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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

$$\mathbb{P}(M_{2}<b\mid M_{1}=a)=\mathbb{P}(M_{2}<b\mid M_{1}=a,M_{1}=X)\mathbb{P}(M_{1}=X)+\mathbb{P}(M_{2}<b\mid M_{1}=a,M_{1}=Y)\mathbb{P}(M_{1}=Y)$$

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

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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 $\frac{\lambda_1\lambda_2}{\lambda_1+\lambda_2}\left(e^{-\lambda_1 x}+e^{-\lambda_2 x }\right)$

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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.

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