"Are you more intelligent than the average of those who are more intelligent than the average?" - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T03:01:36Z http://mathoverflow.net/feeds/question/39408 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/39408/are-you-more-intelligent-than-the-average-of-those-who-are-more-intelligent-than "Are you more intelligent than the average of those who are more intelligent than the average?" Qfwfq 2010-09-20T17:47:53Z 2010-09-20T21:11:16Z <p>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.</p> <p>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! :)</p> <p>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 $f(x)$.</p> <p>Now, define:</p> <p>$M_0:= \int_{-\infty}^{\infty}f(x)dx=1$</p> <p>$\mu_0:=\int_{-\infty}^{\infty}xf(x)dx=0$</p> <p>and, inductively,</p> <p>$M_{n+1}:= \int_{\mu_n}^{\infty}f(x)dx$</p> <p>$\mu_{n+1}:=\frac{1}{M_n}\int_{\mu_n}^{\infty}xf(x)dx$</p> <p>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.</p> <p>So my questions:</p> <ol> <li>How does the sequence $\mu_n$ behave asymptotically? Does it converge? </li> <li>If yes, is there a nice expression for the limit? </li> <li>Is there even a reasonably explicit expression ("closed form") for $\mu_n$ as a function of $n$?</li> </ol> http://mathoverflow.net/questions/39408/are-you-more-intelligent-than-the-average-of-those-who-are-more-intelligent-than/39420#39420 Answer by Nate Eldredge for "Are you more intelligent than the average of those who are more intelligent than the average?" Nate Eldredge 2010-09-20T18:59:19Z 2010-09-20T21:11:16Z <p>We have $\mu_n \uparrow \infty$. Proof: let <code>$$G(y) = \frac{\int_y^\infty x f(x) dx}{\int_y^\infty f(x) dx}$$</code> 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.</p> <p>More generally, this should show that if $X$ is a <s>continuous</s> random variable with essential supremum $M$, and we define $G(y) = E[X | X \ge y]$ for $y &lt; M$, then the iterates $G^n(y) \to M$.</p> http://mathoverflow.net/questions/39408/are-you-more-intelligent-than-the-average-of-those-who-are-more-intelligent-than/39427#39427 Answer by David Speyer for "Are you more intelligent than the average of those who are more intelligent than the average?" David Speyer 2010-09-20T20:08:18Z 2010-09-20T20:55:14Z <p>As in Nate's answer, we are interested in iterating the function <code>$$G(y) := \frac{ \int_{y}^{\infty} x e^{- x^2} dx}{\int_{y}^{\infty} e^{- x^2} }.$$</code></p> <p>The numerator is $e^{-y^2}/2$ (elementary). The denominator is <code>$e^{-y^2}/2 \cdot y^{-1} \left( 1-(1/2) y^{-2} + O(y^{-4}) \right)$</code> (see <a href="http://en.wikipedia.org/wiki/Error_function#Asymptotic_expansion" rel="nofollow">Wikipedia</a>). So <code>$G(y) = y + (1/2) y^{-1} + O(y^{-3})$</code>. </p> <p>Set $z_n = \mu_n^2$. Then <code>$$z_{n+1} = (\mu_n+\mu_n^{-1}/2 + O(\mu_n^{-3}))^2 = \mu_n^2 + 1 + O(\mu_{n}^{-2}) = z_n + 1 + O(z_n^{-1}).$$</code> So $z_n \approx n$ and we see that $\mu_n \to \infty$ like $\sqrt{n}$.</p> <p>I haven't checked the details, but I think you should be able to get something like $\mu_n = n^{1/2} + O(1)$.</p>