Proof. This follows immediately from combining (2.5) with (2.3).

Proof. First, let $p>M^{-d}$ . Since $M^D=M^dp$ , we infer from Theorem 3.1 that for any $n\in{\mathbb N}$ , ${\mathbb E} Z_n^d=p^{-n}{\mathbb E} V_d(F_n)=1$ and

$$ \mathrm{Var}(Z^d_n)=\mathrm{Var}(p^{-n} V_d(F_n))=p^{-2n} \mathrm{Var}(V_d(F_n))=\frac{1-p}{M^dp-1} \Big(1-\Big(M^dp\Big)^{-n}\Big). $$

Moreover, by Theorem 2.1, we have $\overline{\mathcal V}_d(F)=1$ , and thus we conclude from Corollary 6.1 and (2.3) that ${\mathbb E} Z^d_\infty=\overline{\mathcal V}_d(F) {\mathbb E} W_\infty=1$ and $\mathrm{Var}(Z^d_\infty)=\frac{1-p}{M^dp-1}$ . From this the asserted convergence of the expectation and variance are obvious.

If $p\leq M^{-d}$ , then $F=\emptyset$ almost surely. Moreover, it is clear from the definition of F that $F(\omega)=\emptyset$ , for some $\omega\in\Omega$ , if and only if there is some $m=m(\omega)\in{\mathbb N}$ such that $F_m(\omega)=\emptyset$ , and in this case one has $Z^d_n(\omega)\to 0$ as $n\to\infty$ . Hence $Z_n^d\to 0$ almost surely as $n\to \infty$ , as stated. Now the last assertions are obvious from (3.1).

Proof. Since $M^{d-1-D}=(Mp)^{-1}$ , we infer from Theorem 3.2 that

$$ {\mathbb E} Z^{d-1}_n=(Mp)^{-n}{\mathbb E} V_{d-1}(F_n)=\bar c_1 \bigg(1+\frac{M-1}{1-p}\bigg(\frac pM\bigg)^{n+1}\bigg) $$

for any $n\in{\mathbb N}$ . As noted above, Theorem 2.1 and Corollary 6.1 imply that ${\mathbb E} Z^{d-1}_n \to {\mathbb E} Z^{d-1}_\infty$ , as $n\to\infty$ , for any $p>M^{-d}$ , and therefore we conclude that ${\mathbb E} Z^{d-1}_\infty=\bar c_1$ (which can also be deduced directly from Theorem 2.1). The stated formula for the difference ${\mathbb E} Z^{d-1}_\infty - {\mathbb E} Z^{d-1}_n$ is now obvious. Moreover, from (3.3) we infer that

\begin{align*} \mathrm{Var}(Z^{d-1}_n) &=(Mp)^{-2n} \mathrm{Var}(V_{d-1}(F_n)) =\bar c_2 + \begin{cases} O\left(\left(\frac pM\right)^n\right) & \text{if } p^2>1/M^{d-1},\\[5pt] O\left(\left(\frac pM\right)^n \cdot n\right) & \text{if } p^2=1/M^{d-1},\\[5pt] O\left(\left(M^{d}p\right)^{-n}\right) & \text{if } p^2<1/M^{d-1}, \end{cases} \end{align*}

which implies in particular that $\mathrm{Var}(Z^{d-1}_n)\to \bar c_2$ as $n\to\infty$ . From Corollary 6.1 we see that $\mathrm{Var}(Z^{d-1}_\infty)=(\overline{\mathcal V}_k(F))^2\cdot \mathrm{Var}(W_\infty)=\bar c_2$ . Hence $\mathrm{Var}(Z^{d-1}_n)\to \mathrm{Var}(Z^{d-1}_\infty)$ as $n\to\infty$ , and the stated order of convergence for the difference follows at once.

