# Expected value of order statistics $X_{(i+1)}$ conditional on $X_{(i)} < t$

Consider $$N$$ random variables $$X_1, X_2, \ldots, X_N$$ that are i.i.d. distributed according to some cumulative distribution function $$F$$. Assume we receive a signal that says that $$n$$ number of the random variables will have values above some threshold $$t$$ (however we don't know which). To ease notation let $$S_A$$ denote this subset of random variables, and let $$S_B$$ denote the remaining $$N-n$$ variables. Let $$g(S_A) = min(S_A)$$ be the 1st order statistics of $$S_A$$.

1) What is the conditional expected value of $$g(S_A)$$?
$$\mathbb{E}[g(S_A)|t,n]$$

I know that the pdf and expected value corresponding to the 1st order statistics of the entire set, i.e. $$X_{(1)} = min(X_1, X_2, \ldots, X_N)$$, is respectively $$f_{X_{(1)}}(x) = N(1-F(x))^{N-1}f(x)$$ $$\mathbb{E}[X_{(1)}] = N \int_{-\infty}^\infty x \left(1 - F(x)\right)^{N-1} f(x) dx$$ Setting $$N=n$$ in the equation above would not give $$\mathbb{E}[g(S_A)|t,n]$$, since I haven't taken account of the fact that the lowest $$N-n$$ random variables have values below $$t$$. I think I need something like $$\mathbb{E}[g(S_A)|t,n] = \mathbb{E}[X_{(N-n+1)}| X_{(N-n)} < t]$$

Furthermore let $$h(S_B) = h(|S_B|) = h(N-n)$$ be a linear function of the size of $$S_B$$.

2) What is the conditional expected value of $$g(S_A)h(S_B)?$$ $$\mathbb{E}[g(S_A)h(S_B)|t,n]$$ For general functions $$g$$ and $$h$$, $$\mathbb{E}[g(S_A)h(S_B)|t,n] \ne \mathbb{E}[g(S_A)|t,n] \times \mathbb{E}[h(S_B)|t,n]$$, since $$S_A$$ and $$S_B$$ can be considered dependent random variables. But is it the case that $$\mathbb{E}[g(S_A)h(S_B)|t,n] = \mathbb{E}[g(S_A)|t,n] \times \mathbb{E}[h(S_B)|t,n]$$ when $$h$$ is a function of the size of $$S_B$$?

• For given $t$, $h(S_B)=h(N-n)=h(t)$ seems to be a deterministic value, no? I so, then it goes outside the conditional expectation, and we are left with $\mathbb{E}[g(S_A)|t,n]$ Or am I missing something? – leonbloy Nov 14 '18 at 18:34
• @leonbloy $t$ is the threshold, while $N-n$ is the size of $S_B$. But, you might be right that when conditioning on $n$, then $h(N-n)$ can be considered deterministic, thus moved outside the expectation. – bonna Nov 14 '18 at 18:54
• Yes, sorry about the confusion in notation. Anyway, my point applies. If $h$ (conditioned) is deterministic, then the problem is way simpler than stated - actually it reduced to point 1), right? – leonbloy Nov 14 '18 at 18:58
• @leonbloy: Yes. Regarding 1), with theorem 2.4.1 in Arnord, Balakrishnan, Nagaraja (2008) I can calculate $\mathbb{E}[X_{(N-n+1)}| X_{(N-n)} = t] = \int_t^\infty \left[ x \frac{n!}{(n-1)!} \left(\frac{1-F(x)}{1-F(t)}\right)^{n-1} \frac{f(x)}{1-F(t)} \right]dx$ – bonna Nov 14 '18 at 19:09

We are told that exactly $$n$$ rvs have a value greater than $$t$$. It's clear (perhaps not so much?) that the statistic of those $$n$$ variables are only affected by the truncation (but they are still independent). Then, the result for the 1st order statistic applies to the truncated distributions.

Let $$G(x)$$ be cumulative density of the $$n$$ truncated variables, with $$x> t$$. Then $$G(x) = \frac{F(x)-F(t)}{1-F(t)}$$

(Here, and at what follows, we are implicitly assuming conditioning on $$n,t$$).

Letting $$A(x)$$ be the CDF of the minimum, we get

$$A(x)= 1 - (1-G(x))^n=1 - \left(1-\frac{F(x)-F(t)}{1-F(t)}\right)^n=1 - \left(\frac{1-F(x)}{1-F(t)}\right)^n$$

From this you can readily compute the expectation and solve point 1).

$$\mathbb{E}[g(S_A)|t,n] = \int_t^\infty \left[x a(x)\right] dx = \int_t^\infty \left[x n \left(\frac{1-F(x)}{1-F(t)}\right)^{n-1} \frac{f(x)}{1-F(t)}\right] dx$$

The rest is rather trivial, because $$h()$$ conditioned on $$(n,t)$$ is deterministic, hence it goes outside the expectation.