# Finding MLE of this confusing setup of distributions

I have been given two random variables $$X$$ and $$Y$$ following exponential with means $$\lambda_1$$ and $$\lambda_2$$. Let $$Z_1 = min(X,Y)$$ and $$Z_2$$ will be 0 if $$Z_1 = X_1$$ and it will be 1 if $$Z_1 = X_2$$.

It is required to Find mle of $$\lambda_1$$ and $$\lambda_2$$ based on sample of size n on $$Z_1$$ and $$Z_2$$.

I can derive the distribution of $$Z_1$$ as:

$$F(z_1) = 1 - e^{-\frac{\lambda_1 + \lambda_2}{\lambda_1 \lambda_2}z_1}$$.

Also, for deriving the distribution of $$Z_2$$, I can see that:

$$P(Z_2 = 0) = P(X < Y) = \frac{1}{2} = P(Z_2 = 1)$$.

Hence, $$Z_2$$ can be thought of as Binomial distribution with $$p = \frac{1}{2}$$.

Now, I am not sure how can I make the Likelihood equation in this case? Specially, How can I make the likelihood equation in two distinct distributions. I could just multiply they if they were independent but in this case we can see that $$Z_1$$ appears in the distribution of $$Z_2$$ which means we cannot just multiply the likelihoods of $$Z_1$$ and $$Z_2$$ as they are non-independent. Any hints would be appreciated. Thanks in advance.

• The cdf for $Z_1$ is $1-e^{-(\lambda_1+\lambda_2)z_1}$ and the distribution of $Z_2$ will only be a binomial distribution with $p=1/2$ when $\lambda_1=\lambda_2$. – JimB Mar 17 at 5:00
• – StubbornAtom Mar 20 at 6:25