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Consider 3 RVs ($X_1$,$X_2$,$X_3$) IID distributed.

  1. What is the probability that $X_1 > X_2$ and $X_1 > X_3$

  2. Also what is probability of $X_1 > X_3$ given that $X_1 > X_2$.

This is not homework, the answers are counter intuitive and I am looking for a proper reasoning and answer.

Dsp guy sam
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2 Answers2

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I assume the random variables are continues.

Question 1: 1/3 $$X_1<X_2<X_3$$ $$X_1<X_3<X_2$$ $$X_2<X_1<X_3$$ $$\color{red}{X_2<X_3<X_1}$$ $$\color{red}{X_3<X_2<X_1}$$ $$X_3<X_1<X_2$$

Question 2: 2/3 $$\color{red}{X_3<X_2<X_1}$$ $$\color{red}{X_2<X_3<X_1}$$ $$X_2<X_1<X_3$$

Masoud
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  • Since X_1 is independent of X_2 and X_3 then the probability of X_1 > X_2 is ?1/2 I assume...now since they are independent then the simulatneouss event that X_1 > X_3 and X_1 > X_2 is?...why should the probability not be multiplied.. since events are independently...why does this reasoning fail here – Dsp guy sam Apr 20 '20 at 11:54
  • Those two event,X_1 > X_3 and X_1 > X_2 , are not independent! – Masoud Apr 20 '20 at 12:42
  • And why are they not independent? – Dsp guy sam Apr 20 '20 at 12:44
  • When you know that $X_1 >X_2$ the chance that $X_1>X_3$ decrease in compare with the case we did no know $X_1 >X_2$. Since in two permutation $X_1>X_3$ and one permutation $X_1<X_3$. So $P(X_1>X_3|X_1 >X_2)>P(X_1>X_3)$. – Masoud Apr 20 '20 at 14:42
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HINT :

If $X_1, X_2,X_3$ are IID random variables then they can be ordered in any permutation with equal probability, i.e $ P[X_1 > X_2 >X_3] = P[X_2 > X_3 > X_1 ] $.

In total there are 3! such ways. Now since all those probabilities are equal and must sum to 1. Probability of any one of those orders is $\frac{1}{6}$

Vishaal Sudarsan
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  • So your suggesting the answer to the first question is 1/6? – Dsp guy sam Apr 20 '20 at 11:20
  • @Dspguysam no you need to solve from there, it would be 1/3 by the way. – Vishaal Sudarsan Apr 20 '20 at 11:21
  • What about the answer to second question – Dsp guy sam Apr 20 '20 at 11:22
  • Why is it 1/3 by the way...in the answer you mention 1/6, what is the reasoning behind 1/3 as the answer – Dsp guy sam Apr 20 '20 at 11:23
  • @Dspguysam You need to use conditioning of Probabilities, the numerator of that conditioning expression actually turns out to be the answer to the first problem. Use the hint to find the denominator to arrive at the answer 2/3. – Vishaal Sudarsan Apr 20 '20 at 11:25
  • Since X_1 is independent of X_2 and X_3 then the probability of X_1 > X_2 is ?1/2 I assume...now since they are independent then the simulatneouss event that X_1 > X_3 and X_1 > X_2 is?...why should the probability not be multiplied.. since events are independently...why does this reasoning fail here? – Dsp guy sam Apr 20 '20 at 11:53
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    @Dspguysam $X_1$ is independent of $X_2$ and $X_3$, but $(X_1,X_2)$ is not independent of $(X_1,X_3)$, this is because $X_1$ depends on itself. Therefore ${X_1 > X_3}$ is not independent of ${X_1 > X_2}$ – Leander Tilsted Kristensen Apr 20 '20 at 12:55
  • @Dspguysam I hope the comment by Leander solves your query. The pair is not independent as $X_1$ is common in both the pairs. – Vishaal Sudarsan Apr 21 '20 at 11:36