Proof. With the intention of applying Theorem 4.1, we consider the process $\tilde{Y}$ defined by $\tilde{Y}_t\;:\!=\;Y_n$ for $t\in I_n\;:\!=\;[n\log \Lambda, (n+1)\log \Lambda)$ , $n\in{\mathbb N}_0$ , and $\tilde{Y}_t\;:\!=\;0$ for $t< 0$ . Note that for any $t\in I_n$ ,

$$ e^{-Dt} |\tilde{Y}_t|\leq e^{-Dn\log \Lambda} |Y_n|=\Lambda^{-Dn} |Y_n|. $$

Define the function $h\;:\;[0,\infty)\to(0,\infty)$ by $h(t)\;:\!=\;h_n$ for $t\in I_n$ , $n\in{\mathbb N}$ . The assumed properties of the sequence $(h_n)$ imply that h is non-increasing and integrable. Moreover, for any $t\geq 0$ we have

\begin{align*} \frac{e^{-Dt}|\tilde{Y}_t|}{h(t)}\leq \frac{\Lambda^{-Dn}|{Y}_n|}{h_n},\end{align*}

and thus (for any realization $\omega\in\Omega$ )

\begin{align*} \sup_{t\geq 0}\frac{e^{-Dt}|\tilde{Y}_t|}{h(t)}=\sup_{n\in{\mathbb N}_0}\sup_{t\in I_n}\frac{e^{-Dt}|\tilde{Y}_t|}{h(t)}\leq \sup_{n\in{\mathbb N}_0}\frac{\Lambda^{-Dn}|{Y}_n|}{h_n}.\end{align*}

Since, by assumption, the random variable on the right-hand side has a finite expectation, so has the one on the left-hand side. Thus, the process $\tilde{Y}$ satisfies the condition (4.11) in Theorem 4.1, and we can apply this theorem to $\tilde{Y}$ . (Note also that $\tilde Y$ vanishes on $(\!-\!\infty,0)$ and is continuous almost everywhere with probability 1.) Recalling that $\lambda=\log \Lambda$ , we conclude in particular that (for $s=0$ ) the rescaled sum

$$e^{-Dn\lambda} \sum_{\sigma\in\mathcal{T}_*} \tilde{Y}^{(\sigma)}_{n\lambda-S_\sigma}=\Lambda^{-Dn} \sum_{\sigma\in\mathcal{T}_*} \tilde{Y}^{(\sigma)}_{(n-|\sigma|)\log \Lambda}=\Lambda^{-Dn} \sum_{\sigma\in\mathcal{T}_*} Y^{(\sigma)}_{n-|\sigma|}$$

converges almost surely, as $n\to \infty$ , to the random variable

\begin{align*}\frac{\lambda W_\infty}{m(D)} \sum_{n=0}^\infty e^{-Dn\lambda}{\mathbb E}[\tilde{Y}_{n\lambda}]= W_\infty \sum_{n=0}^\infty \Lambda^{-Dn}{\mathbb E}[Y_{n}]. \end{align*}

Proof of Theorem 2.2. Fix $k\in\{0,\ldots,d\}$ . In order to express the functionals $Z^k_n$ in the form of the left-hand side of (4.13), we set $\widehat{Z}_n\;:\!=\;M^{nk} V_k(F_n)$ . We will also write $\widehat{Z}^{(\sigma)}_n\;:\!=\;M^{nk} V_k(F^{(\sigma)}_n)$ for the corresponding functionals of the shifted random set $F^{(\sigma)}$ , $\sigma\in\mathcal{T}_*$ . (To simplify the notation, we suppress the dependence on k here.) Observe that, by the inclusion–exclusion principle, we can decompose the random variables $\widehat{Z}_n$ for any $n\in{\mathbb N}$ as follows:

\begin{align*} \widehat{Z}_n&=M^{nk} V_k(F_n)=M^{nk} V_k\Bigg(\bigcup_{j\in\mathcal{T}_1} {F}_n^{[j]}\Bigg)\\[5pt] &= M^{nk}\sum_{j\in\mathcal{T}_1} V_k(F_n^{[j]})+M^{nk}\sum_{T\subset{\mathcal{T}}_1, |T|\geq 2} (\!-\!1)^{|T|-1} V_k\Bigg(\bigcap_{j\in T} F_n^{[j]}\Bigg).\end{align*}

