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Let me start off by noting that I know this question: Prove that $\mathbb P(X>Y) =\frac{b}{a + b}$ if $X, Y$ are exponentially distributed with parameters $a$ and $b$. and understand how it works.

However, I figured there might be another way to solve it which doesn't seem to work out. So I'm asking this in the hope that someone can tell me why this type of argument fails.

Here's my idea:

Writing $P(X>Y)$ is the same as writing $P(X>a>Y)$, so we might as well split this to $P(X>Y) = P(x>a)P(Y<a)$, as $X$ and $Y$ are independent. But this equals $(1-P(x<a))(P(Y<a)) = (1-(1-e^{ax}))(1-e^{bx}) = e^{-ax}-e^{-(a+b)x}\neq \frac{b}{b+a}$

Mitchell Faas
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2 Answers2

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The problem is in the very beginning: "Writing $P(X>Y)$ is the same as writing $P(X>a>Y)$" This is not true, there are many different ways in which $Y$ can be smaller than $X$: both can be smaller than $a$, both can be bigger than $a$, exactly one of $X$ and $Y$ may even be precisely equal to $a$ (although those are probability zero events) and, as you write $X$ can be bigger than $a$ while $Y$ is smaller than $a$.

The point is that $a$ is a parameter which is determined by the distribution and should be considered a given. It is not just some dummy variable you invented yourself to make the computation easier.

Vincent
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If $x,y\in\mathbb R$ then the statements $x>y$ and $\exists a\in\mathbb R[x>a>y]$ are the same, but for a fixed $a\in\mathbb R$ the statements $x>y$ and $x>a>y$ are not the same (the first does not imply the second).

If $X,Y$ are random variables, i.e. measurable functions $\Omega\to\mathbb R$ then $X>Y$ is true iff $X(\omega)>Y(\omega)$ for every $\omega\in\Omega$, and this is equivalent with: $$\forall\omega\in\Omega\exists a\in\mathbb R[X(\omega)>a>Y(\omega)]$$

This is not equivalent with:$$\exists a\in\mathbb R\forall\omega\in\Omega[X(\omega)>a>Y(\omega)]$$

And certainly not with:$$\forall\omega\in\Omega[X(\omega)>a>Y(\omega)]$$where $a$ denotes some fixed element of $\mathbb R$.

If the possibility would be there then things would get even more worse if $a$ would denote a very special fixed element of $\mathbb R$ (a parameter of an involved distribution).

drhab
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