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$X$ and $Y$ are independent random variables following the exponential distribution such that: $$X,Y\sim \mathrm{exp}(\lambda)$$ $$W=min(X,Y)$$ $$Z=max(X,Y)$$ $$O=Z-W$$ $$M=\mathbf{1}_{X\le Y}=\begin{Bmatrix} 1 & if X\le Y \\ 0 & if X>Y \end{Bmatrix}$$

So far I know that $W\sim \mathrm{exp}(2\lambda)$ and M follows a bernoulli distribution with parameter $\frac{1}{2}$

In addition to this $P(W>w,M=1)=P(W>w)P(M=1)$ (i.e. the events are independent)

I now need to:
a. express the event $(W\le w,M=1,O\le t)$ in terms of $X$ and $Y$
b. find $P(W\le w,M=1,O\le t)$ and $P(W\le w,M=0,O\le t)$

For a. I rewrote the event as $$(W\le w,M=1,O\le t)=(min(X,Y)\le w,X\le Y,max(X,Y)-min(x,y)\le t)$$ However, I wasn't sure how to condense this into one inequality (e.g. $P(W>w,M=1)=P(Y\ge X>w)$)

For b. I had no clue, I think it is because I have been unable to do part a. properly.

I found $f_{W,Z}(w,z)=2\lambda^2e^{\lambda(z+w)}\mathbf{1}_{0\le w\le z}$

Using this I tried to work out $$\begin{align} P(O\le t)&=\int_{0}^{t-w}\int_{w}^{t-w}2\lambda^2e^{-\lambda(z+w)}\mathbf{1}_{0\le w\le z, z-w\le t}dzdw\\ &=1-2\lambda e^{-\lambda t}(t-w)-e^{-2\lambda(t-w)} \end{align}$$

But I wasn't really sure whether this was actually useful or how to connect the different parts of this probability.

Any hints to get me in the right direction would be much appreciated.

EDIT: I believe it may be something along the lines of proving $W\le w$ and $O\le t$ are independent so then the probability would be $$P(W\le w,M=1,O\le t)=(1-e^{-2\lambda w})\frac{1}{2}(1-e^{-\lambda t})$$

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  • $\begingroup$ If $X$ and $Y$ are independent, then you should end up with $O \sim \exp(\lambda)$ thanks to the memoryless property of the exponential distribution so $\mathbb P(O \le t) = 1-e^{-\lambda t}$ for $t \ge 0$ $\endgroup$ – Henry Feb 23 at 16:14
  • $\begingroup$ So $Z-W$ is independent of $Z$ and $W$? $\endgroup$ – m4thsSt00dent Feb 23 at 16:17
  • $\begingroup$ Not at all: $Z-W$ is determined by $Z$ and $W$. What is true is that $Z-W$ is independent of $W$, though the question may be asking you to prove this. My comment was more of a check. $\endgroup$ – Henry Feb 23 at 16:22
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With $w,t\ge 0$ you have

$$\mathbb P(W\le w,M=1,O\le t) \\=\mathbb P(X\le w, X\le Y\le X+t) \\=\int_{x=0}^w\int_{y=x}^{x+t} \lambda^2 e^{-\lambda x}e^{-\lambda y}\, dy\, dx \\=\int_{x=0}^w \lambda(1-e^{-\lambda t}) e^{-2\lambda x}\, dx \\= \frac12 (1-e^{-\lambda t}) (1-e^{-2\lambda w}).$$

The next stage is letting $w \to \infty$ so you get $\mathbb P(M=1,O\le t) =\frac12 (1-e^{-\lambda t})$

and by symmetry $\mathbb P(M=0,O\le t) =\frac12 (1-e^{-\lambda t})$

so adding these together $\mathbb P(O\le t) =(1-e^{-\lambda t})$

meaning $O \sim \exp(\lambda)$

so $O$ is independent of $W$ (and of $M$).

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    $\begingroup$ @m4thsSt00dent When $X\leq Y$, what are $\min(X,Y)$ and $\max(X,Y)$ going to be? Surely, $X$ and $Y$ respectively. So, indeed... $$\begin{align}\{W\leqslant w, M=1, \mathcal O\leqslant t\} &=\{\min(X,Y), X\leqslant Y, \max(X,Y)-\min(X,Y)\leqslant t\} \\[1ex]&=\{X\leqslant w, X\leqslant Y, Y-X\leqslant t\}\\[1ex] &=\{X\leqslant w, X\leqslant Y\leqslant X+t\}\end{align}$$ $\endgroup$ – Graham Kemp 2 days ago

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