Finding critical region of statistic when the distribution of the transform of the statistic is known

Let $X_1, ..., X_n$ be a random sample from $\mathrm{Exp}(\lambda_1)$ and $Y_1, ..., Y_m$ be a random sample from $\mathrm{Exp}(\lambda_2)$. I have found the generalized likelihood ratio test (GLRT) statistic $\Lambda$ for testing $H_0:\lambda_1 =\lambda_2$ vs $H_1:\lambda_1 \neq \lambda_2$ is $\Lambda=\frac{n^n m^m}{(m+n)^{m+n}} T^{-n}(1-T)^{-m}$ where $T = \frac{\sum_i X_i}{\sum_i X_i + \sum_j Y_j}$.

Now I need to find the critical region of the test by the asymptotic distribution of the test $\Lambda$. I know the relation $W = 2 \log(\Lambda) \sim \chi_d^2$ where d is the number of parameters to be tested.

How can I use this relation to find the critical region in the above problem?

• I've changed the title, since from the question it is obvious that the distribution of statistic is known. Apr 4, 2012 at 13:54
• Yes the distribution of the statistics $T$ is $beta(n,m)$ which is known but I don't know the distribution of $\Lambda$ so I need to approximate asymtotically. Apr 4, 2012 at 14:02
• But you say that $2\log\Lambda\sim \chi^2_d$. Why don't you use this relationship? Apr 4, 2012 at 14:08
• This is my question, how to use this to find the critical region? Apr 4, 2012 at 14:13
• it sounds like you're just asking how to find the quantiles of the null distribution - I'd say use a computer (qchisq in R). May 16, 2012 at 14:28

if

$W = 2 \log(\Lambda) \sim \chi_d^2$

then isn't

${e}^{\frac{\chi^2_{(d,0.975)}}{2}}$

an upper bound for $\Lambda$? I could be wrong.

in R this is found with:

exp(qchisq(.975,d)/2)


and for the lower bound:

exp(qchisq(.025,d)/2)

• You are correct (+1). May 16, 2012 at 16:15