# Likelihood ratio test for distribution with two parameters

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?

• Likelihood is the joint density of the $(X_i,Y_i)$'s. Since $X_i$ is independent of $Y_i$, this is just the product of joint densities of $(X_i)$'s and $(Y_i)$'s. MLE is derived as usual. May 11, 2021 at 15:59

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: $${\cal L}_0(x_i, y_j ; \lambda) = \prod_i P(x_i; \lambda) \prod_j P(y_j; 1/\lambda)$$