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.

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 $\langle a,b\rangle$.

share|improve this question
    
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
show 1 more comment

1 Answer 1

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 $$\mathbf{E}[S^p] \neq \sum^{2 m}_{j=0}(\mathbf{E}[z^p_j])$$

share|improve this answer
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.