2.1 The method

Let $\mathbf \Omega$ be the probability space and let $A_1, \ldots , A_n \subset {\mathbf \Omega }$ be events as in Theorem 1. We define a univariate polynomial of a complex variable $z$ by

(2.1) \begin{equation} p(z)=\int _{{\mathbf \Omega }} \prod _{i=1}^n (1-z [A_i]) \ dx, \end{equation}

where $[A_i]$ is the indicator of $A_i$ defined by (1.3) and $dx$ stands for the integration against the product probability measure in $\mathbf \Omega$ . Then

\begin{equation*}p(1)=\int _{{\mathbf \Omega }} \prod _{i=1}^n (1-[A_i]) \ dx = {\mathbb P}\left ( \bigcap _{i=1}^n \overline {A}_i \right )\!.\end{equation*}

On the other hand, $p(0)=1$ and for the $k$ -th derivative of $p$ at $0$ we have

(2.2) \begin{equation} \begin{aligned} p^{(k)}(0)=&(\!-\!1)^k k! \sum _{1 \leq i_1 \lt i_2 \lt \ldots \lt i_k \leq n} \int _{{\mathbf \Omega }} [A_{i_1}] \cdots [A_{i_k}] \ dx \\= &(\!-\!1)^k k! \sum _{1 \leq i_1 \lt i_2 \lt \ldots \lt i_k \leq n} {\mathbb P}\left (A_{i_1} \cap \ldots \cap A_{i_k}\right ) \\= &(\!-\!1)^k k! \sigma _k. \end{aligned} \end{equation}

Clearly, $\deg p \leq n$ . Thus our goal is to approximate $p(1)$ within relative error $\epsilon$ from $p^{(k)}(0)$ with $k =O(\ln (n/\epsilon ))$ .

2.2 Approximating $p(1)$ from $p^{(k)}(0)$

It turns out that for our purposes it suffices to show that $p(z) \ne 0$ in some neighbourhood of the interval $[0, 1]$ in the complex plane $\mathbb C$ . More precisely, if a polynomial $g(z)$ with $\deg g \leq n$ does not have zeros in a $\delta$ -neighbourhood of $[0, 1] \subset {\mathbb C}$ for some $0 \lt \delta \lt 1$ , then for any $0 \lt \epsilon \lt 1$ , the value of $g(1)$ , up to relative error $\epsilon$ (properly defined if $g(1)$ is a non-zero complex number), is determined by $g^{(k)}(0)$ for $k \leq K=e^{O(1/\delta )}\ln (n/\epsilon )$ . Given the values $g(0), g'(0), \ldots , g^{(K)}(0)$ , the approximation can be computed in $O(K^3)$ time. Furthermore, if $g(0)=1$ then an approximation of $\ln g(1)$ within additive error $\epsilon$ can be written as a polynomial $Q$ in $g^{(k)}(0)$ with $k \leq K$ and $\deg Q \leq K$ . Below we briefly describe how it is done, see Section 2.2 of [Reference Barvinok2] and also [Reference Barvinok3] for detail.

Suppose we have a univariate polynomial $g\,:\, {\mathbb C} \longrightarrow {\mathbb C}$ of degree $n \geq 1$ such that $g(z) \ne 0$ in the disc $|z| \lt \beta$ of some radius $\beta \gt 1$ . We can choose a branch of $f(z)=\ln g(z)$ in the disc and consider the Taylor polynomial of $T_m(z)$ of degree $m \geq 1$ of $f(z)$ , computed at $z=0$ :

\begin{equation*}T_m(z)=f(0)+\sum _{k=1}^m \frac {f^{(k)}(0)}{k!} z^k.\end{equation*}

It turns out that $T_m(1)$ approximates $f(1)$ within an additive error that decreases exponentially with $m$ :

\begin{equation*}\left | f(z) - T_m(z)\right | \ \leq \ \frac {n}{(m+1) \beta ^m (\beta -1)} \quad \text{provided} \quad |z| \leq 1.\end{equation*}

Hence to approximate $f(1)$ by $T_m(1)$ , respectively $g(1)$ by $\exp \left \{T_m(1)\right \}$ , within an additive error, respectively relative error, of $0 \lt \epsilon \lt 1$ , one can choose $m=O_{\beta }\left (\ln (n/\epsilon )\right )$ . Moreover, the derivatives $f^{(k)}(0)$ for $k=1, \ldots , m$ are computed from the derivatives $g^{(k)}(0)$ for $k=1, \ldots , m$ by solving an $m \times m$ system of linear equations with an invertible triangular matrix having $g(0) \ne 0$ on the diagonal.

Suppose now the polynomial $g$ satisfies a weaker condition or being non-zero in a $\delta$ -neighbourhood $U$ of the interval $[0, 1] \subset {\mathbb C}$ . Then one can reduce it to the already considered case of a polynomial that is non-zero in a disc, by replacing $g(z)$ with the composition $h(z)=g(\phi (z))$ , where $\phi \,:\, {\mathbb C} \longrightarrow {\mathbb C}$ is an explicit polynomial such that $\phi (0)=0$ , $\phi (1)=1$ , and $\phi$ maps the disc $\{z\,:\, |z| \leq \beta \}$ with some $\beta =1+ e^{-O(1/\delta )}$ inside $U$ . We have $\deg h=\deg g + \deg \phi$ and since $\phi (0)=0$ , the derivatives $h^{(k)}(0)$ for $k=1, \ldots , m$ are computed from the derivatives $g^{(k)}(0)$ for $k=1, \ldots , m$ .

