# Establishing independence between two random variables

I am currently working through some basic exercises in probability and have run into a snag. I am given two independent random variables $$X$$ and $$Y$$ that are both exponentially distributed with respective parameters $$\lambda_{1}$$ and $$\lambda_{2}$$. The exercise is to establish the independence of $$\min(X,Y)$$ and $$\min(X,Y)-\max(X,Y)$$. Denoting $$\min(X,Y)$$ as $$M_{1}$$, $$\max(X,Y)$$ as $$M_{2}$$, and $$M_{1}-M_{2}$$ as $$Z$$, my thought was to show that for $$x_{1},x_{2}\in\mathbb{R}$$ that

$$\mathbb{P}(M_{1}\leq x_{1},Z\leq x_{2})=\mathbb{P}(M_{1}\leq x_{1})\mathbb{P}(Z\leq x_{2})$$

I've computed cdf's for $$M_{1}$$ (which is exponential) and $$Z$$ (which has a cdf I don't recognize), but moving from there is where I'm stuck. I am aware that we can restrict to the case where $$x_{1}\geq 0$$ and $$x_{2}\leq 0$$. Any hints would be appreciated.

Edit: I've looked at Leonbloy's answer to

Independence between maximum and minimum of exponential

and I'm still somewhat confused. His $$C$$ is my $$-Z$$. He has the line

$$\mathbb{P}(M_{2}

I can't determine how this line is justified. It looks like the law of total probability, but I am not sure.

I think memorylessness of the exponential distribution is key here

• The distribution of $$\max(X,Y)- \min(X,Y)$$, the reverse of what you are talking about, is affected by which of $$X$$ and $$Y$$ happens first, i.e. which is smaller, but not when the first happens, i.e. the value of $$\min(X,Y)$$, since the time between the events only depends on which event has not yet happened and not on how long the earlier event took to happen

• This memorylessness also tells you that which happens first is independent of when it happens, so giving the independence you are asking about

I suspect you can also say from memorylessness:

1. $$\mathbb P(X \le Y) = \frac{\lambda_1}{\lambda_1+\lambda_2}$$

2. $$\min(X,Y)$$ has an exponential distribution with parameter $$\lambda_1+\lambda_2$$ so density $$(\lambda_1+\lambda_2)e^{-(\lambda_1+\lambda_2)x}$$

3. $$\max(X,Y)- \min(X,Y)$$ has a mixture distribution, the same distribution as $$X_1$$ with probability $$\frac{\lambda_2}{\lambda_1+\lambda_2}$$ and the same distribution as $$X_2$$ with probability $$\frac{\lambda_1}{\lambda_1+\lambda_2}$$, so combined density $$\frac{\lambda_1\lambda_2}{\lambda_1+\lambda_2}\left(e^{-\lambda_1 x}+e^{-\lambda_2 x }\right)$$

For any RVs $$(X,Y)$$ with density $$f(x,y)$$ the joint density of their minimum $$S$$ and their maximum $$L$$ is equal to $$f_{S,L}(s,l) = f(s,l) + f(l,s)$$ when $$s\le l$$ and zero otherwise. Therefore the joint density of the minimum $$S$$ and the difference $$D=L-S$$ is equal to $$f_{S,D}(s,d) = f(s,s+d)+f(s+d,s)$$ In our case, the expression on the RHS is equal to $$\lambda_1 \lambda_2 e^{\lambda_1 s}e^{\lambda_2 s} \left (e^{\lambda_2 d} + e^{\lambda_1 d}\right)= (\lambda_1 + \lambda_2) e^{-(\lambda_1 + \lambda_2)s}\times \frac{\lambda_1 \lambda_2}{\lambda_1+\lambda_2} \left (e^{-\lambda_1 d} + e^{- \lambda_2 d}\right),$$ a function of $$s$$ times a function of $$d$$, and therefore the marginals are independent. THe RHS expresses it as a product of densities.