crit.val ))/m # proportion of rejections at 0.05 level
print(paste0("Nominal test size = 0.050, simulated test size = ",round(empirical.size,3)))

# Plots illustrating the sampling distribution of -2*log(LR) statistic
par(mfrow=c(2,2),mar=c(2,2,2,1))
plot(logLRs,main="Scatterplot of -2ln(LR) across simulation runs")
abline(h=crit.val,col="red")
abline(h=empirical.crit.val,col="blue")
plot(NA)
plot(density(logLRs),main="Density of -2ln(LR)")
chisq.quantiles=qchisq(p=seq(from=0.001,to=0.999,by=0.001),df=1)
chisq.density=dchisq(x=chisq.quantiles,df=1)
lines(y=chisq.density,x=chisq.quantiles,col="blue",lty="dashed")      # Chi^2(1) overlay
abline(v=crit.val,col="red")                                          # asympt. nominal crit. val. in red
abline(v=empirical.crit.val,col="blue")                               # empirical crit. val. in blue
br=24
hist(logLRs,breaks=br,main="Histogram of -2ln(LR)")
lines(y=chisq.density*m/br,x=chisq.quantiles,col="blue",lty="dashed") # Chi^2(1) overlay
abline(v=crit.val,col="red")                                          # asympt. nominal crit. val. in red
abline(v=empirical.crit.val,col="blue")                               # empirical crit. val. in blue
par(mfrow=c(1,1))


• Jan 27, 2021 at 12:33
• Jan 27, 2021 at 12:33
• You are on the boundary of the parameter space: the result you are relying on doesn't hold in such cases. Typically the distribution of the LR is a mixture of chi-squared distributions.
– whuber
Jan 27, 2021 at 14:31
• @whuber, thank you so much! I was suspecting it, but I got confused about the definition of the parameter space. Is the parameter space defined by $H_0 \cup H_1$, so in my case, $\mu\in [0,\infty)$? Jan 27, 2021 at 14:34
• Yes: this became clear to me when I saw you force $\hat\mu$ to be non-negative in your code at max(0,mean(x)).
– whuber
Jan 27, 2021 at 14:57

Apparently the test statistic not supposed to have the $$\chi^2(1)$$ asymptotic distribution under $$H_0$$. Thanks to @whuber for pointing this out.
The example violates an assumption for the distribution to be $$\chi^2(1)$$. The assumption is that the parameter of interest in not on the boundary of the parameter space. I am not entirely sure how the parameter space is defined, but it seems it is defined by $$H_0\cup H_1$$. In my case this would mean $$\mu\in [0,\infty)$$. Now $$\mu=0$$ due to $$H_0$$ is on the boundary of $$[0,\infty)$$, so that constitutes a violation.