We show that for some $\delta =O(1/\Delta )$ and $p(z)$ defined by (2.1), we indeed have $p(z) \ne 0$ in the $\delta$ -neighbourhood of $[0, 1] \subset {\mathbb C}$ , assuming the hypothesis of Theorem 1. Assuming that the values of $\sigma _k$ defined by (1.6) can be computed in $n^{O(k)}$ time, the straightforward approach of [Reference Barvinok2] and [Reference Barvinok3] allows one to approximate $p(1)={\mathbb P}\left ( \bigcap _{i=1}^n \overline {A}_i\right )$ within relative error $\epsilon$ in quasi-polynomial $n^{O\left ( \ln (n/\epsilon )\right )}$ time, provided we consider $\Delta$ to be fixed in advance.

If we assume that ${\mathbb P}\left (A_{i_1}\cap \ldots \cap A_{i_k}\right )$ can be computed in $2^{O(k)}$ time, there is a possibility of a speed-up to a genuinely polynomial time, using the approach of Patel and Regts [Reference Patel and Regts25], see also [Reference Liu, Sinclair and Srivastava20]. Here, we very briefly sketch how. With events $A_1, \ldots , A_n$ we associate a hypergraph $H=(V, E)$ with set $V=\{1, \ldots , m\}$ of vertices and set $E \subset 2^V$ of edges, where each event $A_i$ is represented by an edge $J_i \subset \{1, \ldots , m\}$ , consisting of the vertices $j$ such that $A_i$ depends on the coordinate $\xi _j$ . From the technique of [Reference Patel and Regts25] and [Reference Liu, Sinclair and Srivastava20], it follows that to compute (2.2), it suffices to compute the probabilities when the associated hypergraph is connected (note that for an arbitrary hypergraph, the required probability is the product of the probabilities over its connected components). Then it turns out that for a fixed $\Delta$ , and for $k=O_{\Delta } \ln (n/\epsilon )$ there are only $\left (n/\epsilon \right )^{O_{\Delta }(1)}$ connected subhypergraphs of $H$ with $k$ edges. Very recently, a different way to accelerate computations was described in [Reference Bencs and Regts7].

We note that a particular version of the polynomial (2.1) was used earlier by Bencs and Regts [Reference Bencs and G. Regts6] to approximate the volume of a truncated independence polytope of a graph and then by Guo and N [Reference Guo and Vishvajeet11] to approximate the volume of a truncated fractional matching polytope.

Thus our goal is to prove that the polynomial $p(z)$ of (2.1) does not have zeros in a neighbourhood of the interval $[0, 1] \subset {\mathbb C}$ . We let

\begin{equation*}U=\{z \in {\mathbb C}\,:\, \operatorname {dist}(z, [0,1]) \leq \delta \} \quad \text{for} \quad \delta =\frac {1}{6\Delta },\end{equation*}

where $\operatorname {dist}$ is the Euclidean distance in the complex plane ${\mathbb C}={\mathbb R}^2$ .

Let us fix a $z \in U$ and let us define $f_i\,:\, {\mathbf \Omega } \longrightarrow {\mathbb C}$ by

\begin{equation*}f_i=1-z[A_i] \quad \quad \text{for} \quad i=1, \ldots , n. \end{equation*}

Then

\begin{equation*}|f_i(x)| \ \leq \ 1+ \frac {1}{6\Delta } \quad \text{for all} \quad x \in {\mathbf \Omega } \quad \text{and} \quad i=1, \ldots , n \end{equation*}

and

\begin{equation*}{\mathbb P}\left (f_i(x) \ne 1 \right ) \ \lt \ (3 \Delta )^{-3\mu _i}\quad \text{for} \quad i=1, \ldots , n. \end{equation*}

Hence we will be proving the following reformulation of Theorem 1.

Theorem 2. Let $\Omega _1, \ldots , \Omega _m$ be probability spaces and let ${\mathbf \Omega }=\Omega _1 \times \cdots \times \Omega _m$ be their product. Let $f_1, \ldots , f_n\,:\, {\mathbf \Omega } \longrightarrow {\mathbb C}$ be random variables, such that each $f_i$ depends on not more than $r_i$ coordinates and shares a coordinate with at most $\Delta _i$ other functions $f_j$ . Let

\begin{equation*}\Delta =\max \left \{5, \ \Delta _i\,:\, i=1, \ldots , n \right \}\end{equation*}

and let

\begin{equation*}\mu _i =\min \left \{ r_i,\ \Delta _i+1\right \} \quad \text{for} \quad i=1, \ldots , n.\end{equation*}

Suppose further that

(2.3) \begin{equation} \left | f_i(x)\right | \ \leq \ 1 + \frac {1}{6 \Delta } \quad \text{for all} \quad x \in {\mathbf \Omega } \quad \text{and} \quad i=1, \ldots , n \end{equation}

and that

(2.4) \begin{equation} {\mathbb P}\left (f_i (x)\ne 1 \right ) \ \lt \ (3 \Delta )^{-3 \mu _i} \quad \text{for} \quad i=1, \ldots , n. \end{equation}

Then

\begin{equation*}\int _{{\mathbf \Omega }} \prod _{i=1}^n f_i(x) \ d x \ne 0.\end{equation*}

We will prove Theorem 2 by proving a stronger statement by induction on the number $n$ of functions.

We will use some simple geometric facts.

Lemma 3. Let $\Omega$ be a probability space and let $f\,:\, \Omega \longrightarrow {\mathbb C}$ be an integrable random variable. Suppose that for every $\omega \in \Omega$ we have $f(\omega ) \ne 0$ and that for any $\omega _1, \omega _2 \in \Omega$ the angle between $f(\omega _1) \ne 0$ and $f(\omega _2) \ne 0$ , considered as vectors in ${\mathbb R}^2 = {\mathbb C}$ , does not exceed $\theta$ for some $0 \leq \theta \lt 2\pi /3$ . Then

\begin{equation*}\left | \int _{\Omega } f(x) \ dx \right | \ \geq \ \left ( \cos \frac {\theta }{2}\right ) \int _{\Omega } |f(x)| \ dx.\end{equation*}

Lemma 4. Let $v \in {\mathbb C} \setminus \{0\}$ be a non-zero complex number and let $u \in {\mathbb C}$ be another complex number such that

\begin{equation*}|u| \ \leq \ (\sin \theta ) |v| \quad \text{for some} \quad 0 \leq \theta \lt \pi /2.\end{equation*}

Then the angle between $v$ and $v+u$ , considered as vectors in ${\mathbb R}^2 ={\mathbb C}$ , does not exceed $\theta$ .

