2
$\begingroup$

I am having trouble of calculating the following probability:

Let $\epsilon_i$, $i=1,\dotsc,N$ be Rademacher random variables. Let $n_i\in \{0, 1, 2, \dotsc, M\}$, $i=1,\dotsc,N$ such that $\sum_{i=1}^Nn_i=M$. I want to calculate $ P\left(\left\{\prod_{i=1}^N\epsilon^{n_i}_i=1\right\}\bigcap\left\{\sum_{i=1}^N\epsilon_i=0\right\}\right). $

Thank you.

  • 0
    Thank you. Yes, $n_i$ are constants and $\epsilon_i$ are i.i.d. But it turns out that counting all combinations that satisfy both conditions is hard part for me...2012-02-11

2 Answers 2

0

I'll give you just my idea. Suppose you have ordered the $\epsilon_i$ such that the first $J$ have $n_i$ odd. This because, when $n_i$ is odd, $\epsilon_i^{n_i}=\epsilon_i$; when $n_i$ is even, $\epsilon_i^{n_i}=1$.

Then $\prod_{i=1}^N\epsilon_i^{n_i}=\prod_{i=1}^J\epsilon_i$.

Now what you want is that in the first $J$ there are an even number of r.v.'s with result -1, in all the N half and half.

So you can have 0 of -1 in the first J (so all +1) and in the N-J remaining N/2 are -1 (of course you need $2N\geq J$) and the number of such combination are $1\cdot C(N-J,N/2)$.

Then you can have 2 of -1 in the first J ($C(J,2)$) and you need N/2-2 of (-1) in the remaining N-J ($C(N-J,N/2-2)$)...

Hope it will help you and it is correct.

I am now reading from the original post that $\sum_{i=1}^Nn_i=M$. I havent read before: I dont think this change the solution; it will maybe open an easier way.

  • 0
    If $\sum_{i=1}^Nn_i=M$, isn't it mean that we can actually represent $J$ in terms of $M$? I mean one can calculate what and how many possibilities for $J$?2012-02-15
0

To begin with, in order to satisfy

$ \prod_{i=1}^{N}\epsilon_{i}^{n_{i}}=1 $

we must have that, for $S=\left\{ i\backepsilon\epsilon_{i}=-1\right\} $, $ \sum_{i\in S}n_{i} $ is even. Since

$ \sum_{i=1}^{N}n_{i}=M $

we want to enumerate the partitions of M in N parts such that a sum of a subset S of its parts is even. The subset is selected at random, selecting a part if $\epsilon_{i}=-1$ and leaving a part out if $\epsilon_{i}=1$. We don't care about the $n_{i}$ for which $\epsilon_{i}=1$.

Furthermore, in order to satisfy

$ \sum_{i=1}^{N}\epsilon_{i}=0 $

there needs to be the same number of $\epsilon_{i}$ such that $\epsilon_{i}=1$ as there are $\epsilon_{i}$ such that $\epsilon_{i}=-1$, so that N has to be even.

Therefore, the partitions of M into N parts (where N is even) have to be such that a subset of its parts is the partition of an even number. If M is odd, this subset must be strict. And we select parts from the partition of M at random, with a coin toss.

The problem is not solved, but it is framed in terms that could lead to a solution.

  • 0
    You are right. I will try to correct this.2012-02-13