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Question: If $X$ and $Y$ are two independent geometric random variables with parameters $a$ and $b$, where $$\forall k \geq 1, P(X = k) = a(1-a)^{k-1} $$ and $$\forall k \geq 1, P(Y = k) = b(1-b)^{k-1}$$ Then are $X - Y$ and $\min(X, Y)$ independent?

I know that $\min(X, Y)$ follows a geometric distribution with parameter $1 - (1-a)(1-b)$. How to prove or disprove their independence?

Yunxuan Zhang
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Mohamed Guendzi
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  • Memorylessness may be helpful. – Henry Jan 27 '25 at 22:16
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    You might start by showing $P(X>Y \mid \min(X,Y)=k)$ and $P(X<Y \mid \min(X,Y)=k)$ and $P(X=Y \mid \min(X,Y)=k)$ do not depend on $k$. Then show $P(X-Y = d \mid X+Y=k)$ does not depend on $k$ either perhaps looking at $d>0,d<0,d=0$ separately. – Henry Jan 27 '25 at 22:23
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    Have you tried proving this from the definition, showing that $P(\min(X,Y)=m,Y-X=d)=P(\min(X,Y)=m)P(Y-X=d)$ for all $m\ge 1,d\in \mathbb Z$? Seems like a good place to start. What goes wrong when you do that? – Mike Earnest Jan 27 '25 at 23:25
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    In fact, nothing is wrong; the result surprises me. They're actually independent. Thanks, @MikeEarnest ! – Mohamed Guendzi Jan 27 '25 at 23:32
  • This property (in the discrete context) is characteristic of the geometric distribution (modulo a translation), by a result of T.S. Ferguson [A characterization of the geometric distribution, Amer. Math. Monthly 72 (1965) 256–260]. – John Dawkins Jan 27 '25 at 23:55
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    The same $X-Y$ and $\min(X,Y)$ being independent property is true for two independent exponentially distributed random variables even if they have different rates, essentially for the same memorylessness reason. – Henry Jan 27 '25 at 23:58
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    @Mohamed Guendzi Since you appear to have now answered the question yourself, I suggest you write up the answer and post it so that the question will no longer be listed as unanswered. – lonza leggiera Jan 28 '25 at 01:20

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