We will also use the classical Markov inequality in the following situation.

Lemma 5. Let ${\mathbf \Omega }_1$ and ${\mathbf \Omega }_2$ be probability spaces with respective probability measures ${\mathbb P}_1$ and ${\mathbb P}_2$ , let ${\mathbf \Omega }={\mathbf \Omega }_1 \times {\mathbf \Omega }_2$ be their product with probability measure ${\mathbb P}={\mathbb P}_1 \times {\mathbb P}_2$ , and let $A \subset {\mathbf \Omega }$ be an event. For $\omega _1 \in {\mathbf \Omega }_1$ we define an event $A_{\omega _1} \subset {\mathbf \Omega }_2$ by

\begin{equation*}A_{\omega _1}=\left \{ \omega _2 \in {\mathbf \Omega }_2\,:\, \quad (\omega _1, \omega _2) \in A \right \}.\end{equation*}

Then for any $\tau \geq 1$ , we have

\begin{equation*}{\mathbb P}_1 ( \omega _1\,:\, {\mathbb P}_2 \left ( A_{\omega _1}\right ) \ \geq \ \tau {\mathbb P}(A) ) \ \leq \ \tau ^{-1}.\end{equation*}

Proof. We define a random variable $f\,:\, {\mathbf \Omega }_1 \longrightarrow [0, 1]$ by

\begin{equation*}f(\omega _1) = {\mathbb P}_2 (A_{\omega _1} ) \quad \text{for} \quad \omega _1 \in {\mathbf \Omega }_1.\end{equation*}

Then

\begin{equation*}\int _{{\mathbf \Omega }_1} f(\omega _1) \ d \omega _1 = {\mathbb P}(A),\end{equation*}

where $d\omega _1$ stands for the integration against the probability measure ${\mathbb P}_1$ . Since $f$ is non-negative, the result follows by applying the standard Markov inequality to $f$ .

3.1 The induction hypothesis

Let

\begin{equation*}\theta =\frac {1}{\Delta ^2} \quad \text{and} \quad \beta = 1+\frac {1}{\Delta }.\end{equation*}

The induction hypothesis consists of the following two claims.

Claim 1. Let $f_i\,:\, {\mathbf \Omega } \longrightarrow {\mathbb C}$ , $i=1, \ldots , n$ , be random variables as in Theorem 2 . Then

(3.1) \begin{equation} \int _{{\mathbf \Omega }} \prod _{i=1}^n f_i(x) \ dx \ne 0, \quad \int _{{\mathbf \Omega }} \prod _{i=1}^{n-1} f_i(x) \ dx \ne 0 \end{equation}

and the angle between two non-zero complex numbers ( 3.1 ), considered as vectors in ${\mathbb R}^2 = {\mathbb C}$ , does not exceed $\theta$ . For $n=1$ , we agree that the second number is 1, following the general convention that the product of the numbers from an empty set is 1, while the sum of the numbers from an empty set is 0.

Claim 2. Let $f_i\,:\, {\mathbf \Omega } \longrightarrow {\mathbb C}$ , $i=1, \ldots , n$ , be random variables as in Claim 1 , and let $k \leq n$ . Let $\widehat {f}_{n-k+1}, \ldots , \widehat {f}_n\,:\, {\mathbf \Omega } \longrightarrow {\mathbb C}$ be random variables such that

(3.2) \begin{equation} \left | \widehat {f}_i(x)\right | \ \leq \ 1+ \frac {1}{6\Delta } \quad \text{for all} \quad x \in {\mathbf \Omega } \quad \text{and} \quad i=n-k+1, \ldots , n, \end{equation}

and such that each $\widehat {f}_i$ depends on a subset of the coordinates that $f_i$ depends on (note, however, that $\widehat {f}_i$ do not have to satisfy ( 2.4 )). Then

\begin{equation*}\left | \int _{{\mathbf \Omega }} \left (\prod _{i=1}^{n-k} f_i(x) \prod _{i=n-k+1}^n \widehat {f}_i(x) \right )\ dx \right | \ \leq \ \beta ^k \left | \int _{{\mathbf \Omega }} \prod _{i=1}^n f_i(x) \ dx \right |\!. \end{equation*}

In words: if we replace any $k$ of the functions $f_i$ by some functions $\widehat {f}_i$ , depending on the same or a smaller set of coordinates, and still satisfying ( 2.3 ) but not necessarily ( 2.4 ), then the absolute value of the integral of the product can increase by a factor $\beta ^k$ at most.

Remark 1. Suppose that Claim 1 holds, and let $\widehat {f}_n\,:\, {\mathbf \Omega } \longrightarrow {\mathbb C}$ be yet another random variable that depends on a subset of coordinates that $f_n$ depends on and such that

\begin{equation*}\left | \widehat {f}_n(x)\right | \ \leq \ 1 + \frac {1}{6 \Delta } \quad \text{and} \quad {\mathbb P}\left ( \widehat {f}_n(x) \ne 1 \right ) \ \lt \ (3 \Delta )^{-3\mu _n}.\end{equation*}

