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I am lost with the signs cancellation. Please help me to calculate this inner pruduct.

Let $a$ and $b$ be two $2m$ dimentional vectors such that their entries are Rademacher random variables and such that the sum of the variables for each vector is zero. i.e. $$P(a_i=1)=P(a_i=-1)=P(b_i=1)=P(b_i=-1)=\frac{1}{2}$$ and $$ \sum\limits_{i=0}^{2m}a_i=\sum\limits_{i=0}^{2m}b_i=0 $$ Find the inner product $\langle a,b\rangle$.

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As far as I understand you need to find expcted value of the inner product. Right? – Norbert Jul 18 '12 at 18:39
Yes, my real question is a p-th moment of the inner product. Thats why I am started from the consudering the inner product itself. – AlexK Jul 18 '12 at 18:53
Are this random vector independent? – Norbert Jul 18 '12 at 18:54
The vectors a and b are independetnt. Dependence only between the coordinates of the vector a and between the coordinates of the vector b. – AlexK Jul 18 '12 at 18:56
So you should mention all this in your question – Norbert Jul 18 '12 at 18:56

The inner product of two $2m+1$-dimensional vectors is: $$S = a_0 b_0 + a_1 b_1 + \ldots + a_{2m} b_{2m} = \sum^{2 m}_{j=0} z_j$$ If the entries are independent and identically distributed random variables (the $a_j$ and $b_j$ are all mutually independent) then the sequence of $z_j$ is i.i.d. itself, and the $p$-th moment is trivially $$\mathbf{E}[S^p] = \sum^{2 m}_{j=0}(\mathbf{E}[z^p_j])$$ using the independence of all $z_j$, whose distribution is known since the PDF of a product of independent RVs $a_j, b_j$ (in fact $P(z_j = 1) = 1/2, P(z_j = -1) = 1/2$).

If not, then each $z_j$ would be in general dependent to $z_{k \neq j}$ and in order to express $P(S = s)$ in closed form you would require the joint PDF $P(z_0, z_1, \ldots, z_{2 m})$ since $$\mathbf{E}[S^p] \neq \sum^{2 m}_{j=0}(\mathbf{E}[z^p_j])$$

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