Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $r_i, i=1,\ldots,m$ be random variables with $P(r_i=1)=P(r_i=-1)=1/2$. let $b_i, i=1,\ldots,m$ be real numbers. I should calculate $E\left(|\sum_{i=1}^m b_ir_i|^4| \sum_{i=1}^m r_i=0\right)$ using the following hint:

Let $$X=\left\{r\in\{1, -1\}^m \quad | \quad r_i= \begin{cases} 1,&\quad \text{if} \quad i \quad\text{is in }Y\\[4mm] -1,&\quad \text{if}\quad i \quad\text{is not in }Y \end{cases} \right\},$$ where $Y\subset\{1,\ldots,m\}, \operatorname{card}(Y)=m/2$.

For $X$ we put into correspondence the group $\Pi_m$ of all permutations of the set $\{1,\ldots,m\}$ as follows $$ \pi(\cdot)\longleftrightarrow r_i= \begin{cases} 1,&\quad \text{if} \quad \pi(i)\leq \frac m2\\[4mm] -1,&\quad \text{if} \quad \pi(i)>\frac m2 \end{cases}. $$

On the group $\Pi_m$ I consider the normalized counting measure $\mu_m(A)=\operatorname{card}(A)/m!$ for $A\subset \Pi_m$ and the normalized metric $d_m(\pi_1, \pi_2)=\frac 1m \#\{i:\pi_1(i)\neq \pi_2(i), \quad \pi_1, \pi_2 \in \Pi_m\}$.

It is known that $\Pi_m$ is a normal Levy family and for $A_\epsilon=\{\pi\quad| \exists \pi'\in A: d_m(\pi, \pi')\leq \epsilon\}$ we have $$ \inf_{\mu_m(A)\geq 1/2}\mu_m(A_\epsilon)\geq 1-2\exp(-c\epsilon^2m), \quad \text{$c>0$ is a constant}. $$ It is known that in a Levy family we have phenomenon of concentration of measure around one value of a function. So, if $f:\Pi_m\longrightarrow R$ is a function with modulus of continuity $\omega_f(\epsilon)=\sup_{d_m(\pi_1, \pi_2)}|f(\pi_1)-f(\pi_2)|$ and with median $M_f$, then $$ \mu\left(|f-M_f|\leq \omega_f(\epsilon)\right)\geq 2\inf_{\mu_m(A)\geq 1/2}\mu_m(A_\epsilon)-1. $$

But now I am confused with the next step. What can I say about expectation which I need to find?

Thank you for your help.

share|cite|improve this question
Unless I'm missing something, it seems there is a much easier answer without all the mess above. Namely, $\sum r_i = 0$ means $\sum -r_i = 0$ so every $\sum r_ib_i$ has a corresponding $\sum -r_ib_i$ and hence the expected value must be zero. – Thomas Andrews Mar 9 '12 at 17:32
@Thomas: are you missing the absolute value vertical bars or the power of 4? – Henry Mar 9 '12 at 22:32
up vote 1 down vote accepted

when you condition on the $\sum r_i = 0$ you change from $r_i$ being i.i.d. to sampling without replacement from a population of $m/2$ 1's and $m/2$ -1'2. Expand the power. It's not that bad, e.g. by symmetry $E(r_1r_2r_3r_4 \vert \sum r_i = 1) = 0$

share|cite|improve this answer
You meant to condition on $\sum r_i=0$, not $\sum r_i=1$. – Byron Schmuland Mar 31 '12 at 14:22

Your Answer


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.