For any $n\in{\mathbb N}$ , we define $Y_n$ to be the second of the two summands above, and we set $Y_0\;:\!=\;0$ (which is consistent). Using the relation $V_k(F_n^{[j]})=M^{-k}V_k(F^{(j)}_{n-1})$ (recall that $F^{[j]}_n=\phi_j(F^{(j)}_{n-1})$ , where $\phi_j$ has contraction ratio $1/M$ , and that $V_k$ is homogeneous of order k), we infer that

\begin{align*} \widehat{Z}_n&=\sum_{j\in\mathcal{T}_1} M^{(n-1)k} V_k(F^{(j)}_{n-1})+ Y_n=\sum_{j\in\mathcal{T}_1} \widehat{Z}^{(j)}_{n-1}+ Y_n.\end{align*}

Now each of the variables $\widehat{Z}^{(j)}_{n-1}$ , $j\in\mathcal{T}_1$ , can be decomposed in the very same manner, which yields $\widehat{Z}^{(j)}_{n-1}=\sum_{\sigma\in\mathcal{T}_2, \sigma|1=j} \widehat{Z}^{(\sigma)}_{n-2} + Y_{n-1}^{(j)}$ for $n\in{\mathbb N}_{\geq 2}$ . For $n=1$ we simply get $\widehat{Z}^{(j)}_{n-1}=V_k(F^{(j)}_0)=V_k(J)=q_{d,k}$ for any $j\in\mathcal{T}_1$ . Iterating this procedure, we end up with one term $Y^{(\sigma)}_\ell$ for each finite word $\sigma\in\mathcal{T}_*$ of length $|\sigma|$ up to $n-1$ , i.e. we obtain

(5.2) \begin{align} \widehat{Z}_n&=\sum_{\sigma\in\mathcal{T}_n} \widehat{Z}^{(\sigma)}_0+\sum_{\sigma\in\mathcal{T}_*, |\sigma|<n} Y^{(\sigma)}_{n-|\sigma|}.\end{align}

Here the first summand simplifies to $q_{d,k}|\mathcal{T}_n|=q_{d,k} N_n$ , since, for each $\sigma\in\mathcal{T}_n$ , $F^{(\sigma)}_0$ equals the unit cube J and so $\widehat{Z}^{(\sigma)}_0= V_k(J)=q_{d,k}$ .

We wish to study the limit of $Z^k_n=M^{-Dn} \widehat{Z}_n$ as $n\to\infty$ , which we can do by studying separately the limits of the two sequences given by the two summands in (5.2). The limiting behavior of the first resulting sequence $(M^{-Dn} q_{d,k}N_n)$ is obvious. Since $M^{Dn}=(M^{d}p)^{n}={\mathbb E} N_n$ , we conclude that

(5.3) \begin{align} M^{-Dn} q_{d,k}N_n=q_{d,k} W_n \to q_{d,k} W_\infty \quad \text{ almost surely, as } n\to\infty.\end{align}

Thus the first summand shows the desired factorization, with the first factor being deterministic and the second one being $W_\infty$ . The second sequence is of a form which allows us to employ Proposition 4.1. In order to do so we need to verify that our process $(Y_n)_n$ satisfies the assumptions of this proposition. More precisely, we need to verify that there is a non-increasing and summable sequence $(h_n)_{n\in{\mathbb N}}$ such that

(5.4) \begin{align} {\mathbb E}\Bigg[\sup_{n\in{\mathbb N}} \frac{M^{-Dn}|Y_n|}{h_n}\Bigg]<\infty.\end{align}

This can be derived from the following result, Proposition 5.1, whose proof is postponed to the end of the section.

Proposition 5.1. Let F be a fractal percolation in $[0,1]^d$ with parameters $M\in{\mathbb N}_{\geq 2}$ and $p\in(0,1]$ . Then there exists a non-increasing and summable sequence $(h_n)$ such that, for any subset $T\subset\{1,2,\ldots, M^d\}$ with $|T|\geq 2$ ,

(5.5) \begin{align} {\mathbb E}\Bigg[\sup_{n\in{\mathbb N}} \frac{M^{(k-D)n}}{h_n}\Bigg|V_k\Bigg(\bigcap_{j\in T} F^{[j]}_n\Bigg)\Bigg|\Bigg]<\infty.\end{align}

