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This link shows that that $P(X_i = x) = 0$ so can we say, \begin{equation} P\left(X_1 < X_2 < X_3\right) = P\left(X_1 \le X_2 \le X_3\right) \end{equation}

Assumptions

  • $X_i, X_j$ are random uniform and continuous between 0 and 1
  • $X_i, X_j$ are independent
  • Lets just look at the non-trivial case of $i \ne j$
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Alexander McFarlane
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  • Yes. Rewrite the right hand side as a union of ${ X_1 < X_2 < X_3 }$ and every event contained in ${ X_1 \leq X_2 \leq X_3 }$ that involves at least one equality to see why. – dsaxton Jan 19 '16 at 15:14
  • "This link shows that that P(Xi=Xj)=0" No, the link shows that P(X=x)=0 for every x, if the random variable X is continuous. As already pointed to you explicitely, this is not the same statement. – Did Jan 20 '16 at 07:55
  • Right you are. Edited. – Alexander McFarlane Jan 20 '16 at 08:04
  • Funny, you accepted an answer stating as a given that $P(X_i=X_j)=0$ although, if I understand correctly what you write, this is the step which is blocking you and which led you to ask several questions on the site. Or maybe you now know why indeed $P(X_i=X_j)=0$? – Did Jan 20 '16 at 17:20
  • Apparently not that many people seem to know explicitly why! It's greatly helped me in my grasp of probability. So often, learning little things as givens can trip you up later! Would you suggest that the before mentioned is not an answer in fact? – Alexander McFarlane Jan 20 '16 at 17:22

1 Answers1

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Yes.

The events $A \equiv X_1 \le X_2 \le X_3$ and $B \equiv X_1 < X_2 < X_3$ are related by $A = B \cup (X_1 = X_2 \cap X_2 < X_3) \cup (X_1 < X_2 \cap X_2 = X_3) \cup (X_1 =X_2=X_3) $

Because the events $X_i = X_j$ ($i\ne j$) have zero probability, and $P(C\cap D)\le P(C)$, this leads to $P(A)=P(B)$

leonbloy
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