Then the angle between two complex numbers

(3.3) \begin{equation} \int _{{\mathbf \Omega }} \prod _{i=1}^n f_i(x) \ dx \ne 0 \quad \text{and} \quad \int _{{\mathbf \Omega }} \widehat {f}_n(x) \prod _{i=1}^{n-1} f_i(x) \ dx \ne 0 \end{equation}

does not exceed $\displaystyle 2 \theta =\frac {2}{\Delta ^2}$ . Indeed, both numbers (3.3) form an angle of at most $\theta$ with

\begin{equation*}\int _{{\mathbf \Omega }} \prod _{i=1}^{n-1} f_i(x) \ dx \ne 0.\end{equation*}

Iterating, we conclude that if $\widehat {f}_{n-k+1}, \ldots , \widehat {f}_n\,:\, {\mathbf \Omega } \longrightarrow {\mathbb C}$ are random variables satisfying (3.2) and also such that

\begin{equation*}{\mathbb P}\left (\widehat {f}_i(x) \ne 1 \right ) \ \lt \ (3\Delta )^{-3\mu _i} \quad \text{for} \quad i=n-k+1, \ldots , n\end{equation*}

then the angle between two complex numbers

\begin{equation*}\int _{{\mathbf \Omega }} \prod _{i=1}^n f_i(x) \ dx \ne 0 \quad \text{and} \quad \int _{{\mathbf \Omega }} \left ( \prod _{i=1}^{n-k} f_i(x) \prod _{i=n-k+1}^n \widehat {f}_i(x) \right ) \ dx \ne 0\end{equation*}

does not exceed $2 k \theta = 2k/\Delta ^2$ .

In words: if we replace any $k$ of the random variables $f_i$ satisfying (2.3) and (2.4) by some other random variables on the same or smaller sets of coordinates satisfying the same conditions, then the value of the integral rotates by at most an angle of $2k/\Delta ^2$ .

Finally, for the sake of transparency, in the proof below we replace condition (2.4) by

(3.4) \begin{equation} {\mathbb P}(\,f_i(x) \ne 1) \ \lt \ (\gamma\! \Delta )^{-3\mu _i}, \end{equation}

where $\gamma \gt 0$ is some absolute constant. In the course of the proof it then becomes clear that the induction can be carried through for a sufficiently large $\Delta$ and $\gamma$ , and numerical computations show that we can indeed choose $\Delta \geq 5$ and $\gamma =3$ , as claimed.

3.2 Base $n=1$

In this case, we have just one function $f_1$ in some $r \leq r_1$ coordinates, satisfying (2.3) and (3.4).

If $r=0$ then $\mu _1=0$ . Since $r=0$ , the function $f_1$ is a constant and by (3.4), we must have $f_1 \equiv 1$ , so that

\begin{equation*}\int _{{\mathbf \Omega }} f_1(x) \ dx =1.\end{equation*}

Claim 1 then holds tautologically. By (3.2),

\begin{equation*}\left | \int _{{\mathbf \Omega }} \widehat {f}_1(x) \ d x \right | \ \leq \ 1+ \frac {1}{6\Delta } \ \leq \ \beta \left | \int _{{\mathbf \Omega }} f_1(x) \ dx \right |\!,\end{equation*}

and hence Claim 2 also holds.

Suppose now that $r \geq 1$ , so that $\mu _1=1$ . Let

\begin{equation*}B=\{x \in {\mathbf \Omega }\,: \quad f_1(x)=1 \}, \quad \text{so that} \quad {\mathbb P}(\overline {B}) \ \lt \ (\gamma \!\Delta )^{-3}.\end{equation*}

From (2.3) and (3.4),

(3.5) \begin{equation} \begin{aligned} \left |1-\int _{{\mathbf \Omega }} f_1(x) \ dx \right | = &\left | \int _{\overline {B}} (1- f_1(x)) \ dx \right | \ \leq \ \left (2 + \frac {1}{6 \Delta }\right ) (\gamma \!\Delta )^{-3}. \end{aligned} \end{equation}

It is clear now that for $\gamma \geq 1$ sufficiently large, we have

(3.6) \begin{equation} \left (2 + \frac {1}{6 \Delta }\right ) (\gamma \!\Delta )^{-3} \ \leq \ \sin \frac {1}{\Delta ^2} =\sin \theta , \end{equation}

so that by Lemma 4 the angle between complex numbers $1$ and

\begin{equation*}\int _{{\mathbf \Omega }} f_1(x) \ dx \ne 0\end{equation*}

does not exceed $\theta$ and Claim 1 holds. Numerical computations show that (3.6) indeed holds for $\Delta \geq 3$ and $\gamma \geq 1$ .

To check Claim 2, from (3.5), we have

(3.7) \begin{equation} \left | \int _{{\mathbf \Omega }} f_1(x)\ dx \right | \ \geq \ 1 - \left (2 + \frac {1}{6\Delta }\right ) (\gamma \!\Delta )^{-3}. \end{equation}

From (3.2), we have

(3.8) \begin{equation} \left | \int _{{\mathbf \Omega }} \widehat {f}_1(x) \ d x \right | \ \leq \ 1+ \frac {1}{6\Delta }. \end{equation}

It is now clear that for $\gamma \gt 1$ sufficiently large, we have

(3.9) \begin{equation} 1+ \frac {1}{6\Delta } \ \leq \ \left (1+\frac {1}{\Delta }\right ) \left (1 - \left (2 + \frac {1}{6\Delta }\right ) (\gamma \!\Delta )^{-3}\right )\!, \end{equation}

and so by (3.7) and (3.8),

\begin{equation*}\left | \int _{{\mathbf \Omega }} \widehat {f}_1(x) \ dx \right | \ \leq \ \beta \left | \int _{{\mathbf \Omega }} f_1(x) \ dx \right |\!,\end{equation*}

and Claim 2 holds too.

Numerical computations show that (3.9) holds provided $\Delta \geq 3$ and $\gamma \geq 1$ .

