# Is this argument about probability correct?

Let's say, we have $i$ continuous IID random variables $X_1, X_2, \ldots,X_i$ whose domain is $\mathbb R$. This sequence divides the real axis into $i + 1$ intervals. Now, if we have another random $X_j$ who also have the same distribution, is correct that the probability of this $X_j$ falling into these $i+1$ intervals are equal?

(For me, it seems to be intuitively correct since there are no difference between these $X$'s. )

## 2 Answers

Yes. Think of the $(i+1)!$ orderings of the $i+1$ variables $\{X_j\}_{j=1}^{i+1}$. Each occurs with an equal probability of $\dfrac{1}{(i+1)!}$. $X_{i+1}$ is in position $j$ for exactly $i!$ of those arrangements. Thus, it is in each position with probability $\dfrac{i!}{(i+1)!}=\dfrac{1}{i+1}$.

• I should mention that I did use the continuity of the random variables by assuming that the probability that any two variables were equal is $0$. – robjohn Sep 20 '11 at 22:20

On the basis that you have no more information about the distribution or indeed the values actually taken, then yes.

To see this, instead consider the $i+1$ random variables $X_1, X_2, \ldots X_i$ and $X_j$. Since they are identically and independently distributed, the $(i+1)!$ possible orders are equally likely, and the values will be distinct with probability $1$.

So it is equally likely that $X_j$ takes any of the $i+1$ possible positions in the order, and so it is equally likely that $X_j$ is in any of the $i+1$ intervals created by the other $i$ random variables.