Timeline for Brain-teaser: What is the expected length of an iid sequence that is monotonically increasing when drawn from a uniform [0,1] distribution?
Current License: CC BY-SA 4.0
6 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jun 13, 2018 at 18:07 | comment | added | Sextus Empiricus | Yes, you are right. I used the uniform case to deduce my statement but falsely used $ce^{1-x}-1$ instead of $ce^{-x}-1$ | |
| Jun 13, 2018 at 16:55 | comment | added | Matthew Towers | @MartijnWeterings I think $C=e$, not 1, e.g in the uniform case we get $e e^{-x} -1$ | |
| Jun 12, 2018 at 18:33 | comment | added | Matthew Towers | @amoeba I agree $F(0)$ shouldn't depend on the distribution of the $X$s, but other values of $F$ should: the general solution of that DE is $F=Ce^{-\int \pi}-1$ | |
| Jun 12, 2018 at 18:21 | comment | added | amoeba | +1 Very clever indeed. But since the final answer does not depend on the distribution (as the other answer discusses), this computation should also somehow not depend on $\pi(y)$. Is there any way to see it? CC to @m_t_. | |
| Jun 12, 2018 at 18:11 | comment | added | Matthew Towers | This is very clever. Just to spell it out a bit: your observations are that 1) if $L$ is the length of the longest initial increasing sequence minus one then it's enough to determine $E(L|X_0=x)=:F(x)$ and set $x=0$, and 2) $E(L|X_0=x,X_1=y)$ is zero if $y<x$ and $1+E(L|X_0=y)$ otherwise. Since $E(L|X_0=x)=E(E(L|X_0=x,X_1)) = \int_\mathbb{R} f_X(y) E(L|X_0=x, X_1=y) dy = \int_x^1 f_X(y)(1+E(L|X_0=y))dy= \int_x^1 f_X(y)(1+F(y))dy$ we get $F'(x)=-f_X(x)(1+F(x))$, which in the uniform case can be solved directly. | |
| Jun 12, 2018 at 12:07 | history | answered | jf328 | CC BY-SA 4.0 |