# Show that $P(X_k \le x) = P\{N(x) \ge k\}$

Suppose $$x_1 ... x_n$$ are the order statistics of an iid sample from a continuous distribution $$F(x)$$. Show that $$P(X_k \le x) = P\{N(x) \ge k\}$$ where $$N(x)$$, the number of sample values less than x, is binomial with parameters $$n$$ and probability $$p = F(x)$$.

Given the statement looks (i believe it to be) like this: $$P(X_k \le x) = \sum_{k=0}^n\binom{n}{k}F(x)^k(1-F(x))^{n-k}$$

However, I do not see how $$\sum_{k=0}^n\binom{n}{k}F(x)^k(1-F(x))^{n-k} = P\{N(x) \ge k\}$$

Use the Q above to show that the density of $$X_k$$ is $$p(x) = \binom{n}{k}F(x)^{k-1}(1-F(x))^{n-k}f(x)$$ where $$f(x)$$ is the density from $$F(x)$$. Verify the density using the multinomial argument:

$$n!\epsilon^n\prod_ip_{\theta}(x_i)$$

which is a general heuristic to deal with order statistics from an iid sample from a continuous density $$p_{\theta}(x)$$.

• If $X_{(k)}\le x$ then at least $k$ of the variates in the $n$ sample are less than $x$ since $$X_{(1)}\le\cdots\le X_{(k)}$$ Mar 9, 2022 at 14:36
• The sum should read $\sum_{i=k}^n {n \choose i} F(x)^i (1-F(x))^{n-i}$ which is clearly equal to $P(N(x)\ge k)$. Mar 9, 2022 at 15:21
• @JarleTufto I cannot understand why this is true? For example, $X_k$ takes a completely different distribution. Mar 9, 2022 at 15:25
• $N(x)$ is the total number of observations $\le x$. Each observation $X_i \le x$ with probabiity $P(X_i\le x)=F(x)$ and hence (because of indepence), $N(x)\sim \operatorname{bin}(n,F(x))$. And the event that $X_{(k)}\le x$ is the same event that $N(x)\ge k$. Mar 9, 2022 at 15:46
• In any case, in the continuous case, the alternative derivation in stats.stackexchange.com/a/78559/77222 is much more illuminating (differentiating the above sum of binomial point masses is a rather tedious exercise). Mar 9, 2022 at 15:53

Show that $$P(X_{(k)} \le x) = P(N(x) \ge k)$$

You can proof this by showing that the events on the left side and the right side $$X_{(k)} \le x \qquad \text{and} \qquad N(x) \ge k$$ are the same event.

There is however a tricky detail which is that they are not exactly the same event.

You have on the left side an inequality that is not strict (less than or equal) whereas on the right side you have a strict inequality with the definition of $$N(x)$$ being 'the number of sample values less than $$x$$'.

### Discrete distribution

For discrete distributions this difference in the events will result in the probabilities not being equal. Take for instance a sample of size one drawn from a Bernoulli distribution. Then $$P(X_{(1)} \leq 1) = 1 \qquad \text{and} \quad P(N(1) \geq 1) = 1-p$$

### Continuous distribution

To still make the proof for continuous probabilities, we could use as starting point one of the following alternative equations instead. These are made by replacing either the $$\leq$$ sign on the left by a $$<$$ sign, or the $$<$$ sign on the right by a $$\leq$$ sign such that the events are the same.

$$\begin{array}{}P(X_{(k)} < x) &=& P(N(x) \ge k) \\ P(X_{(k)} \le x) &=& P(N^\prime(x) \ge k) \\ \end{array}\\$$

with $$N^\prime(x)$$ meaning the number of sample values less than or equal to $$x$$.

For these two expressions, you can easily see that the events are the same because they imply each other (and also the events will be the same for the discrete case). Take for example the second statement:

• If there the number of values smaller than or equal to $$x$$ is larger than $$k$$ then at least the first $$k$$ order statistics must be smaller than or equal to $$x$$ and therefore the $$k$$-th order statistic must be smaller than or equal to $$x$$.
• If the $$k$$-th order statistic is smaller than or equal to $$x$$, then so must be at least the $$k-1$$ order statistics with a lower order, and therefore there are at least $$k$$ values smaller than or equal to $$x$$.

The trick to complete the proof for the continuous distribution is that the probability of the different events (with signs $$\leq$$ or $$<$$) are the same. We have

$$P(X_{(k)} \leq x) = P(X_{(k)} < x) \\ P(N(x) \ge k) = P(N^\prime(x) \ge k)$$

The reason is because the probability of the complement of the events is zero. For instance $$\begin{array}{} P(X_{(k)} \leq x) - P(X_{(k)} < x) &=& P(\lbrace X_{(k)} \leq x \rbrace \setminus \lbrace X_{(k)} < x \rbrace)\\ &=& P(X_{(k)} = x) \\ &=& 0 \end{array}$$

### Summarizing

$$\begin{array}{ccc} \rlap{\overbrace{\phantom{P(X_{(k)} \leq x ) = P(X_{(k)} < x)}}^{\substack{\text{different events} \\ \text{but same probability} \\ \text{for continuous distributions}}}} P(X_{(k)} \leq x ) &=& \underbrace{P(X_{(k)} < x) = P(N(x) \ge k)}_{\text{same events}}\\ && \overbrace{P(X_{(k)} \le x) = P(N^\prime(x) \ge k)}_{} &=& P(N(x) \ge k \llap{\underbrace{\phantom{P(N^\prime(x) \ge k) = P(N(x) \ge k}}_{\substack{\text{different events} \\ \text{but same probability} \\ \text{for continuous distributions}}}}) \end{array}$$

So we have $$P(X_{(k)} \le x) = P(N(x) \ge k)$$ not entirely because the events are the same, but because they are events with the same probability, which is because the difference between the events has probability zero for continuous distributions.

• Thanks for the breakdown this makes things clearer; for there $\exists X_k(X_k \le x)$ therefore $N(x)$ represents those values of $x$ greater than them values $k \in X_k$? Mar 9, 2022 at 19:22
• @SextusEmpiricus While not done by the OP, it makes more sense to define $N(x)$ as the number of observations less or equal to $x$ (as is usually done to put it on an equal footing with the definition of the cdf of $X_{(k)}$), in which case the statement of the OP would hold also for discrete random variables. So I think your answer just adds to the confusion here. Mar 10, 2022 at 9:45
• @JarleTufto I have updated it, but i wonder whether it makes it less confusing. When you define N(x) differently then you do not answer the question directly and you still need to show why the equality for the differently defined N(x) also implies the equality for the N(x) from the original question. Mar 14, 2022 at 10:06