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I have a question about the following from Introduction to Probability by Blitzstein: enter image description here

I was able to show $L \sim$Expo(2) and use $M-L= \vert X-Y \vert$ to perform a double integral to show $M-L \sim$Expo(1), but got stuck on showing $M-L,L$ are independent. I didn't use the memoryless property in the 1st 2 parts so I suspect it will come in for this last part. So in my current approach I was forced to show all 3 statements separately, so I was also wondering if there's a way to do it all at once, of it at least 2 statements can be shown simultaneously. I also want to add I asked some friends who know probability better than I do, and they suggested order statistics. But order statistics have not come up yet in this point of Blitzstein's book.

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Let's think about how $M$ is distributed conditionally on $L=l$. We know that there was another exponential variable $L=l$ that it is greater than, but $X$ and $Y$ are independent, so it will be conditionally distributed like $X$ given $X>l.$ So we can write $$ P(M>m|L=l) = P(X>m|X>l).$$

Next let's look at the distribution of $Z=M-L$ conditional on $L=l.$ We have $$P(Z>z|L=l) = P(M>z+l|L=l) = P(X>z+l|X>l) = P(X>z)$$ where the last equality used the memoryless property. Since $P(Z>z|L=l)$ does not depend on $l$, $Z$ and $L$ are independent. We have $P(X>z) = e^{-z}$ since $X$ is a standard exponential, so $Z$ is a standard exponential.

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    Thanks a lot. I think the step $P(M>m|L=l)=P(X>m|X>l)$ was clever and I probably wouldn't have come up with that on my own. – 2017-02-18
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    Why is that first step true? I'm not following. Referring to $$P(M > m | L = l) = P(X > m | X > l).$$ don't see why that's true because of X and Y being independent... – 2017-05-06
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    @ClarkKent If they weren't independent it wouldn't be true but I wouldn't say it's true because they're independent. The first part of the sentence was the main logical step, not the independence part. Think, 'what do you know about $M$ given that $L=l?'$ The answer is that you know it's exponential and that it's greater than or equal to $l$ and that's it. Independence is important here cause if they weren't independent $L=l$ would give more information than that. – 2017-05-07
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    @ClarkKent If that seems overly hand wavy and you want something more formal, the joint PDF of $M$ and $L$ is $f_{M,L}(m,l) = 2e^{-m}e^{-l}1_{m\ge l}$ (this is from a standard formula of order statistics). From that the conditional PDF can be computed to be $f_{M|L}(m|l) = e^{-(m-l)}1_{m\ge l}$ which can be integrated to give $P(M>m|L=l) = e^{-(m-l)}1_{m\ge l}$. – 2017-05-07
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    Your formal argument is correct but $M = \max(X,Y)$ is not distributed exponentially. $X$ and $Y$ are. – 2017-05-08
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    The only way I've thought of verifying your first step is by noting $P(M > m | L = l) = P(M > m | L = l, X > Y)$ and then you see $P(M > m | L = l) = P(X > m | Y = l, X > l) = P(X > m | X > l)$ by independence of $X$ and $Y$. That's probably what you meant. – 2017-05-08
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    @ClarkKent I never said $M$ was distributed exponentially. I was reasoning conditionally. Your verification is correct but it isn't what I 'meant'. Like you, I used symmetry without saying so explicitly. I meant 'You have an RV in front of you and you are told it's the larger of two iids and the smaller has value $l$. What have you been told about the RV in front of you? Answer: That it is larger than $l$.' That coupled with the info that its prior distribution is the same as $X$'s (or $Y$'s) gives the answer. I'll cop to that not being much more of an answer than 'it's intuitive'. – 2017-05-09
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    @ClarkKent Think of 'expo' as prior, and the fact that 'it's largest of two iid and the smaller one has value $l$' as additional info. I'm just translating the additional info to 'it is bigger than $l$.' I think you're thinking in terms of 'largest of two expos' as the prior and 'the smaller one has value $ l$' as the additional info, which would make sense given the form of the left-hand-side of the equation, but wasn't how I was thinking about it. Not sure if that helps but anyway it seems you found a way to verify it on your own. – 2017-05-09
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    Oh okay that makes more sense. – 2017-05-09