# Hypothesis test for composite null hypothesis of exponential parameter

I'm having trouble defining the reject region based on the generalized likelihood ratio test. This is from a question of past exam I'm self-studying and still have the doubt, given that I got it wrong.

Let $$X_{1},\ldots,X_{n}$$ a sample from an exponential distribution $$X \sim \text{Exp}(\theta)$$ with $$\theta = \mathbb{E}[X]$$. The parameter space is $$\Omega = \{\theta \mid\theta>0\}.$$

I need to contrast the next hypothesis: $$H_{0}:\theta \geq \theta_{0}$$ vs $$H_{1}: \theta< \theta_{0}$$.

The set of the null hypothesis is: $$\Omega_{0} = \{\theta \mid\theta \geq \theta_{0}\}$$ and the set of the alternative hypothesis is: $$\Omega_{1} = \{\theta \mid\theta < \theta_{0}\}$$. So that $$\Omega = \Omega_{0} \cup\Omega_{1}.$$

The likelihood function for $$\theta$$ is given by: $$L(\theta) = \theta^{-n} \exp\{ -\theta^{-1} \bar{X} \}$$

I know the generalized likelihood ratio is given by $$\Lambda = \frac{\sup_{\theta \in \Omega_{0}} L (\theta)}{\sup_{\theta \in \Omega} L (\theta)}$$

and serves as a test statistic. For the denominator, we have that the MLE of $$\theta$$ is given by $$\hat{\theta} = \bar{X}$$, and with this the denominator is $$L(\hat{\theta}) = \bar{X}^{-n}\exp\{ -\bar{X}^{-1}n\bar{X} \} = \bar{X}^{-n}\exp\{ -n \}$$

My problem is: how do I solve for the numerator $$\sup_{\theta \in \Omega_{0}} L (\theta)$$ ? Because I already know how to solve the test given a simple null hypothesis $$H_{0}: \theta = \theta_{0}$$ and I'm really stucked with this problem.

Thank you for your help.

• See these notes Exmp 4.11 p78, where the same problem is discussed. There $\theta$ is the exponential rate, not mean. – BruceET May 30 at 21:08

Unrestricted MLE of $$\theta$$ is as you say $$\hat\theta=\overline X$$, the sample mean.

Now under the restriction $$\theta\ge\theta_0$$, argue that MLE of $$\theta$$ must be $$\hat{\hat\theta}=\begin{cases}\hat\theta&,\text{ if }\hat\theta\ge\theta_0 \\ \theta_0&,\text{ if }\hat\theta<\theta_0\end{cases}$$

So depending upon whether $$\overline X\ge \theta_0$$ or $$\overline X<\theta_0$$, the likelihood ratio statistic takes the form

\begin{align} \Lambda=\frac{\sup_{\theta\ge\theta_0} L(\theta)}{\sup_{\theta}L(\theta)}&=\frac{L(\hat{\hat\theta})}{L(\hat\theta)} \\&=\begin{cases}1&,\text{ if }\hat\theta\ge\theta_0 \\\\ \frac{L(\theta_0)}{L(\hat\theta)}&,\text{ if }\hat\theta<\theta_0\end{cases} \end{align}

Now it is a matter of studying this ratio as a function of $$\overline X$$ when $$\hat\theta<\theta_0$$. Remember to reject $$H_0$$ for small values of $$\Lambda$$. The case corresponding to $$\hat\theta\ge\theta_0$$ leads to trivial acceptance of $$H_0$$.

• Thank you @StubbornAtom. Your explanation was very clear. – hugdelcur96 May 31 at 0:51

Test. Once you figure out that you want to reject for small $$\bar X,$$ the task remains to find the critical value $$c$$ for a test of $$H_0: \theta \ge \theta_0$$ vs $$H_a: \theta < \theta_0$$ at level $$\alpha = 0.05.$$ That is, $$P(\bar X \ge c\,|\,\theta_0) = 0.05.$$

Null distribution Suppose that $$\bar X$$ is the mean $$\bar X$$ of a random sample $$X_1, X_2, \dots, X_n$$ from $$\mathsf{Exp}(\theta_0),$$ so that $$E(X_i) = \theta.$$ Then you can use moment generating functions to show $$\bar X \sim \mathsf{Gamma}(\text{shape} = n, \text{scale}=\theta/n).$$ Thus $$E(\bar X) = \theta$$ and $$SD(\bar X) = \theta/\sqrt{n}).$$

The following simulation in R of a million samples of size $$n =25$$ from $$\mathsf{Exp}(\text{mean} = \theta = 50)$$ $$\equiv$$ $$\mathsf{Exp}(\text{rate} = \lambda = 0.02)$$ compares a histogram of the simulated distribution of means $$\bar X$$ with the density function of $$\mathsf{Gamma}(n, \text{scale} = \theta/n) \equiv \mathsf{Gamma}(n, \text{rate} = n/\theta).$$ (Perhaps see Wikipedia on gamma distributions; R uses the rate parameter.)

set.seed(2019)
n = 25;  th = 50;  lam = 1/th
a = replicate( 10^6, mean(rexp(n, lam)) )
lbl = "Simulated Sums of Exponential Data with Gamma Density"
hist(a, prob=T, br = 50, col="skyblue2", main = lbl)
curve(dgamma(x, n, n/th), 0, 120, add=T, lwd=2)


Critical value. Then, testing $$H_0: \theta \ge \theta_0 = 50$$ vs. $$H_a: \theta < 50,$$ based on $$n = 25$$ exponential observations, we can use R to find the critical value as $$c = 34.764.$$ That is, $$P(\bar X \le 34.764\, |\, \theta_0 = 50) = 0.05.$$

th.0 = 50;  n = 25
c = qgamma(.05, n, n/th.0);  c
[1] 34.76425


Power. If we happen to sample from an exponential distribution with mean $$\theta_a = 40$$ (rate $$\lambda_a = 0.025),$$ what is the probability of rejecting $$H_0?$$ The result from R is only $$P(\bar X \le 34.764\, |\, \theta_a = 40) = 0.268.$$ So using this test, $$n = 25$$ observations is not enough reliably to discover if the true population mean is $$\theta_a = 40$$ instead of $$\theta_0 = 50.$$ However, the power against the alternative value $$\theta_a = 30$$ is almost 80%.

n = 25; th = 40; c = 34.764
pwr = pgamma(c, n, n/th);  pwr
[1] 0.2682961

n = 25; th = 30; c = 34.764
pwr = pgamma(c, n, n/th);  pwr
[1] 0.7942828

• Thank you @BruceET. Your answer was tremendously useful and super clear. – hugdelcur96 May 31 at 0:58
• Glad it was helpful. – BruceET May 31 at 1:00