Let us now show that this statement implies the condition (5.4). First observe that, since $\mathcal{T}_1$ is a random subset of $\Sigma=\{1,\ldots,M^d\}$ , we have for any $n\in{\mathbb N}$

\begin{align*} |Y_n|&=M^{nk}\Bigg|\sum_{T\subset{\mathcal{T}}_1, |T|\geq 2} (\!-\!1)^{|T|-1} V_k\Bigg(\bigcap_{j\in T} F_n^{[j]}\Bigg)\Bigg|\\[5pt] &\leq M^{nk}\sum_{T\subset{\mathcal{T}}_1, |T|\geq 2} \Bigg|V_k\Bigg(\bigcap_{j\in T} F_n^{[j]}\Bigg)\Bigg| \leq\sum_{T\subset \Sigma, |T|\geq 2} M^{nk} \Bigg|V_k\Bigg(\bigcap_{j\in T} F_n^{[j]}\Bigg)\Bigg|.\end{align*}

Therefore, Proposition 5.1 implies that there is some non-increasing and summable sequence $(h_n)$ such that

\begin{align*} {\mathbb E}\Bigg[\sup_{n\in{\mathbb N}} \frac{M^{-Dn}}{h_n}|Y_n|\Bigg] &\leq {\mathbb E}\Bigg[\sup_{n\in{\mathbb N}} \frac{M^{-Dn}}{h_n} \sum_{T\subset \Sigma, |T|\geq 2} M^{nk} \Bigg|V_k\Bigg(\bigcap_{j\in T} F_n^{[j]}\Bigg)\Bigg|\Bigg]\\[5pt] &\leq {\mathbb E}\Bigg[\sum_{T\subset \Sigma, |T|\geq 2} \sup_{n\in{\mathbb N}} \frac{M^{(k-D)n}}{h_n} \Bigg|V_k\Bigg(\bigcap_{j\in T} F_n^{[j]}\Bigg)\Bigg|\Bigg]\\[5pt] &\leq \sum_{T\subset \Sigma, |T|\geq 2} {\mathbb E}\Bigg[\sup_{n\in{\mathbb N}} \frac{M^{(k-D)n}}{h_n} \Bigg|V_k\Bigg(\bigcap_{j\in T} F_n^{[j]}\Bigg)\Bigg|\Bigg]<\infty,\end{align*}

verifying the condition (5.4). Hence we can apply Proposition 4.1 to the sequence of the second summands in (5.2). We infer that, as $n\to\infty$ ,

(5.6) \begin{align} M^{-Dn} \sum_{\sigma\in\mathcal{T}_*} Y^{(\sigma)}_{n-|\sigma|}\to {W_\infty} \cdot \sum_{n=1}^\infty M^{-Dn}{\mathbb E}[Y_{n}],\end{align}

where, for any $n\in{\mathbb N}$ ,

\begin{align*} {\mathbb E}[Y_{n}]&=M^{kn}{\mathbb E}\Bigg[\sum_{T\subset{\mathcal{T}}_1, |T|\geq 2} (\!-\!1)^{|T|-1} V_k\Bigg(\bigcap_{j\in T} F_n^{[j]}\Bigg)\Bigg]\\[5pt] &=M^{kn}{\mathbb E}\Bigg[\sum_{T\subset\Sigma, |T|\geq 2} (\!-\!1)^{|T|-1} V_k\Bigg(\bigcap_{j\in T} F_n^{j}\Bigg)\Bigg].\end{align*}

Here the last equality is due to (5.1). Using this last representation, we get

\begin{align*} \sum_{n=1}^\infty M^{-Dn}{\mathbb E}[Y_{n}] &=\sum_{n=1}^\infty M^{(k-D)n}\sum_{T\subset\Sigma, |T|\geq 2} (\!-\!1)^{|T|-1} {\mathbb E}\Bigg[V_k\Bigg(\bigcap_{j\in T} F_n^{j}\Bigg)\Bigg]\\[5pt] &= \sum_{T\subset\Sigma, |T|\geq 2} (\!-\!1)^{|T|-1} \sum_{n=1}^\infty M^{(k-D)n} {\mathbb E}\Bigg[V_k\Bigg(\bigcap_{j\in T} F_n^{j}\Bigg)\Bigg],\end{align*}

where the interchange of the summations in the last equality is justified as long as all the (finitely many) series in the last expression converge. But this convergence is shown in [Reference Klatt and Winter18, Proposition 5.1] for any $p\in(0,1]$ . Recall that this is an expression for the limit of the second summands in (5.2). Combining it with the limit of the first summands in (5.3), we see by comparison with (2.4) that

\begin{align*} Z^k_\infty= \lim_{n\to\infty} Z_n^k= \overline{\mathcal V}_k(F)\cdot W_\infty \text{ almost surely},\end{align*}

as asserted in Theorem 2.2. This completes the proof.

