# Indicator of two exponential random variables

I am working on the following problem from a statistics qualifying exam. I have attempted progress at the 3 parts but do not feel certain, or satisfied, with my work. I am not sure if $$\textit{I}$$ am making errors or if there are typos in the problem to begin with. Here is the set-up:

Suppose $$X$$ and $$Y$$ are independent exponentially distributed random variables with scale parameters $$\lambda$$ and $$\mu$$ respectively, i.e. $$f(x;\lambda)=\frac{1}{\lambda}\mathrm{exp}(-x/\lambda)$$ and similarly for $$Y$$. Let $$Z=\mathrm{min}(X,Y)$$, and $$W=1$$ when $$Z=X$$ and $$W=0$$ otherwise.

I need to do the following: (1) Find the joint distribution of $$Z$$ and $$W$$. (2) Prove that $$Z$$ and $$W$$ are independent. (3) In the case that $$(X,Y)$$ are not observable but instead $$(Z_1,W_1),\ldots,(Z_n,W_n)$$ are observed (i.e. in a random sample of size n$$\geq$$2), find the maximum likelihood estimates of $$(\lambda,\mu)$$ in terms of $$\bar{Z}$$ and $$\bar{W}$$, the sample averages. Then determine if $$\hat{\lambda}_{MLE}^{-1}$$ is unbiased for $$\lambda^{-1}$$.

I have determined the following:

(i) $$Z$$ is exponentially distributed with scale parameter $$\frac{\lambda\mu}{\lambda+\mu}$$.

(ii) $$W=\mathcal{1}_{X\leq Y}$$.

(iii) The joint p.d.f. of $$X$$ and $$Y$$ is $$f(x,y;\lambda,\mu)=\frac{1}{\lambda\mu}\mathrm{exp}(-x/\lambda-y/\mu)$$.

For the joint distribution, I have computed

$$\mathbb{P}(Z\leq t,W=1)=\int_{0}^{t}\int_{0}^{y}f(x,y;\lambda,\mu)\,\mathrm{d}x\,\mathrm{d}y=1-\mathrm{exp}(-t/\mu)+\frac{\lambda}{\lambda+\mu}(\mathrm{exp}(-t(\frac{1}{\lambda}+\frac{1}{\mu}))-1)$$, $$\mathbb{P}(Z\leq t,W=0)=1-\mathrm{exp}(-t/\lambda)+\frac{\mu}{\lambda+\mu}(\mathrm{exp}(-t(\frac{1}{\lambda}+\frac{1}{\mu}))-1)$$.

However, this leads me to the conclusion that $$Z$$ and $$W$$ are not independent since I can compute $$\mathbb{P}(Z\leq t)=1-\mathrm{exp}(-t(\frac{1}{\lambda}+\frac{1}{\mu}))$$ and $$\mathbb{P}(W=1)=\frac{\mu}{\lambda+\mu}$$. (That is, multiplying things out I find $$\mathbb{P}(Z\leq t,W=1)\neq\mathbb{P}(Z\leq t)\mathbb{P}(W=1))$$. I suspect I have not properly computed the joint distribution but am unsure how to check whether it is correct.

Secondarily, when it comes to finding the M.L.E., I know that if I observe $$\bar{W}$$, then $$n\bar{W}$$ times we have $$X\leq Y$$ (so $$Z=X$$) and the other $$n(1-\bar{W})$$ times we have $$Y\leq X$$ (so $$Z=Y$$). Using this, I write the likelihood function in terms of $$\bar{Z}$$ and $$\bar{W}$$ as

$$L(\lambda,\mu;\bar{Z},\bar{W})=(\lambda\mu)^{-n}\mathrm{exp}(-n^2\bar{Z}\bar{W}/\lambda)\mathrm{exp}(-n^2\bar{Z}(1-\bar{W})/\mu)$$.

From this I derived that $$\hat{\lambda}_{MLE}=n\bar{Z}\bar{W}$$ and $$\hat{\mu}_{MLE}=n\bar{Z}(1-\bar{W})$$. But now I am certainly unsure how to compute $$\mathbb{E}\hat{\lambda}_{MLE}^{-1}$$. (If it is a typo, I can compute $$\mathbb{E}\hat{\lambda}_{MLE}$$.)

$$\mathbb P(Z < t, W=1) = \mathbb P(X < t, Y > X) = \int_{x=0}^{t}\int_{y=x}^{\infty}f(x,y;\lambda,\mu)\,\mathrm{d}y\,\mathrm{d}x = \frac{\mu}{\lambda+\mu}\left(1 - e^{-t\left(\frac1\lambda+\frac1\mu\right)}\right)$$
$$\mathbb P(Z < t, W=0) = \mathbb P(Y < t, X > Y) = \int_{y=0}^{t}\int_{x=y}^{\infty}f(x,y;\lambda,\mu)\,\mathrm{d}x\,\mathrm{d}y = \frac{\lambda}{\lambda+\mu}\left(1 - e^{-t\left(\frac1\lambda+\frac1\mu\right)}\right)$$