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$X$ and $Y$ are two independent variables with an exponential distribution with parameters $\lambda$ and $\mu$; $A = \min(X,Y)$, $B = \max(X,Y)$ and $C = B-A$. I want to prove that $A$ and $C$ are independent. I have provided two different demonstrations, but they are both wrong and i don't know why:

1° Demonstration:
$ \begin{eqnarray} \mathbb{P}\left(CX\right) \\ &=& 1-\mathbb{P}\left(Y>2A-X+t \Big| Y>X\right) \\ &=& 1-\mathbb{P}\left(Y>2A-2X-t\right) \end{eqnarray} $ because exponential is memoryless. So: $ = \mathbb{P}\left(2X+Y-2A So $C$ and $A$ are not independent...Something's wrong!

2° Demonstration:
$\begin{eqnarray} \mathbb{P}\left(C by conditioning $ \ldots = \mathbb{P}\left(A+B-2A < t\right) = \mathbb{P}\left(B-A < t\right) = \mathbb{P}\left(C This seems correct, but that would mean that this property is true for every $X$, $Y$. But this is not true. So where's the mistake?

Thanks

2 Answers 2

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The errors has been pointed by by joriki's aswer. I don't think that starting with $P(C is the best way to prove independence, you want to show that $P(C is the same as $P(C for any value of $a$.

Then $P(C

But the last term does not depends on $a$ (and this proves independence).

Because (here we used memorylessness) :

$$\begin{align} P(B

So $P(B which does not depend on $a$ - note that we don't need to compute $P(A=X)$.

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In both cases, the step interrupted by text is wrong.

In the first one, it should be $-t$ on the right-hand side of the inequality.

In the second one, I don't know what you mean by "by conditioning"; it seems you used the equations in the condition to rewrite the inequality, which is OK, and then simply dropped the condition, for which I see no justification.