3.3 Induction step $n-1 \Longrightarrow n$ , special case

Thus we assume that $n \geq 2$ . Suppose that $f_n$ depends on the coordinates $\xi _j \ : \ j \in J$ , so $|J| =r \leq r_n$ . If $r=0$ , then $f_n$ is a constant, and by (3.4), we have $f_n \equiv 1$ . Then

\begin{equation*}\prod _{i=1}^n f_i(x)= \prod _{i=1}^{n-1} f_i(x)\end{equation*}

and Claim 1 holds trivially.

Since $\widehat {f}_n$ depends on a subset of the set of coordinates that $f_n$ depends on, the function $\widehat {f}_n$ is also a constant. Then by (3.2)

\begin{align*} \begin{aligned} &\left | \int _{{\mathbf \Omega }} \left ( \prod _{i=1}^{n-k} f_i(x) \prod _{i=n-k+1}^n \widehat {f}_i(x)\right ) \ d x\right | \\ \leq \ &\left (1+ \frac {1}{6\Delta }\right ) \left | \int _{{\mathbf \Omega }} \left ( \prod _{i=1}^{n-k} f_i(x) \prod _{i=n-k+1}^{n-1} \widehat {f}_i(x) \right ) \ d x\right | \\ \leq \ & \beta ^k \left | \int _{{\mathbf \Omega }} \prod _{i=1}^n f_i(x)\ dx \right |, \end{aligned} \end{align*}

where the last inequality follows by the induction hypothesis. Thus Claim 2 also holds.

3.4 Induction step $n-1 \Longrightarrow n$ , preparations for the general case

Without loss of generality, we assume now that the set $\{\xi _j\,:\, j \in J\}$ of coordinates that $f_n$ depends on is non-empty, so $f_n$ depends on some $r$ coordinates, $1 \leq r \leq r_n$ . We form two product probability spaces

(3.10) \begin{equation} {\mathbf \Omega }_1 = \times _{j \in J} \Omega _j \quad \text{and} \quad {\mathbf \Omega }_2 = \times _{j \notin J} \Omega _j, \end{equation}

so that

\begin{equation*}{\mathbf \Omega } = {\mathbf \Omega }_1 \times {\mathbf \Omega }_2.\end{equation*}

We then write $x \in {\mathbf \Omega }$ as $x=(u, y)$ , where $u \in {\mathbf \Omega _1}$ and $y \in {\mathbf \Omega _2}$ . We denote the probability measure in $\mathbf \Omega$ by $\mathbb P$ , in $\mathbf \Omega _1$ by ${\mathbb P}_1$ and in ${\mathbf \Omega }_2$ by ${\mathbb P}_2$ . We use the notation $dx$ , $du$ , and $dy$ to denote integration against $\mathbb P$ , ${\mathbb P}_1$ , and ${\mathbb P}_2$ , respectively.

Let $I \subset \{1, \ldots , n-1\}$ be the set of indices $i$ such that $f_i(x)$ shares a coordinate with $f_n(x)$ . Then

(3.11) \begin{equation} |I| \leq \Delta . \end{equation}

For $i \in I$ and $u \in {\mathbf \Omega _1}$ , we define a random variable $h_i(\!\cdot |u)\,:\, {\mathbf \Omega _2} \longrightarrow {\mathbb C}$ by

\begin{equation*}h_i(y| u) =\, f_i(u, y) \quad \text{for} \quad u \in {\mathbf \Omega _1}, \quad y \in {\mathbf \Omega }_2 \quad \text{and} \quad i \in I. \end{equation*}

Similarly, for $i \in I$ such that $i \gt n-k$ and random variables $\widehat {f}_i$ as in Claim 2, we define a random variable $\widehat {h}_i(\!\cdot | u)\,:\, {\mathbf \Omega _2} \longrightarrow {\mathbb C}$ by

\begin{equation*}\widehat {h}_i(y| u) = \widehat {f}_i(u, y) \quad \text{for} \quad u \in {\mathbf \Omega _1}, \quad y \in {\mathbf \Omega }_2 \quad \text{and} \quad i \in I. \end{equation*}

We note that each $h_i(y|u)$ depends on at most $r_i-1$ coordinates and shares a coordinate with at most $\Delta _i-1$ of functions $h_j(y|u)$ for $j \in I \setminus \{i\}$ and $f_j(y)$ for $j \notin I \cup \{n\}$ .

We further define functions $\phi , \widehat {\phi }\,:\, {\mathbf \Omega _1} \longrightarrow {\mathbb C}$ by

(3.12) \begin{equation} \begin{aligned} &\phi (u) = \int _{{\mathbf \Omega _2}} \left ( \prod _{i \in I} h_i(y|u) \prod _{\substack {i \notin I \\ i \lt n }} f_i(y) \right )\ dy \quad \text{and} \\ &\widehat {\phi }(u) =\int _{{\mathbf \Omega _2}} \left (\prod _{\substack {i \in I \\ i \leq n-k}} h_i(y|u) \prod _{\substack { i \notin I \\ i \leq n-k}} f_i(y) \prod _{\substack { i \in I \\ i \gt n-k}} \widehat {h}_i(y|u) \prod _{\substack { i \notin I \\ n-k \lt i \lt n}} \widehat {f}_i(y) \right )\ dy. \end{aligned} \end{equation}

Then by Fubini’s Theorem

(3.13) \begin{equation} \begin{aligned} & \int _{{\mathbf \Omega }} \prod _{i=1}^n f_i(x) \ dx = \int _{{\mathbf \Omega _1}} f_n(u) \phi (u) \ du \\ &\int _{{\mathbf \Omega }} \prod _{i=1}^{n-1} f_i(x) \ dx = \int _{{\mathbf \Omega _1}} \phi (u) \ du \quad \text{and} \\ &\int _{{\mathbf \Omega }} \left ( \prod _{i=1}^{n-k} f_i(x) \prod _{i=n-k+1}^n \widehat {f}_i(x)\right ) \ dx = \int _{\mathbf \Omega _1} \widehat {f}_n(u) \widehat {\phi }(u) \ du. \end{aligned} \end{equation}

