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.

  • $\begingroup$ 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. $\endgroup$ Mar 9, 2012 at 17:32
  • $\begingroup$ @Thomas: are you missing the absolute value vertical bars or the power of 4? $\endgroup$
    – Henry
    Mar 9, 2012 at 22:32

1 Answer 1


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 $E(r_1r_2r_3r_4 \vert \sum r_i = 1) = 0$

  • $\begingroup$ You meant to condition on $\sum r_i=0$, not $\sum r_i=1$. $\endgroup$
    – user940
    Mar 31, 2012 at 14:22

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .