Question

I'm sure that many MO users would answer "Oh, yes, I'm more intelligent than the average intelligence of the population that has an intelligence greater than the (absolute) average". And someone, less modestly: "Even more than the average of those who are more intelligent than the average of those who are more intelligent than the average". And so on.

Anyway, the silly title was created just to attract curiosity a bit: my question is not about the intelligence of MO users, on which I have no doubts! :)

So, take a quantity $X$ that we suppose normally distributed (b.t.w., I have no deep knowledge of probability theory), i.e. it's described by a gaussian distribution that we suppose standardized and call $f(x)$.

Now, define:

$M_0:= \int_{-\infty}^{\infty}f(x)dx=1$

$\mu_0:=\int_{-\infty}^{\infty}xf(x)dx=0$

and, inductively,

$M_{n+1}:= \int_{\mu_n}^{\infty}f(x)dx$

$\mu_{n+1}:=\frac{1}{M_n}\int_{\mu_n}^{\infty}xf(x)dx$

I think this describes the situation in which your $X$ (tallness? Weight?...) has the value $\mu_n$ precisely when you're as $X$ as the average of those who are more $X$ than the average of those who are more $X$ than...... (n times). If not, please explain why.

So my questions:

1. How does the sequence $\mu_n$ behave asymptotically? Does it converge?
2. If yes, is there a nice expression for the limit?
3. Is there even a reasonably explicit expression ("closed form") for $\mu_n$ as a function of $n$?

Answer 1

As in Nate's answer, we are interested in iterating the function $$G(y) := \frac{ \int_{y}^{\infty} x e^{- x^2} dx}{\int_{y}^{\infty} e^{- x^2} }.$$

The numerator is $e^{-y^2}/2$ (elementary). The denominator is $e^{-y^2}/2 \cdot y^{-1} \left( 1-(1/2) y^{-2} + O(y^{-4}) \right)$ (see Wikipedia). So $G(y) = y + (1/2) y^{-1} + O(y^{-3})$.

Set $z_n = \mu_n^2$. Then $$z_{n+1} = (\mu_n+\mu_n^{-1}/2 + O(\mu_n^{-3}))^2 = \mu_n^2 + 1 + O(\mu_{n}^{-2}) = z_n + 1 + O(z_n^{-1}).$$ So $z_n \approx n$ and we see that $\mu_n \to \infty$ like $\sqrt{n}$.

I haven't checked the details, but I think you should be able to get something like $\mu_n = n^{1/2} + O(1)$.

Answer 2

We have $\mu_n \uparrow \infty$. Proof: let $$G(y) = \frac{\int_y^\infty x f(x) dx}{\int_y^\infty f(x) dx}$$ so that $\mu_{n+1} = G(\mu_n)$. Clearly $G$ is a continuous function and $G(y) > y$ for all $y$. But if $\mu_n \to \mu$ for some finite $\mu$ we must have $G(\mu) = \mu$, a contradiction.

More generally, this should show that if $X$ is a continuous random variable with essential supremum $M$, and we define $G(y) = E[X | X \ge y]$ for $y < M$, then the iterates $G^n(y) \to M$.