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

2° Demonstration:
$$\begin{eqnarray} \mathbb{P}\left(C<t \big| A=X\right) &=&\mathbb{P}\left(X+Y-2A<t \big| A=X\right) \\ &=& \mathbb{P}\left(X+Y-2A<t \big| A=X, B=Y\right) \\ &=& \ldots \end{eqnarray}$$ by conditioning $$ \ldots = \mathbb{P}\left(A+B-2A < t\right) = \mathbb{P}\left(B-A < t\right) = \mathbb{P}\left(C<t\right)$$ 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

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2 Answers 2

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.

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

Then $P(C<t∣A=a) = P(B-A<t|A=a) =P(B<t+a|A=a) $

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

Because (here we used memorylessness) :

$$P(B<b|A=a) = P(B<b | A=a \cap A=X) P(A=X)+ P(B<b | A=a \cap A=Y) P(A=Y)= P(Y< b-a) P(A=X) + P(X<b-a) P(A=Y)$$

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

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