Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Good day to everyone.

While solving some problem of studying character I obtained some statement to prove, which is like following (this is my internal interest to prove it rigorously).

Assume that $X_1, \ldots, X_k, X_{k+1}, \ldots, X_\ell, X_{\ell + 1}, \ldots, X_n$ are i.i.d. random variables of unknown distribution ($1 \le k < \ell < n$). See also EDIT1 for crucial assumtions.

What is really intuitively understandable, but pretty unclear how to write down, is the next statement: prove that the events $$ \max(X_1, \ldots, X_k) \le \max(X_{k+1}, \ldots, X_\ell) $$ and $$ \max(X_1, \ldots, X_{\ell}) \le \max(X_{\ell+1}, \ldots, X_n) $$ are independent.

Qualitatively and intuitively it's ok: the second one "forgets" about the configuration of the maximums in $X_1, \ldots, X_\ell$.

It's also pretty clear how to depict it "with picture": the algebra of events of the probability subspace of the first $\ell$ coordinates generates the cylindric algebra, which is subalgebra of events of the whole ($n$-coordinate) probability space.

Somehow I have the feeling that conditioning on such cylindric subalgebra should not change the distribution, but I have a lack of imagination (and knowledge of the results) now, to get something out of that.

Can anybody explain that? Thanks.

EDIT1 Due to the comments I got a clear understanding that this is not true in general, so assume that the distribution is continuous, according to the comments. Is it possible to say anything in this case?

share|improve this question
1  
Have you tried to see whether it's the case for $k=1$, $l=2$ and $n=3$? –  Davide Giraudo Sep 30 '13 at 16:22
2  
The crucial missing hypothesis is that the common distribution is continuous. The counterexamples (in my answer, say) are based on ties happening with positive probability. –  Did Sep 30 '13 at 17:10
    
@Did modified the setting. –  gron Sep 30 '13 at 17:25
add comment

2 Answers

up vote 4 down vote accepted

Not sure the result holds...

Try $(k,\ell,n)=(1,2,3)$ then the events are $A=[X_1\leqslant X_2]$ and $B=[\max(X_1,X_2)\leqslant X_3]$. Assume that the random variables $X_k$ are Bernoulli with parameter $p$, that is, $P[X_k=1]=p$ and $P[X_k=0]=1-p$.

Then $P[A]=(1-p)+p^2$, $P[B]=(1-p)^2+(1-(1-p)^2)p$, and $A\cap B=[X_1\leqslant X_2\leqslant X_3]$ hence $P[A\cap B]=(1-p)((1-p)+p^2)+p^3$. Since $P[A]$, $P[B]$ and $P[A\cap B]$ are polynomials in $p$ of degree $2$, $3$ and $2$ respectively, the identity $P[A\cap B]=P[A]\cdot P[B]$ cannot hold for every $p$ in $[0,1]$.

To be specific, assume that $p=\frac12$, then $(X_1,X_2,X_3)$ is uniformly distributed on the cube $\{0,1\}^3$, $P[A\cap B]=\frac12$ and $P[A]\cdot P[B]=\frac34\cdot\frac58\ne\frac12$.

However, if the common distribution is continuous, then the result holds when $(k,\ell,n)=(1,2,3)$ since, by exchangeability of the random vector $(X_1,X_2,X_3)$ and because ties have probability zero, then $P[A]=\frac12$, $P[B]=\frac13$ and $P[A\cap B]=\frac16$.

In the general continuous setting, write $A=[M_1\lt M_2]$, $B=[\max(M_1,M_2)\lt M_3]$ and $A\cap B=[M_1\lt M_2\lt M_3]$, with hopefully obvious notations. Then $A$ means that the record amongst the $\ell$ first random variables happens in the $\ell-k$ last ones. Thus, $P[A]=(\ell-k)/\ell$. Likewise $P[B]=(n-\ell)/n$. To study $A\cap B$, condition on $B$ and draw one by one the ranks of the sample $(X_k)$ in decreasing order. The first rank is in the $(\ell,n]$ range, by hypothesis. The first rank not in the $(\ell,n]$ range is uniform in $[1,\ell]$, by exchangeability. Note that $A$ happens if and only if this rank is in $(k,\ell]$ hence $P[A\mid B]=(\ell-k)/\ell$. QED.

share|improve this answer
    
Thanks and please see the addition edit. I'm starting to get confused :( –  gron Sep 30 '13 at 16:47
    
Modified the last paragraph. –  Did Sep 30 '13 at 17:09
    
Well, having assumed the distribution to be continuous how to build the proving of their independency then? –  DimG Sep 30 '13 at 17:17
1  
@DimG Are you in a hurry? :-) See edited answer. –  Did Sep 30 '13 at 17:27
    
Yes @Did it's great and it's the solution, thanks! –  gron Sep 30 '13 at 17:46
show 1 more comment

Take $X,Y,Z$ i.i.d with Bernoulli(0.5) distribution. Then $P(X \leq Y) = \frac{3}{4}$ and $P(\max(X,Y)=1) = \frac{3}{4}$ so $P(\max(X,Y) \leq Z) = \frac{5}{8}$. But $$ P(X\leq Y, \max(X,Y) \leq Z) = P(X \leq Y, Y \leq Z) = \frac{1}{2} \neq \frac{15}{32}. $$

share|improve this answer
1  
Nice example... :-) –  Did Sep 30 '13 at 16:33
    
Thanks and please see the addition edit. I'm starting to get confused :( –  gron Sep 30 '13 at 16:47
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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