0
$\begingroup$

I'm probably really overcomplicating things but I want to specify the likelihood ratio test with significance level $\alpha = 0.05$

I have three random samples (sample sizes: $n_1, n_2$ and $n_3$), with sample means $X$ and $Y$ and $Z$, respectively. $X_i$ has an $\exp(\mu_1)$ distribution, $Y_i$ has an $\exp(\mu_2)$ distribution, and $Z_i$ has an $\exp(\mu_3)$ distribution.

I understand how to do it for one random sample but I don't understand how to approach it for three samples.

MLE of $\lambda$ is the reciprocal of the sample mean.
H0 :$\mu_1=\mu_2 =\mu_3$ versus H1 : H0 is not true.

$\endgroup$
  • $\begingroup$ Welcome to Mathematics Stack Exchange! A quick tour will enhance your experience. Here are helpful tips to write a good question and write a good answer. For typesetting, please use MathJax. $\endgroup$ – dantopa May 24 '19 at 5:04
  • $\begingroup$ The method is the same. Given the sample $(x_1,\ldots,x_{n_1},y_1,\ldots,y_{n_2},z_1,\ldots,z_{n_3})$ write down the likelihood function $L(\mu_1,\mu_2,\mu_3)=\prod_{i=1}^{n_1}f_{X_1}(x_i)\prod_{i=1}^{n_2}f_{Y_i}(y_i)\prod_{i=1}^{n_3}f_Z(z_i)$. Form the likelihood ratio statistic $\Lambda=\frac{\sup_{H_0}L(\mu_1,\mu_2,\mu_3)}{\sup_{H_o\cup H_1}L(\mu_1,\mu_2,\mu_3)}$ and remember to reject $H_0$ for small values of $\Lambda$. So start by finding the maximum likelihood estimate of $(\mu_1,\mu_2,\mu_3)$ under $H_0$ and under $H_0\cup H_1$. $\endgroup$ – StubbornAtom May 24 '19 at 19:24
  • $\begingroup$ Welcome to Mathematics Stack Exchange! A quick tour will enhance your experience. Here are helpful tips to write a good question and write a good answer. For equations, use MathJax. $\endgroup$ – dantopa May 25 '19 at 3:49
0
$\begingroup$

If $X$, $Y$, and $Z$ are independent and have the same mean, then the sum $T=X+Y+Z$ is Gamma distributed with $\alpha=3$ and $\beta=\lambda$. So, under the zero hypothesis, the sum is a Gamma random variable; then, you use a test (for instance, a qq plot or a histogram) to see if the data match a Gamma random variable with the required confidence.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.