Applying Lemma 5 with $\tau =(\gamma \!\Delta )^3$ , from (3.4), we conclude that

(3.14) \begin{equation} {\mathbb P}_1 \Bigl (u \in {\mathbf \Omega }_1\,:\, {\mathbb P}_2 (h_i(\!\cdot | u) \ne 1 ) \ \geq \ (\gamma \!\Delta )^{-3(\mu _i-1)} \Bigr ) \ \leq \ (\gamma \!\Delta )^{-3} \quad \text{for} \quad i \in I. \end{equation}

We define an event $A \subset {\mathbf \Omega _1}$ by

(3.15) \begin{equation} A=\Bigl \{u \in {\mathbf \Omega }_1\,:\, {\mathbb P}_2(h_i(\!\cdot | u) \ne 1 ) \ \lt \ (\gamma \!\Delta )^{-3(\mu _i-1)} \quad \text{for all} \quad i \in I \Bigr \}. \end{equation}

Then by (3.11), (3.14), and a union bound,

(3.16) \begin{equation} {\mathbb P}_1(\overline {A}) \ \leq \ \frac {1}{\gamma ^3 \Delta ^2}. \end{equation}

We define an event $B \subset {\mathbf \Omega _1}$ by

(3.17) \begin{equation} B= \{u \in {\mathbf \Omega }_1\,:\,\ f_n(u) =1 \}. \end{equation}

so that by (3.4),

(3.18) \begin{equation} {\mathbb P}_1(\overline {B}) \ \lt \ (\gamma \!\Delta )^{-3\mu _n} \ \leq \ (\gamma \!\Delta )^{-3}. \end{equation}

Combining (3.16) and (3.18), we conclude that

(3.19) \begin{equation} \begin{aligned} &{\mathbb P}_1\left ( \overline {A} \cup \overline {B}\right ) \ \leq \ \frac {2}{\gamma ^3 \Delta ^2}, \\ &{\mathbb P}_1 \left (A \cap B \right ) =1 - {\mathbb P}_1\left (\overline {A} \cup \overline {B}\right ) \ \geq \ \frac {\gamma ^3 \Delta ^2 -2}{\gamma ^3 \Delta ^2} \quad \text{and} \\ &\frac {{\mathbb P}_1 \left (\overline {A} \cup \overline {B}\right )}{{\mathbb P}_1\left (A \cup B\right )} \ \leq \ \frac {2}{\gamma ^3 \Delta ^2-2}. \end{aligned} \end{equation}

It is the last inequality that we will be using.

Since each function $h_i(\!\cdot | u)$ depends on at most $r_i-1$ coordinates and shares a coordinate with at most $\Delta _i-1$ functions $h_j(\!\cdot | u)$ for $j \in I \setminus \{i\}$ and $f_j(y)$ for $j \in \{1, \ldots , n-1\}\setminus I$ , by (3.15) for all $u \in A$ the $n-1$ functions $h_i(\!\cdot | u)$ for $i \in I$ and $f_i$ for $i \in \{1, \ldots , n-1\} \setminus I$ satisfy Claim 1 of the induction hypothesis. Let us choose any two $u_1, u_2 \in A$ . When we switch from $u=u_1$ to $u=u_2$ in the integral (3.12) for $\phi$ , we change the $|I|$ functions $h_i(\!\cdot | u_1)$ to $h_i(\!\cdot | u_2)$ , and hence applying Claim 1 of the induction hypothesis and Remark 1, we conclude that $\phi (u_1) \ne 0$ , $\phi (u_2) \ne 0$ and the angle between these two non-zero complex numbers is at most $2 |I| \theta \ \leq \ 2/\Delta$ by (3.11). Hence by Lemma 3,

(3.20) \begin{equation} \left | \int _{A \cap B} \phi (u) \ du \right | \ \geq \ \left (\cos \frac {1}{ \Delta }\right ) \int _{A \cap B} |\phi (u)| \ du \ \gt \ 0. \end{equation}

Let us now pick any $u_1 \in A$ and any $u_2 \in {\mathbf \Omega _1}$ . If in the formula (3.12) for $\phi (u)$ , we switch $u=u_1$ to $u=u_2$ , we switch at most $|I| \leq \Delta$ of functions $h_i(\!\cdot |u_1)$ to $h_i(\!\cdot |u_2)$ . Applying Claim 2 of the induction hypothesis, we conclude that

(3.21) \begin{equation} \begin{aligned} |\phi (u_2)| \ \leq \ &\beta ^{|I|} |\phi (u_1)| \ \leq \ \left (1 + \frac {1}{\Delta }\right )^{\Delta } |\phi (u_1)| \ \leq \ 3 |\phi (u_1)| \\ &\text{for all} \quad u_1 \in A, \ u_2 \in {\mathbf \Omega _1}. \end{aligned} \end{equation}

It follows then that

(3.22) \begin{equation} \int _{\overline {A} \cup \overline {B}} |\phi (u)| \ du \ \leq \ \frac {3 {\mathbb P}_1\left (\overline {A} \cup \overline {B}\right )}{ {\mathbb P}_1\left (A \cap B\right )} \int _{A \cap B} |\phi (u)| \ du. \end{equation}

Indeed, (3.22) obviously holds if ${\mathbb P}_1\left (\overline {A} \cup \overline {B}\right )=0$ , and if ${\mathbb P}_1\left (\overline {A} \cup \overline {B}\right ) \gt 0$ then (3.22) states that the average value of $|\phi (u)|$ on $\overline {A} \cup \overline {B}$ does not exceed three times the average value of $|\phi (u)|$ on $A \cap B$ , which follows from (3.21).

