Question 11.15: Let X1, . . . , Xn be a random sample of size n from the Neg......

Here

L(\theta\mid \pmb x)=\theta^{n}e^{-\theta t},\quad{\mathrm{where}}\quad \pmb x=(x_{1},\dots,x_{n})\mathrm{~and~}t=\sum\limits_{i=1}^{\displaystyle n}x_{i}.

We also know that the MLE of θ is \hat{\theta}=1/\bar{x}=n/t. Therefore the LR λ is given by:

\lambda=\theta_{0}^{n}e^{-\theta_{0}t}\bigg/\!\left(\frac{n}{t}\right)^{n}e^{-n}=\left(\frac{e\theta_{0}}{n}\right)^{n}t^{n}e^{-\theta_{0}t},

and hence \lambda\lt \lambda_{0}, if and only if t e^{-{\frac{\theta_{0}}{n}}t}\lt C_{0}(={\frac{n\lambda_{0}^{1/n}}{e \theta_{0}}}). We wish to determine the cutoff point C_{0}. To this end, set g(t)=t e^{-d t}\left(d=\theta_{0}/n\right) and observe that g(t) is increasing for 0\lt t\lt \frac{1}{d}=\frac{n}{\theta_{0}}, decreasing for t\gt {\frac{n}{\theta_{0}}}, and attains its maximum at t=n/\theta_{0} (see Figure 11.7). It follows that t e^{-d t}{\lt }C_{0}, if and only if t\lt C_{1} or t\gt C_{2}. Therefore, by setting T=\textstyle\sum_{i=1}^{n}X_{i}, we have: P_{\theta_{0}}(T e^{-{\frac{\theta_0}{n}}T}\lt C_{0})=\alpha, if and only if P_{\theta_{0}}(T\lt C_{1}\ \mathrm{or}\ T\gt C_{2})={{{\alpha}}}. For simplicity, let us take the two-tail probabilities equal. Thus,

P_{\theta_{0}}(T\lt C_{1})=P(T\gt C_{2})={\frac{\alpha}{2}}.

By the fact that the independent X_{i}’s have the f(x;\,\theta_{0})\,=\,\theta_{0}e^{-\theta_{0}x} p.d.f. (under \textstyle H_{0}.), it follows that T is distributed as Gamma with \alpha=n\,{\mathrm{and}}\;\beta=\frac{1}{\theta_{0}}. Therefore its p.d.f. is given by:

f_{T}(t)={\frac{\theta_{0}^{n}}{\Gamma(n)}}t^{n-1}e^{-\theta_{0}t},\quad t\gt 0.

Then C_{1}\operatorname{and}C_{2} are determined by:

\int_{0}^{c_{1}}f_{T}(t)d t=\int_{C_{2}}^{\infty}f_{T}(t)d t={\frac{\alpha}{2}}.

In order to be able to proceed further, take, e.g., n = 2. Then

f_{T}(t)=\theta_{0}^{2}t e^{-\theta_{0}t},\quad t\gt 0,

and

\int_{0}^{C_{1}}\theta_{0}^{2}t e^{-\theta_{0}t}\,d t=1-e^{-\theta_{0}C_{1}}-\theta_{0}C_{1}e^{-\theta_{0}C_{1}},

\int_{C_{2}}^{\infty}\theta_{0}^{2}t e^{-\theta_{0}t}\,d t=e^{-\theta_{0}C_{2}}+\theta_{0}C_{2}e^{-\theta_{0}C_{2}}.

Thus, the relations P_{\theta_{0}}(T\lt C_{1})=P_{\theta_{0}}(T\gt C_{2})={\frac{\alpha}{2}} become, equivalently, for α = 0.05:

0.975e^{p}-p=1,\quad p=\theta_{0}C_{1};\qquad0.025e^{q}-q=1,\quad q=\theta_{0}C_{2}.

By trial and error, we find: p = 0.242 and q = 5.568, so that C_{1}=0.242/\theta_{0} and C_{2}=5.568/\theta_{0}. Thus, for n = 2 and by splitting the error α = 0.05 equally between the two tails, the LR test rejects \textstyle H_{0} when t(=x_{1}+x_{2})\lt 0.242/\theta_{0} or t\gt 5.568/\theta_{0}. For example, for \theta_{0}=1, the test rejects \textstyle H_{0} when t\ \lt 0.242 or t\gt 5.568.