Probability that the dot product of two binary vectors is k and sum of vectors equals a and b respectively I have two binary vectors $x$, and $y$, each with $n$ elements; Each element of $x$ and $y$ belongs to {-1, 1}, and is drawn from uniform random distribution.
How would I compute the probability of $x \cdot y == k$ and $\sum x == a$ and $\sum y == b$? where $k$, $a$, and $b$ are all intergers ranging from $-n$ to $n$.
I am able to compute the probability of $x \cdot y == k$ from combinatorial analysis, assuming (1) $-n \le k \le n$ and (2) $k \text{ mod } 2 == n \text{ mod } 2$, then $P(x \cdot y == k) = $ $n \choose {(n+k)/2}$ $  2^{-n}$.
From here, I tried to derive the probability for $x \cdot y == k$ and $\sum x == a$ and $\sum y == b$, but I was stuck due to coupled relations between $x \cdot y$ and $\sum x$ and $\sum y$.
 A: Let's put
$$
\left\{ \matrix{
  {\bf u} = \left( {1,1, \cdots ,1} \right)^T \quad {\bf u}^T {\bf u} = n \hfill \cr 
  {\bf z} = {{{\bf x} + {\bf u}} \over 2}\quad {\bf w}
 = {{{\bf y} + {\bf u}} \over 2} \hfill \cr}  \right.
$$
so that $\bf z , \bf w$ are actual binary vectors.
Then
$$
\eqalign{
  & S_z  = {\bf u}^T {\bf z} = {1 \over 2}S_x  + {n \over 2}\quad S_w  = {1 \over 2}S_y  + {n \over 2}  \cr 
  & S_{zw}  = {\bf z} \cdot {\bf w} = {\bf z}^T {\bf w}
 = {1 \over 4}\left( {{\bf x}^T  + {\bf u}^T } \right)\left( {{\bf y} + {\bf u}} \right) =   \cr 
  &  = {1 \over 4}\left( {{\bf x}^T {\bf y} + {\bf x}^T {\bf u} + {\bf u}^T {\bf y}
 + {\bf u}^T {\bf u}} \right) =   \cr 
  &  = {1 \over 4}\left( {{\bf x}^T {\bf y} + S_x  + S_y  + n} \right)  \cr 
  & S_{zw}  = {\bf z}^T {\bf w} = {\rm N}{\rm .}\,{\rm of}\,{\rm coincident}\,{\rm ones} \cr} 
$$
where $S_z, S_w$ are the respective number of ones and are integers.
So given two binary strings, with $S_z$ and $S_w$ ones, the number of strings with
$S_{zw}$ coincident ones is readily computed.
We can rearrange $\bf z$ as to have all the $S_z$  ones on the same side (e.g. at the lower indices)
and then consider the combinations of  $\bf w$'s  with a number $0 \le S_{zw} \le S_w$ of coincident ones.
There are totally
$$\binom{n}{S_w}$$
different $\bf w$'s with $S_w$ ones.
Of these, those which have $S_{zw}$ ones in the first $S_z$ places and the remaining
$n-S_{zw}$ in the last $n-S_z$ places are
$$
N(S_{zw} \;\left| {\,n,S_z ,S_w } \right.)
 = \left( \matrix{  S_z  \cr   S_{zw}  \cr}  \right)
\left( \matrix{  n - S_z  \cr  S_w  - S_{zw}  \cr}  \right)
$$
and the probability will be
$$
p(S_{zw} \;\left| {\,n,S_z ,S_w } \right.)
 = {{\left( \matrix{  S_z  \cr   S_{zw}  \cr}  \right)
\left( \matrix{  n - S_z  \cr   S_w  - S_{zw}  \cr}  \right)}
 \over {\left( \matrix{  n \cr   S_w  \cr}  \right)}}
$$
wihich is a Hypergeometric distribution