Similarly, to pass from $\phi (u)$ to $\widehat {\phi }(u)$ for $u \in A$ , we change at most $k-1$ in total functions $h_i(\!\cdot | u)$ to $\widehat {h}_i(\!\cdot | u)$ and $f_i(y)$ to $\widehat {f}_i(y)$ in (3.12), and hence applying Claim 2 of the induction hypothesis, we obtain

\begin{equation*}|\widehat {\phi }(u)| \ \leq \ \beta ^{k-1} |\phi (u)| \quad \text{for all} \quad u \in A \end{equation*}

and hence

(3.23) \begin{equation} \int _{A \cap B} |\widehat {\phi }(u)| \ du \ \leq \ \beta ^{k-1} \int _{A \cap B} | \phi (u)|\ du. \end{equation}

On the other hand, in $\widehat {\phi }(u_2)$ for $u_2 \in {\mathbf \Omega _1}$ compared to $\phi (u_1)$ for $u_1 \in A$ , we replace additionally up to $|I| \leq \Delta$ functions $h_i(\!\cdot |u_1)$ by $h_i(\!\cdot | u_2)$ , and so from Claim 2 of the induction hypothesis, we get

(3.24) \begin{equation} \begin{aligned} |\widehat {\phi }(u_2)| \ \leq \ &\beta ^{k-1} \beta ^{|I|} |\phi (u_1)| \ \leq \ 3\beta ^{k-1} |\phi (u_1)| \\ &\text{for all} \quad u_1 \in A, \ u_2 \in {\mathbf \Omega _1}. \end{aligned} \end{equation}

It follows then that

(3.25) \begin{equation} \int _{\overline {A} \cup \overline {B}} |\widehat {\phi }(u)| \ du \ \leq \ 3 \beta ^{k-1} \frac {{\mathbb P}_1(\overline {A} \cup \overline {B})}{{\mathbb P}_1(A \cap B)} \int _{A \cap B} |\phi (u)| \ du. \end{equation}

Indeed, (3.25) obviously holds if ${\mathbb P}_1(\overline {A} \cup \overline {B})=0$ , and if ${\mathbb P}_1\left (\overline {A} \cup \overline {B}\right ) \gt 0$ then (3.25) states that the average value of $|\widehat {\phi }(u)|$ on $\overline {A} \cup \overline {B}$ does not exceed $3\beta ^{k-1}$ times the average value of $|\phi (u)|$ on $A \cap B$ , which follows from (3.24).

3.5 Induction step $n-1 \Longrightarrow n$ , Claim 1

We compare the first two integrals in (3.13) and argue that the bulk of both integrals come from integrating over the event $A \cap B$ , for $A$ defined by (3.15) and $B$ defined by (3.17), where the integrals obviously coincide. Indeed, by (3.17) we have

\begin{equation*}\int _{A \cap B} f_n(u) \phi (u) \ du= \int _{A \cap B} \phi (u)\ du.\end{equation*}

From (3.13),

(3.26) \begin{equation} \left | \int _{\mathbf{\Omega }} \prod _{i=1}^n f_i(x) \ d x - \int _{A \cap B} \phi (u) \ du \right | \ \leq \ \int _{\overline {A} \cup \overline {B}} | f_n(u) \phi (u)| \ du \end{equation}

and similarly,

(3.27) \begin{equation} \left | \int _{\mathbf{\Omega }} \prod _{i=1}^{n-1} f_i(x) \ d x - \int _{A \cap B} \phi (u) \ du \right | \ \leq \ \int _{\overline {A} \cup \overline {B}} | \phi (u)| \ du. \end{equation}

Using (3.2), (3.22), (3.19), and (3.20) in that order, we conclude that

(3.28) \begin{equation} \begin{aligned} &\int _{\overline {A} \cup \overline {B}} |f_n(u) \phi (u)| \ d u \ \leq \ \left (1 + \frac {1}{6 \Delta }\right ) \int _{\overline {A} \cup \overline {B}} |\phi (u)| \ du \\ &\qquad \leq \ \left (1+ \frac {1}{6 \Delta }\right ) \frac { 3{\mathbb P}_1(\overline {A} \cup \overline {B})}{{\mathbb P}_1(A \cap B)} \int _{A \cap B} |\phi (u)| \ du \\ &\qquad \leq \ \left (1 + \frac {1}{6\Delta }\right ) \left (\frac {6}{\gamma ^3 \Delta ^2- 2 }\right ) \int _{A \cap B} |\phi (u)| \ du \\ &\qquad \leq \ \left (1 + \frac {1}{6\Delta }\right ) \left (\frac {6}{\gamma ^3 \Delta ^2 -2}\right )\left ( \frac {1}{\cos (1/\Delta )}\right ) \left | \int _{A \cap B} \phi (u) \ du \right |. \end{aligned} \end{equation}

and, similarly,

(3.29) \begin{equation} \int _{\overline {A} \cup \overline {B}} |\phi (u)| \ d u \ \leq \ \left (\frac {6}{\gamma ^3 \Delta ^2 -2}\right )\left ( \frac {1}{\cos (1/\Delta )}\right ) \left | \int _{A \cap B} \phi (u) \ du \right |. \end{equation}

In view of (3.26)–(3.29) and Lemma 4, to make sure that

\begin{equation*}\int _{{\mathbf \Omega }} \prod _{i=1}^n f_i(x) \ dx \ne 0, \quad \int _{{\mathbf \Omega }} \prod _{i=1}^{n-1} f_i(x) \ dx \ne 0\end{equation*}

and that the angles between each of these two integrals and

\begin{equation*} \int _{A \cap B} \phi (u) \ du \ \ne 0,\end{equation*}

cf. (3.20), do not exceed $\theta /2 = \frac {1}{2 \Delta ^2}$ each, we should make sure that

(3.30) \begin{equation} \left (1 + \frac {1}{6\Delta }\right ) \left (\frac {6}{\gamma ^3 \Delta ^2 -2}\right )\left ( \frac {1}{\cos (1/\Delta )}\right ) \ \leq \ \sin \frac {1}{2 \Delta ^2}. \end{equation}

Clearly, (3.30) holds for a sufficiently large $\gamma \gt 1$ . Computations show that (3.30) holds when $\Delta \geq 3$ and $\gamma \geq 3$ , so Claim 1 holds.

