# Finding likelihood ratio of two separate distributions

9.34. A random sample $X_1,X_2,\dots , X_n$ arises from a distribution given by $$H_0 : f(x;\theta) = \frac{1}{\theta} , \qquad 0 < x < \theta , \quad\text{ zero elsewhere},$$ or $$H_1 : f(x;\theta) = \frac{1}{\theta}e^{-x/\theta} , \qquad 0 < x < \infty , \quad\text{ zero elsewhere}.$$ Determine the likelihood ratio ($\lambda$) test associated with the test of $H_0$ against $H_1.$

A random sample arises from a distribution either uniform or exponential. The uniform distribution is stated under the null hypothesis whilst the alternative is stated as an exponential distribution. How can we determine likelihood ratio test associated with these two distributions?

• Your responses to my question will be much appreciated if you could view the image file attached to this problem. I mean prior to your suggested solution(s). – Abubakr Taylor May 20 '18 at 9:39
• It would still help to have your question in the question, rather than just in a picture. – Lord Shark the Unknown May 20 '18 at 10:09
• @LordSharktheUnknown I hope u saw my edit. In light of it, do u think your suggested solution needs modification? – Abubakr Taylor May 20 '18 at 14:07

The likelihood of a sample $(x_1,\dots, x_n)$ for the null hypothesis $H_0$ is $$L_0 = \frac{1}{\theta^n} \mathbf{1}_{0< x_1,\dots, x_n < \theta} \, .$$ Similarly, the likelihood for $H_1$ is $$L_1 = \frac {1}{\theta^n} e^{-(x_1+\dots + x_n)/\theta}\mathbf{1}_{0< x_1,\dots, x_n} \, .$$ The test statistic is the ratio of both likelihoods $R = L_1/L_0$, i.e. $$R= e^{-(x_1+\dots + x_n)/\theta} \frac{\mathbf{1}_{0< x_1,\dots, x_n}}{\mathbf{1}_{0< x_1,\dots, x_n <\theta}} \, .$$ The rejection region of the null hypothesis writes $R\geq k$. Thus, if all $0< x_1,\dots, x_n < \theta$, the rejection region has the form $$\frac{x_1+\dots + x_n}{n} \leq k' \, ,$$ for some constant threshold $k'>0$. The comparison of the value of the sample mean with the threshold gives the decision. Otherwise, if one $x_i$ is larger than $\theta$, then $R\to\infty$, and we reject $H_0$.