# Probability of a zero product given one previous zero product

Consider two random vectors $v=(v_1,\dots, v_n)$ and $w=(w_1,\dots, w_{n+1})$. Each element of $v$ is independently $\pm1$ with prob $1/2$. Each element of $w$ is independently $1$ with probability $1/4$, $-1$ with probability $1/4$ and $0$ with probability $1/2$.

Let $X$ and $Y$ denote the inner product of $v$ and $(w_1,\dots,w_n)$ and the inner product of $v$ and $(w_2,\dots,w_n, w_{n+1})$ respectively, that is, $$X=\sum_{i=1}^nv_iw_i,\qquad Y=\sum_{i=1}^nv_iw_{i+1}.$$ We know that for large $n$,

$$P(X = 0)=P(Y=0) \sim \frac1{\sqrt{\pi n}}.$$

What is $P(Y=0\mid X=0)$?

• $v$ and $w$ have different dimensions and their inner product is not well-defined. – Julian Wergieluk Dec 9 '14 at 15:49
• @JulianWergieluk Thank you. I just fixed that. – user66307 Dec 9 '14 at 15:49
• Why the restriction to $n$ even? – Did Dec 10 '14 at 9:35
• @Did It just happens to be all I need. I am happy to get rid of it if you think it makes no difference. Thank you for the edits to the question! – user66307 Dec 10 '14 at 9:36
• This looks related to this question. – robjohn Dec 10 '14 at 12:47

$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\dsc}[1]{\displaystyle{\color{red}{#1}}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[1]{\,{\rm Li}_{#1}} \newcommand{\norm}[1]{\left\vert\left\vert\, #1\,\right\vert\right\vert} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ Lets $\ds{{\cal V}_{j}\pars{m}}$ the probability of any $\ds{v_{j}}$ take the values $\ds{m = \pm 1}$ and $\ds{{\cal W}_{j}\pars{m}}$ the probability of any $\ds{w_{j}}$ take the values $\ds{m = -1,0,1}$ such that $${\cal V}_{j}\pars{m} = \half\,,\qquad {\cal W}_{j}\pars{m}= {1 \over 4}\,\delta_{m,-1} + \half\,\delta_{m0} + {1 \over 4}\,\delta_{m1} ={1 \over 4}\,\delta_{m^{2},1} + \half\,\delta_{m0}$$
With the identity $\delta_{a0}=\ds{\oint_{\verts{z}\ =\ 1}{1 \over z^{1 - a}} \,{\dd z \over 2\pi\ic}}$: \begin{align}&\color{#66f}{\large{\rm P}\pars{Y=0\mid X=0}} ={1 \over 2^{n}}\sum_{v_{1}\ =\ \pm 1}\ldots\sum_{v_{n}\ =\ \pm 1} \sum_{w_{1}\ =\ -1}^{1}{\cal W}_{1}\pars{w_{1}}\ldots \sum_{w_{n + 1}\ =\ -1}^{1}{\cal W}_{1}\pars{w_{n}}\times \\&\oint_{\verts{z}\ =\ 1}{1 \over z^{1 - \sum_{j\ =\ 1}^{n}v_{j}w_{j}}} \,{\dd z \over 2\pi\ic} \oint_{\verts{s}\ =\ 1}{1 \over s^{1 - \sum_{k\ =\ 1}^{n}v_{k}w_{k + 1}}} \,{\dd s \over 2\pi\ic} \\[5mm]&={1 \over 2^{n}}\oint_{\verts{z}\ =\ 1}\oint_{\verts{s}\ =\ 1} {1 \over zs}\sum_{v_{1}\ =\ \pm 1}\ldots\sum_{v_{n}\ =\ \pm 1} \sum_{w_{1}\ =\ -1}^{1}{\cal W}_{1}\pars{w_{1}}\ldots \sum_{w_{n + 1}\ =\ -1}^{1}{\cal W}_{1}\pars{w_{n}}\times \\& z^{\sum_{j\ =\ 1}^{n}v_{j}w_{j}}s^{\sum_{k\ =\ 1}^{n}v_{k}w_{k + 1}} \,{\dd z \over 2\pi\ic}\,{\dd s \over 2\pi\ic} \\[5mm]&={1 \over 2^{n}}\oint_{\verts{z}\ =\ 1}\oint_{\verts{s}\ =\ 1} {1 \over zs}\sum_{v_{1}\ =\ \pm 1}\ldots\sum_{v_{n}\ =\ \pm 1} \sum_{w_{1}\ =\ -1}^{1}{\cal W}_{1}\pars{w_{1}}\ldots \sum_{w_{n + 1}\ =\ -1}^{1}{\cal W}_{1}\pars{w_{n}}\times \\&\pars{z^{w_{1}}s^{w_{2}}}^{v_{1}}\ldots\pars{z^{w_{n}}s^{w_{n + 1}}}^{v_{n}} \,{\dd z \over 2\pi\ic}\,{\dd s \over 2\pi\ic} \\[5mm]&={1 \over 2^{n}}\oint_{\verts{z}\ =\ 1}\oint_{\verts{s}\ =\ 1} {1 \over zs}\sum_{w_{1}\ =\ -1}^{1}{\cal W}_{1}\pars{w_{1}}\ldots \sum_{w_{n + 1}\ =\ -1}^{1}{\cal W}_{1}\pars{w_{n}}\times \\&\pars{z^{w_{1}}s^{w_{2}} + z^{-w_{1}}s^{-w_{2}}}\ldots \pars{z^{w_{n}}s^{w_{n + 1}} + z^{-w_{n}}s^{-w_{n + 1}}} \,{\dd z \over 2\pi\ic}\,{\dd s \over 2\pi\ic} \\[1cm]&={1 \over 2^{n}}\oint_{\verts{z}\ =\ 1}\oint_{\verts{s}\ =\ 1} \\[2mm]&{{\mathbb E}\bracks{% \pars{z^{w_{1}}s^{w_{2}} + z^{-w_{1}}s^{-w_{2}}}\ldots \pars{z^{w_{n}}s^{w_{n + 1}} + z^{-w_{n}}s^{-w_{n + 1}}}} \over zs} \,{\dd z \over 2\pi\ic}\,{\dd s \over 2\pi\ic} \end{align}
$\ds{\tt\mbox{So far, I couldn't go any further}}$.