# 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

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}
$\ds{\tt\mbox{So far, I couldn't go any further}}$.