3.6 Induction step $n-1 \Longrightarrow n$ , Claim 2

Here we compare the first and the last integrals in (3.13), and once again argue that the main contribution to both come from the integral over the event $A \cap B$ , for $A$ defined by (3.15) and $B$ defined by (3.17). From (3.13) and (3.2), we have

\begin{equation*} \begin{aligned} &\left | \int _{{\mathbf \Omega }} \left ( \prod _{i=1}^{n-k} f_i(x) \prod _{i=n-k+1}^n \widehat {f}_i(x) \right ) \ dx\right | = \left | \int _{{\mathbf \Omega _1}} \widehat {f}_n(u) \widehat {\phi }(u) \ du \right | \\ &\qquad \leq \ \left (1+ \frac {1}{6\Delta }\right ) \int _{{\mathbf \Omega _1}} |\widehat {\phi }(u)| \ du. \end{aligned} \end{equation*}

As before, the last integral we split into two cases: over $A \cap B$ and over $\overline {A} \cup \overline {B}$ . Applying (3.2), (3.23), (3.25), and (3.20) in that order, we get

\begin{equation*} \begin{aligned} &\left | \int _{{\mathbf \Omega _1}} \widehat {f}_n(u) \widehat {\phi }(u) \ du \right | \ \leq \ \left (1+ \frac {1}{6 \Delta }\right ) \left ( \int _{A \cap B} |\widehat {\phi }(u)| \ du + \int _{\overline {A} \cup \overline {B}} |\widehat {\phi }(u)| \ du \right ) \\ &\qquad \leq \ \beta ^{k-1} \left (1 + \frac {1}{6 \Delta }\right ) \left (1 + 3 \frac {{\mathbb P}_1(\overline {A} \cup \overline {B})}{{\mathbb P}_1(A \cap B)} \right ) \int _{A \cap B} |\phi (u)| \ du \\ &\qquad \leq \ \beta ^{k-1}\left (1+ \frac {1}{6 \Delta }\right ) \left (1+\frac {6}{\gamma ^3 \Delta ^2 -2}\right )\left ( \frac {1}{\cos (1/\Delta )}\right ) \left | \int _{A \cap B} \phi (u) \ du \right |\!. \end{aligned} \end{equation*}

It is clear now that for sufficiently large $\Delta \geq 1$ and $\gamma \gt 1$ , we have

(3.31) \begin{equation} \left (1+ \frac {1}{6 \Delta }\right ) \left (1+\frac {6}{\gamma ^3 \Delta ^2 -2}\right )\left ( \frac {1}{\cos (1/\Delta )}\right ) \ \leq \ 1 + \frac {1}{3 \Delta } \end{equation}

and computations show that (3.31) holds for $\Delta \geq 5$ and $\gamma \geq 3$ . Hence,

(3.32) \begin{equation} \begin{aligned} &\left | \int _{{\mathbf \Omega _1}} \widehat {f}_n(u) \widehat {\phi }(u) \ du \right | \ \leq \ \beta ^{k-1} \left (1 + \frac {1}{3 \Delta }\right ) \left | \int _{A \cap B} \phi (u) \ du\right | \\ &\qquad \text{provided} \quad \Delta \geq 5 \quad \text{and} \quad \gamma \geq 3. \end{aligned} \end{equation}

On the other hand, from (3.13) and (3.28),

\begin{equation*} \begin{aligned} &\left | \int _{\Omega } \prod _{i=1}^n f_i(x) \ dx\right | = \left | \int _{{\mathbf \Omega _1}} f_n(u) \phi (u) \ du \right | \\ &\qquad \geq \ \left | \int _{A \cap B} \phi (u) \ du \right | - \left | \int _{\overline {A} \cup \overline {B}} f_n(u) \phi (u) \ du \right |\\ &\qquad \geq \ \left (1- \left (1 + \frac {1}{6\Delta }\right ) \left (\frac {6}{\gamma ^3 \Delta ^2 -2}\right )\left ( \frac {1}{\cos (1/\Delta )}\right )\right ) \left | \int _{A \cap B} \phi (u) \ du \right |. \end{aligned} \end{equation*}

It is now clear that if $\Delta \geq 1$ and $\gamma \geq 1$ are sufficiently large, then

(3.33) \begin{equation} \left (1+\frac {1}{6\Delta }\right ) \left (\frac {6}{\gamma ^3 \Delta ^2 -2}\right )\left ( \frac {1}{\cos (1/\Delta )}\right ) \ \leq \ \frac {1}{15 \Delta }. \end{equation}

Computations show that (3.33) indeed holds for $\Delta \geq 5$ and $\gamma \geq 3$ , so we have

(3.34) \begin{equation} \begin{aligned} &\left | \int _{{\mathbf \Omega _1}} f_n(u) \phi (u) \ du \right | \ \geq \ \left (1 - \frac {1}{15 \Delta }\right ) \left | \int _{A\cap B} \phi (u) \ du \right | \\ &\quad \text{provided} \quad \Delta \geq 5 \quad \text{and} \quad \gamma \geq 3. \end{aligned} \end{equation}

Since for $\Delta \geq 5$ , we have

\begin{equation*}\left ( 1+ \frac {1}{3 \Delta }\right ) \left (1- \frac {1}{15 \Delta }\right )^{-1} \ \lt \ 1+ \frac {1}{\Delta } =\beta ,\end{equation*}

combining (3.13), (3.32), and (3.34), we conclude that Claim 2 holds.