We let $X$ and $Y$ be independent and exponentially distributed random variable with $E(X)=\lambda_1$ and $E(Y)=\lambda_2$ where $\lambda=(\lambda_1,\lambda_2) \in {R}_+^2$ and we let $(X_1,Y_1),...,(X_n,Y_n)$ be a sample from this distribution.
Now I have to determinate the log ratio statistic for the composite hypotesis $H_0: \lambda_1\lambda_2=1$
I know that the likelihood ratio test value is given by $2(L_U-L_R)$ where $L_U$ is the log likelihood value and $L_R$ is the log likelihood value where we have used the MLE.
But I'm a bit confused here when we have two parameters for $\lambda=(\lambda_1,\lambda_2)$. How will write the loglikelihood function and MLE?
The alternate likelihood is: $${\cal L}_1(x_i, y_j ; \lambda_1, \lambda_2) = \prod_i P(x_i; \lambda_1) \prod_j P(y_j; \lambda_2)$$
and the null likelihood is: $${\cal L}_0(x_i, y_j ; \lambda) = \prod_i P(x_i; \lambda) \prod_j P(y_j; 1/\lambda)$$