1

I am having lots of trouble with a question that seems at firs quite elementary.

Let $X_1,X_2,\dots,X_n$ be independent and identically distributed random variables, $X_i:(\Omega,\mathcal{F},P')\rightarrow(\mathbb{R},\mathcal{B}(\mathbb{R}),P_X)$ such that $P'$ determines a continuous cumulative distribution function $F_X(x)=P[X\leq x]=P'[\{\omega\in\Omega:X(\omega)\leq x\}]$ which is not necessarily absolute continuous wrt the Lebesgue measure. Then, given the order statistics $X_{i_1}<X_{i_2}<\dots<X_{i_n}$ the conditional distribution of $X_1,X_2,\dots,X_n$ is discrete, and assigns probability $\frac{1}{n!}$ to each "point" $X_{i_1},X_{i_2},\dots,X_{i_n}$ where $i_1.i_2,\dots,i_n$ is a permutation of $1,2,\dots,n$.

The questions does not seem to refer to the distribution of $X_1,X_2,\dots,X_n$ given a particular realization of the order statistics $X_{i_1}=x_{i_1},X_{i_2}=x_{i_2},\dots,X_{i_n}=x_{i_n}$, but the general conditioning.

I have found rather difficult to compute the conditional probabilities as they do not reduce to the simplest cases of conditional distributions.

Any insights or references?

Best Regards,

JM

Julio
  • 143

1 Answers1

1

This is an obvious statement (esp if you ignore the portion about absolutely continuous).

You should be able to show that $P(X_1 < X_2) = P(X_2 < X_1)$, which is the 2 variable case.

And similar, for any permutation $\sigma \in S_n$, we have

$$P(X_1 < X_2 < \cdots <X_n) = P(X_{\sigma(1)} < X_{\sigma(2)} < \cdots < X_{\sigma(n)})$$

Hence, each of these $n!$ events are equally likely to occur, hence occur with probability $\frac{1}{n!}$ (with the disclaimer that they sum to 1).

Note that you are restricting your cases to when all the order statistics are distinct, which means that all the original RV's have distinct values. This might restrict your probability space, if it is not absolutely continuous.

The exact distribution of $X_i$ does not matter. All that matters is that they are IID, and also that it is not the constant RV, in which case, $P(X_1< X_2) = 0$.

Calvin Lin
  • 77,541
  • Dear Calvin, I I have already proved in the general case that. Am I to understand that this unconditional probability is the conditional one? My question goes right to this point. – Julio May 29 '13 at 02:15
  • 2
    @Julio The conditional part is on the probability space which satisfies that the RV have distinct values. This is not unconditional, as I've already conditioned or it (and mentioned it). Put another way, if $U=U[0,1]$ and $X_1 (u) = 1 $ if $u<\frac{1}{2}$, 0 otherwise, and $X_2 (u) = 1$ if $\frac{1}{4} < u < \frac{3}{4}$, then the condition of $X_1 \neq X_2$ means that we're restricting our probability space to the set where $u\in [0, \frac{1}{4}]\cup [\frac{1}{2}, \frac{3}{4} ]$. – Calvin Lin May 29 '13 at 02:23
  • I have defined a random vector containing the random sample, lets say $\tilde{X}$ running from $(\Omega^n,\sigma(\mathcal{F}^n),P=P'^n)$ to $(\mathbb{R}^n,\mathcal{B}^n)$, and since $F_X$ is continuous, $P[X_i=X_j]=0$ for $i\neq j$ arbitrary. Going back to the point, given the order statistics $X_{i_1},X_{i_2},…,X_{i_n}$, $\tilde{X}$ may assume only one of $n!$ values corresponding to the possible permutations ${i_1,i_2,\dots,i_n}$ of ${1,2,\dots,n}$. Thus, $\tilde{X}$ is discrete. Therefore, $P[\tilde{X}=x/X_{i_1},X_{i_2},…,X_{i_n}]=P[X_{i_1}<X_{i_2}<\dots<X_{i_n}]$ right? – Julio May 29 '13 at 14:26
  • I have also proved that, because $X_1,X_2,\dots,X_n$ is a random sample, the probability measure $P=P'^n$ in the product space is symmetric in the sense that if the coordinates of $(\omega_1,\omega_2,\dots,\omega_n)$ are permuted to $(\omega_{i_1},\omega_{i_2},\dots,\Omega_{i_n})$, the probability of any event $A\in\sigma(\mathcal{F}^n)$ does not change. Therefore, it is true for the event ${X_1<X_2<\dots<X_n}$, which gives the required result. – Julio May 29 '13 at 14:36