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I am reading Introduction to Probability by Joseph K. Blitzstein and Jessica Hwang, which states that Continuous r.v.s that are i.i.d have all possible ranking equally likely. In the proof, it is mentioned that

$\forall\ i$ and $j$ with $i\neq j$, the probability of the tie $X_{i} = X_{j}$ is $0$ since $X_{i}$ and $X_{j}$ are independent continuous r.v.s.

I am unable to understand why this statement has to be true. Can you please explain?

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  • $\begingroup$ I think the term "all possible ranking equally likely doesn't make sense. I think this question requires a precise statement of the hypothesis along with the proof to give context to the step in the proof that you provide. $\endgroup$ – Michael Chernick Dec 25 '16 at 16:13
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    $\begingroup$ Why is symmetry in the title? $\endgroup$ – Michael Chernick Dec 25 '16 at 16:37
  • $\begingroup$ @MichaelChernick I suggest taking a look to page 225 of the original source books.google.es/… . For me, the symmetry and ranking issue in this question just gives context and I think it's explained enough in the book. The obscure passage for the OP is the quoted statement, that the author of the book didn't explain. $\endgroup$ – Pere Dec 25 '16 at 20:23
  • $\begingroup$ I think the OP should explain this in the problem so that we can interpret it properly.. The source you referenced is diffficult to go through. The burden for making a sensible sounding question should be with the OP. If your answer is adequate I don't yet see it. @Pere. $\endgroup$ – Michael Chernick Dec 26 '16 at 2:55
  • $\begingroup$ If the source weren't difficult to go through the OP wouldn't need to ask here. Anyway, the question is clearly why that highlighted statement has to be true, and that is what I answered - that is, why ties have zero probability. The OP didn't ask why that implies that all possible ranking are equally likely, that's just context. $\endgroup$ – Pere Dec 26 '16 at 10:33
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According to comments, it seems that here several questions arise from a single statement.

Why is the probability of a tie zero? (assuming independent continuous variables).

In a continuous random variable individual results have zero probability. Only measurable sets (for example intervals) have non zero probability.

The probability of a tie is the probability of the second random variable yielding the same value than the first one. That would be the probability of a single value and therefore it would be zero.

If random variables were discrete, some individual values would have non zero probability and therefore the probability of a tie would be greather than zero.

The requeriment of the two random variables being independent is needed because two non independent continuous variables could have discrete conditional probabilities. For example, if both variables yielded always the same result they would be continuous but ties would have probability 1.

If continuous random variables have 0 probability for individual points, how does an ordering makes sense?

In continuous random variables, individual points have zero probability, but intervals have probabilities larger than zero.

For example, for uniform random variable in [0, 1], any point (for example 0.4, 0.45 or 0.5) has zero probability but the interval [0.4, 0.5] has 10% probability.

Furthermore, two one-side unbounded intervals $(-\infty,x)$ and $(x,+\infty)$ have probabilities different than zero and that probabilities sum 1, that is, the probability of the whole real line $(-\infty,+\infty)$.

When two iid continuous variables $X_1$ and $X_2$ are considered, $X_1=X_2$ has probability zero, as seen as a realization of the second variable it's just a point. However $X_1<X_2$ nad $X_1>X_2$ have probabilities larger than zero and they sum up to 1.

How this makes all orderings equally likely?

The source cited by the OP sees it as rather evident.

The idea behind the symmetry reasoning is that there is no reason to think that $X_1<X_2$ is more likely than $X_1>X_2$ or vice-versa, because we could even change the names of $X_1$ and $X_2$ without changing anything.

This can be generalised to ordering of $n$ iid variables.

The book uses the fact that orderings with ties have probability zero, the only cases with probability different than zero are orderings without ties, all of them with the same probability. Now you just need to count how many of such orderings exist (permutations of n elements) and divide 1 between the number of orderings to get the probability of each one.

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  • $\begingroup$ This is nice but I am puzzled by the OPs question because it is not clear exactly what is assumed and what needs to be proved. It seems to be a sequence of random variables that are iid continuous but what does "have all possible rankings equally likely" mean? $\endgroup$ – Michael Chernick Dec 25 '16 at 16:20
  • $\begingroup$ @MichaelChernick That's just context and doesn't affect why the probability of a tie is zero - that is the actual question. If you don't understand what "have all possible rankings equally likely" mean you can post a question. It would be an interesting question - but a different one. $\endgroup$ – Pere Dec 26 '16 at 10:38
  • $\begingroup$ I am not interested in what the OP meant. He could be confused or it could be wrong but he is the one who needs to clarify it. $\endgroup$ – Michael Chernick Dec 27 '16 at 0:38
  • $\begingroup$ @Pere You got the context right. I know that individual values a continuous random variable assumes a 0 probability. However, I just get confused when I think along this line of thought 'If continuous random variables have 0 probability for individual points, how does an ordering makes sense?' Also, I contradict myself when I think 'continuous random variables have a support and have to assume values from the support and hence can take an individual value' Can you help build the understanding please? $\endgroup$ – Srini Vas Dec 27 '16 at 14:27
  • $\begingroup$ @SriniVas Does the answer address your concerns now? $\endgroup$ – Pere Dec 28 '16 at 12:24
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For i.i.d. continuous rv's $X_1,X_2$, $P(X_1 \leq X_2)= P(X_2 \leq X_1) = 1/2$ is rigorously proven as follows:

For $f_X$ being the common density and $F_X$ the commmon CDF, we have

$$P(X_1 \leq X_2) = \int_{-\infty}^{\infty}\int_{-\infty}^{x_2}f_{X_1,X_2}(x_1,x_2)dx_1 dx_2 $$ $$= \int_{-\infty}^{\infty}f_X(x_2) \int_{-\infty}^{x_2}f_X(x_1) dx_1 dx_2 = \int_{-\infty}^{\infty}f_X(x_2) F_X(x_2) dx_2$$ $$= E[F_X(X_2)]$$

Viewed as a random variable, $F_X(X_2)$ follows a Uniform $U(0,1)$ (by the probability integral transform) and so $E[F(X_2)] = 1/2$. We obtain the same result for $P(X_2 \leq X_1)$.

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I would say it is because the probability that a continuous random variable takes a particular value is 0 (the point being of measure 0 on the real line). And since they r.v.s are independent, the probability of $X_{j}$ taking any particular value $X_{i}$ might already have is 0 as it is a random draw form the same distribution again.

You can see a discussion with example here.

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  • $\begingroup$ The post you cite is very good for supporting your answer and I think it might partly answer the OPs question but I have yet to see what symmetry of iid continuous random variables has to do with and there is not a clear description of the theorem in question. $\endgroup$ – Michael Chernick Dec 26 '16 at 3:05
  • $\begingroup$ @jan-sila I know that individual values a continuous random variable assumes a 0 probability. However, I just get confused when I think along this line of thought 'If continuous random variables have 0 probability for individual points, how does an ordering makes sense?' Also, I contradict myself when I think 'continuous random variables have a support and have to assume values from the support and hence can take an individual value' Can you help build the understanding please? $\endgroup$ – Srini Vas Dec 27 '16 at 14:27

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