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Suppose four numbers $\{a,b,c,d\}$ where $a$ and $c$ random variables from a continuous distribution with support on $\mathbb{R}$.

Does $b\neq d$ imply $|a−b|−|c−d|+a−c\neq 0$ almost surely?

I really want to believe this proposition is true but that is typically not enough to make it so.

Edit:

MickMack's counter example made me realize I need to refine the proposition to make it:

Does $|b|\neq|d|$ imply $|a−b|−|c−d|+a−c\neq 0$?

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2 Answers 2

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Are there any constraint on b and d? If we take $b=2$ and $d=-2$, then:

$|a-b|-|c-d|+a-c=|a-2|-|c+2|+a-c$

If we only suppose that a and c are sampled from a continuous distribution, then : $P(a<2,c<-2)$ might be $>0$

In this case :

$|a-b|-|c-d|+a-c=|a-2|-|c+2|+a-c = 2-a-(-2-c)+a-c=0$

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    $\begingroup$ Maybe you only need to say that : $|b|\neq|d|$ $\endgroup$
    – MickMack
    Sep 12, 2014 at 13:02
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I am not sure why you say that $a$ and $c$ are sampled.

For a continuous distribution, the probablility you get an exact value is always zero. Idea is that for a continuous distribution, you have an uncountable infinity of possible values, so that the probability of an exact realization is zero. It is the same thing with the probability that you get a value related to a value you previously got.

It is different for a non continuous distribution.

And if you take only numbers, you have counterexamples, for instance ex $\{1,1,0,-1\}$ you get $0$

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  • $\begingroup$ $b$ and $d$ are not random variables. $\endgroup$
    – user603
    Sep 12, 2014 at 12:53

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