For a sequence $(X_n)_{n \geq 1}$ of i.i.d. real RV with $P(X_1=x)=0$, the event $\lbrace X_1<X_2< \dots < X_n \rbrace$ is independent of permutation

Problem: Let $(X_n)_{n \geq 1}$ be a sequence of i.i.d real RV such that $P(X_1=x)=0$, then for any permutation $\sigma \in \lbrace 1, \dots, n \rbrace$ we have the equality $$P(X_{ \sigma(1)} < X_{\sigma(2)} < \dots < X_{\sigma(n)})=P(X_1< X_2< \dots < X_n )$$

My approach: I thought that the best way to start this is to show it for $n=2$ and then hopefully see a pattern that can be easily applied/generalized via induction over $n \in \mathbb{N}$.

So let $X,Y$ be two real i.i.d. random variables, with $P(X=x)=0$ for all $x \in \mathbb{R}$, it then easily follows that $P(X=Y)=0$ i.e. $P(X>Y)+P(X<Y)=1$. I want to show that $P(X<Y)=P(X>Y)$, thanks to the above, that means I want to show that $P(X<Y)=1/2$

I haven't made any use of the fact that $X,Y$ have the same distribution yet.

I want to write $P(X<Y)=P(X=x<y=Y)$ for $x,y \in \mathbb{R}: x<y$ but I am stuck on how to continue here, maybe someone can provide me with some hints.

• Because the variables are iid, it is immediate (from the definition of independence) that the distribution of $(X_1, \ldots, X_n)$ is the same as the distribution of $(X_{\sigma(1)}, \ldots, X_{\sigma(n)})$ for any permutation $\sigma$. You may directly apply this result to the problem. – whuber Apr 18 '16 at 22:28
• Thanks a lot @whuber. To reason your statement we have for every Borel set $A_1, \dots, A_n$ the equation $P(X_1 \in A_1, \dots , X_n \in A_n)=P(X_1 \in A_1) \cdot ... \cdot P(X_n \in A_n)$ on the right side I can perform multiplicative permutations ($\cdot$ is commutative) and also I have $n!$ possible permutations for $X_1 \in A_1 , \dots , A_n$. I would conclude that in the case of $n=2$ (as described in my question) we have $P(X<Y)=P(X>Y)$ because $X,Y$ are equally distributed. – Spaced Apr 19 '16 at 16:11

Because the value of $P(X\lt Y)$ is a fact about the random variables $X$ and $Y$, it is fundamental to all of mathematics that this fact remains the same no matter what you happen to name those variables. Now, since $X$ and $Y$ have the same distribution and are independent, you may name them $X$ and $Y$ (in that order) or $Y$ and $X$, and it does not matter. Accordingly,

$$P(X \lt Y) = P(Y \lt X)$$

in all such cases.

The same reasoning shows that the chances do not change whenever we interchange two of the variables

\eqalign{ P(X_1 \lt \cdots \lt X_i \lt \cdots \lt X_j \lt \cdots \lt X_n) \\ = P(X_1 \lt \cdots \lt X_j \lt \cdots \lt X_i \lt \cdots \lt X_n), }

simply because the two probabilities refer to precisely the same thing, merely renamed.

Because every permutation can be written as a product of such interchanges, the result follows immediately.

Notice that this result is more general than the one requested in the question, which was limited to continuous variables.