Let $X$ and $Y$ be two independent random variables with respective pdfs:

$$f \left(x;\theta_i \right) = \begin{cases} \frac{1}{\theta_i} e^{-x/ {\theta_i}} \quad 0<x<\infty, 0<\theta_i< \infty \\ 0 \quad \text{elsewhere} \end{cases} $$

for $i=1,2$. Two indepedent samples are drawn in order to test $H_0: \frac{\theta_1}{\theta_2} = k$ against $H_1 : \frac{\theta_1}{\theta_2} \neq k $ of sizes $n$ and constant $k$.

$(1)$ Find the likelihood ratio test statistic $\lambda = \lambda(X_1, \dots, X_n, Y_1, \dots, Y_n)$

(2) Use large sample approximation for the null distribution of $2 \log \Lambda$ , and duality of testing and interval estimation for LRT with $\alpha = 0.95$ and $n=100, \overline{X} =2, \overline{Y} =1$ to describe a confidence set for $H_0$.

Attempted answer $(1)$: With the help of Likelihood Ratio for two-sample Exponential distribution and $\theta_1=k\theta_2$,

$$L_{H_0} = k^{n}\theta^{-2n}\cdot \exp\left\{-\theta^{-1}\left(\sum x_i+k\sum y_i\right)\right\},$$

and the MLE is

$$\hat \theta_0 = \frac {\sum x_i+k\sum y_i}{2n} = \frac {1}{2}({\overline{x} + k\overline{y}}) $$

So$$ L_{H_0}(\hat \theta_0) = (\hat \theta_0)^{-2n}\cdot e^{-2n}$$

Under the alternative, the likelihood is

$$L_{H_1} = \theta_1^{-n}\cdot \exp\left\{-\theta_1^{-1}\left(\sum x_i\right)\right\}\cdot \theta_2^{-n}\cdot \exp\left\{-\theta_2^{-1}\left(k\sum y_i\right)\right\}$$

and the MLE's are

$$\hat \theta_1 = \frac {\sum x_i}{n_1} = \bar x, \qquad \hat \theta_2 = \frac {\sum y_i}{n_2} = \bar y$$

So $$L_{H_1}(\hat \theta_1,\,\hat \theta_2) = (\hat \theta_1)^{-n}(\hat \theta_2)^{-n}\cdot e^{-2n}$$

To get the LRT:

$$\frac {L_{H_0}(\hat \theta_0)}{L_{H_1}(\hat \theta_1,\,\hat \theta_2)} =\frac{k^{n}\cdot(\hat \theta_0)^{-2n}}{(\hat \theta_1)^{-n}(\hat \theta_2)^{-n}}=\frac{k^{n} \cdot (\hat \theta_1)^{n}(\hat \theta_2)^{n}}{(\hat \theta_0)^{2n}}=\frac{k^{n} \cdot (\overline x)^{n}(\overline y)^{n}}{(\frac{1}{2}(\overline x + k\overline y))^{2n}}=\frac{k^{n} \cdot 2^{2n} \cdot (\overline x)^{n}(\overline y)^{n}}{(\overline x + k\overline y)^{2n}}=\Lambda$$ Is this correct?

attempted answer $(2)$: with the help of (1): large sample approximation says $-2\ln(\Lambda)$ converges in distribution to $\chi^{2}_{k}$ where $k= \lvert 2-1 \rvert = 1$. Is the test one-sided? Then, for $\alpha=0.95 \rightarrow Z=3.841$. We have $-2 \ln \Lambda \le 3.841 \iff -2 \ln\left(\dfrac{k^{100}\cdot \:2^{200}\cdot \:2^{100}}{\left(2+k\right)^{200}}\right)\le 3.841$ Which is not easy to solve and since its a text-book problem - I think I have a mistake somewhere.

  • $\begingroup$ I believe you have the inequality the wrong way round, you should reject the null hypothesis is $-2 \log \Lambda \ge \chi^2_{r-s}$ $\endgroup$ Mar 14, 2022 at 16:31


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