独立随机变量落在相邻顺序统计量之间的概率求解问题

Great question! Let's break this down step by step—first, I'll confirm a standard assumption here: I'm assuming we have $n$ independent and identically distributed (iid) random variables $Y_1, Y_2, ..., Y_n$, all independent of $X$, with $Y_{(i)}$ and $Y_{(i+1)}$ being the $i$-th and $(i+1)$-th order statistics of this sample (so $Y_{(1)} \le Y_{(2)} \le ... \le Y_{(n)}$ almost surely).

Method 1: Probability Difference (Builds on Your Existing Calculation) #

Since you already know how to compute $P(X \ge Y_{(i)})$, we can leverage the monotonicity of order statistics to split the target probability:
$$P(Y_{(i)} \le X \le Y_{(i+1)}) = P(X \ge Y_{(i)}) - P(X \ge Y_{(i+1)})$$
This works because the event ${X \ge Y_{(i)}}$ fully contains ${X \ge Y_{(i+1)}}$ (since $Y_{(i)} \le Y_{(i+1)}$), so subtracting the two probabilities leaves exactly the probability that $X$ falls between the two order statistics.

You can simplify the double integral you mentioned for $P(X \ge Y_{(k)})$ (where $k$ is $i$ or $i+1$) using conditional expectation. Since $X$ and $Y$ are independent, we can write:
$$P(X \ge Y_{(k)}) = \mathbb{E}\left[ P(Y_{(k)} \le X \mid X) \right]$$
Given $X=x$, $P(Y_{(k)} \le x)$ is just the CDF of the $k$-th order statistic at $x$, which is:
$$F_{Y_{(k)}}(x) = \sum_{m=k}^n \binom{n}{m} F_Y(x)^m (1-F_Y(x))^{n-m}$$
Substituting this in, we get a single integral (easier to compute than the double integral you noted):
$$P(X \ge Y_{(k)}) = \int_{-\infty}^{\infty} \sum_{m=k}^n \binom{n}{m} F_Y(x)^m (1-F_Y(x))^{n-m} f_X(x) dx$$
Calculate this for both $k=i$ and $k=i+1$, then subtract the results to get your target probability.

Method 2: Counting Logic (Simpler for Continuous Distributions) #

If $X$ and $Y$ are continuous random variables (so $P(X=Y_j)=0$ for any $j$), we can reframe the problem with counting: the event ${Y_{(i)} \le X \le Y_{(i+1)}}$ is equivalent to exactly $i$ of the $Y_j$ being less than or equal to $X$ (since $Y_{(i)}$ is the $i$-th smallest, $X$ sitting between it and $Y_{(i+1)}$ means exactly $i$ samples are to the left of $X$, and $n-i$ are to the right).

Since each $Y_j$ is independent of $X$, $P(Y_j \le X \mid X=x) = F_Y(x)$. The probability of exactly $i$ successes (where a "success" is $Y_j \le X$) follows a binomial distribution, so given $X=x$, this probability is:
$$\binom{n}{i} F_Y(x)^i (1-F_Y(x))^{n-i}$$
Integrate this over the entire distribution of $X$ to get the final probability:
$$P(Y_{(i)} \le X \le Y_{(i+1)}) = \int_{-\infty}^{\infty} \binom{n}{i} F_Y(x)^i (1-F_Y(x))^{n-i} f_X(x) dx$$
This formula is often more efficient to compute than the sum-based integral from Method 1.

Adjustment for Discrete Distributions #

If $X$ or $Y$ is discrete, we need to account for non-zero probabilities where $X$ equals $Y_{(i)}$ or $Y_{(i+1)}$. The target probability becomes:
$$P(Y_{(i)} < X < Y_{(i+1)}) + P(X=Y_{(i)}) + P(X=Y_{(i+1)})$$

• $P(Y_{(i)} < X < Y_{(i+1)})$ uses the same counting logic as above, but with strict inequalities (counting exactly $i$ $Y_j < X$).
• $P(X=Y_{(i)})$ can be calculated by summing over all possible values of $x$:
  $$P(X=Y_{(i)}) = \sum_x P(X=x) \cdot P(Y_{(i)}=x)$$
  where $P(Y_{(i)}=x)$ is the probability mass function of the $i$-th order statistic for the discrete $Y$ distribution.

内容的提问来源于stack exchange，提问作者Cain Rose