I have a statistic question which I can intuitively guess the answer but I cannot find a good way to prove it.

Suppose $N$ IID random variables $X_1, ..., X_N$ and their order statistics $X_{(1)}\leq ...\leq X_{(N)}$. Consider the probability

$P(X_{(1)}\leq \alpha_1, X_{(2)}<\alpha_2, ...,X_{(k)}<\alpha_{k}, X_1>\alpha_k)$

where $\alpha_1\leq \alpha_2\leq ... \leq \alpha_k$ are constants and $k\leq N$ is an integer.

I know how to compute the probability of the order statistic, so the only annoying part is $X_1$. Using the conditional probability, we can rewrite the above probability as

$P(X_{(1)}\leq \alpha_1, X_{(2)}<\alpha_2, ...,X_{(k)}<\alpha_{k})$ $=P(X_{(1)}\leq \alpha_1, X_{(2)}<\alpha_2, ...,X_{(k)}<\alpha_{k})*P(X_1>\alpha_k|X_{(1)}\leq \alpha_1, X_{(2)}<\alpha_2, ...,X_{(k)}<\alpha_{k})$

That is the place where I am stuck. From the conditional probability, we observe that $X_1$ has to be at least $(k+1)$th largest value. since the random variables are IID and the information of the order statistics do not make any $X$ "special", so $X_1$ has $\frac{N-k}{N}$ chance to meet the requirement. Therefore, purely based on intuition, The probability can be rewritten as

$P(X_{(1)}\leq \alpha_1, X_{(2)}<\alpha_2, ...,X_{(k)}<\alpha_{k}, X_1>\alpha_k)$ $=\frac{N-k}{N}P(X_{(1)}\leq \alpha_1, X_{(2)}<\alpha_2, ...,X_{(k)}<\alpha_{k})$

I do not know if my guess is correct or not, and I would like to see a formal proof of the question. I will appreciate it if anyone can shed light on it.


A simulation shows that the conditional probability $P(X_1>\alpha_k|X_{(1)}\leq \alpha_1, X_{(2)}<\alpha_2, ...,X_{(k)}<\alpha_{k})$

is not equal to

$P(X_1$ is at least $(k+1)$th largest value$|X_{(1)}\leq \alpha_1, X_{(2)}<\alpha_2, ...,X_{(k)}<\alpha_{k})$

where the latter one is equal to $\frac{N-k}{N}$, so my guess is not correct. I'll appreciate anyone's help.

  • $\begingroup$ What is exactly the question? Do you need to express your probability in terms only of $X_{(1)}$ and $X_{(2)}$? Because I guess in your case it would be simpler to just use $X_1$ and $X_2$. As you noted, $X_1$ must be the greater, implying that $$P\left(X_{(1)} \leq \alpha_1, X_{(2)} > \alpha_2, X_1> \alpha_2\right) = P\left(X_1 > \alpha_2, X_2 \leq \alpha_1\right).$$ Am I correct? $\endgroup$ – dfnu Aug 14 '19 at 5:46
  • 1
    $\begingroup$ @dfnu You are right, the exact question is for the case where you have more than two random variables. This example might be too simple since we can find an equivalent probability in the form of $X_1$ and $X_2$. I'll update my question. Thanks. $\endgroup$ – Jeff Aug 14 '19 at 14:18
  • $\begingroup$ The simpler case can anyhow be used to test whether your intuition is correct. Have you tried? $\endgroup$ – dfnu Aug 14 '19 at 15:01
  • $\begingroup$ @dfnu Good point! For the simplest example, the answer is yes. $\endgroup$ – Jeff Aug 14 '19 at 15:36
  • $\begingroup$ you might want to test also the three-variable case, where some interdependence might arise. $\endgroup$ – dfnu Aug 14 '19 at 15:48


I just want to better specify the reasoning the brought me to the expression in the original comment, and correct some mistakes.

Let us start, for example, with $k=2$. We have $N$ independent identically distributed random variables $X_i$ with CDF $F(x)$, and $\alpha_1 < \alpha_2 $. If $X_{(i)}$ is the $i$th random variable after ordering, then we aim at computing

$$\mathcal P_{N,2}(\boldsymbol{\alpha})=P\left(X_{(1)} < \alpha_1 \land X_{(2)} < \alpha_2 \land X_1 > \alpha_2 \right).$$

Using conditional probabilities we have

$$\mathcal P_{N,2}(\boldsymbol{\alpha}) = P\left(X_1 > \alpha_2\right)P\left(X_{(1)} < \alpha_1 \land X_{(2)} < \alpha_2 | X_1 > \alpha_2 \right).\tag{1}\label{1}$$

Conditioning on $X_1$ implies that neither $X_{(1)}$ nor $X_{(2)}$ correspond to $X_1$. All the other variables, independently of the value assumed by $X_1$ can assume first and second position in the ordering. Therefore we can expand the second term in RHS of \eqref{1} by noting that we can select any of the $i=2,3,\dots,N$ variables as $X_{(1)}$ (it must assume a value smaller than $\alpha_1$) and then any of the other $N-2$ variables as $X_{(2)}$ (it must assume a value between $\alpha_1$ and $\alpha_2$). All the remaining $N-3$ variables need to have value greater than $\alpha_2$. We get

\begin{eqnarray} \mathcal P_{N,2}(\boldsymbol{\alpha}) &=& \left[1-F(\alpha_2)\right] \cdot (N-1)(N-2)\cdot F(\alpha_1) \left[F(\alpha_2)-F(\alpha_1)\right]\cdot\left[1-F(\alpha_2)\right]^{N-3}=\\ &=&(N-1)(N-2) F(\alpha_1) \left[F(\alpha_2)-F(\alpha_1)\right]\left[1-F(\alpha_2)\right]^{N-2}. \end{eqnarray}

If this is correct, then we can extend the reasoning to any $k$:

$$\mathcal P_{N,k}(\boldsymbol{\alpha})=\frac{(N-1)!}{(N-k-1)!}F(\alpha_1)\left[1-F(\alpha_k)\right]^{N-k} \prod_{i=2}^{k}\left[F(\alpha_i)-F(\alpha_{i-1})\right]$$


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