1
$\begingroup$

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 $S_B$?

$\endgroup$
  • $\begingroup$ 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? $\endgroup$ – leonbloy Nov 14 '18 at 18:34
  • $\begingroup$ @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. $\endgroup$ – bonna Nov 14 '18 at 18:54
  • $\begingroup$ 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? $\endgroup$ – leonbloy Nov 14 '18 at 18:58
  • $\begingroup$ @leonbloy: Yes. Regarding 1), with theorem 2.4.1 in Arnord, Balakrishnan, Nagaraja (2008) I can calculate $\mathbb{E}[X_{(N-n+1)}| X_{(N-n)} = t] = \int_t^\infty \left[ x \frac{n!}{(n-1)!} \left(\frac{1-F(x)}{1-F(t)}\right)^{n-1} \frac{f(x)}{1-F(t)} \right]dx$ $\endgroup$ – bonna Nov 14 '18 at 19:09
2
+50
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.