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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?

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  • $\begingroup$ 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. $\endgroup$ May 11, 2021 at 15:59

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

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