Proof of Proposition 5.1. First observe that, for any index set $T\subset\Sigma$ , the set $C\;:\!=\;\bigcap_{j\in T} J_j$ is a cube with side length $1/M$ and some dimension $u\leq d-1$ . If $u=0$ the estimate in (5.5) is obviously satisfied. So assume $u\geq 1$ . For any $n\in{\mathbb N}$ , the set $\bigcap_{j\in T} F_n^{[j]}$ is a subset of C and thus at most u-dimensional. Note that in each of the sets $F_n^{[j]}$ , only those level-n cubes which intersect C are relevant for the intersection $\bigcap_{j\in T} F_n^{[j]}$ . This means we can also model the independent random sets $F^{[j]}$ , $j\in T$ , by independent u-dimensional fractal percolations $K^{\{j\}}$ , $j\in T$ , constructed on the cube C (with the same parameters p and M as F). That is, we have $F^{[j]}\cap C\overset{d}{=} K^{\{j\}}$ for each $j\in T$ , and therefore $F_n^{[j]}\cap C\overset{d}{=} K_n^{\{j\}}$ for each $n\in{\mathbb N}$ . In particular, this implies $\bigcap_{j\in T} F_n^{[j]}\overset{d}{=} \bigcap_{j\in T} K_n^{\{j\}}$ . Choose some index $j_1$ from T and denote by $N^u_n$ the number of level-n cubes contained in $K_n^{\{j_1\}}$ . Recall that the sequence $(N^u_n)$ forms a Galton–Watson process with binomial offspring distribution $\text{Bin}(M^u, p)$ . Now we can apply [Reference Klatt and Winter18, Lemma 7.2], according to which there is some constant $c_{u,k}$ (independent of n) such that

\begin{align*} \Bigg|V_k\Bigg(\bigcap_{j\in T} F_n^{[j]}\Bigg)\Bigg|\overset{d}{=} \Bigg|V_k\Bigg(\bigcap_{j\in T} K_n^{\{j\}}\Bigg)\Bigg|\leq C_k^{\textrm{var}}\Bigg(\bigcap_{j\in T} K_n^{\{j\}}\Bigg) \leq c_{u,k} M^{-kn} N^u_n. \end{align*}

Note that we are dealing here with unions of level-n subcubes of C and not of the unit cube. However, this only changes the constant of the lemma by a factor of $M^{-k}$ , since we can scale the whole situation by a factor of M to take place in a unit-size cube. Multiplying by $M^{(k-D)n}$ and setting $W^{u}_n\;:\!=\;N^{u}_n/{\mathbb E} N^{u}_n$ , where ${\mathbb E} N^{u}_n=({\mathbb E} N^{u}_1)^n= M^{un}p^n$ , we get

\begin{align*} M^{(k-D)n} \Bigg|V_k\Bigg(\bigcap_{j\in T} F_n^{[j]}\Bigg)\Bigg|\leq c_{u,k} M^{(u-D)n}p^n W^{u}_n= c_{u,k} M^{(u-d)n} W^{u}_n. \end{align*}

Now, choosing e.g. $h_n\;:\!=\;M^{-n/2}$ , $n\in{\mathbb N}$ (which obviously forms a non-increasing and summable sequence), we conclude from the above inequality that

(5.7) \begin{align}{\mathbb E}\Bigg[\sup_{n\in{\mathbb N}} \frac{M^{(k-D)n}}{h_n} \Bigg|V_k\Bigg(\bigcap_{j\in T} F_n^{[j]}\Bigg)\Bigg|\Bigg]\leq c_{u,k} {\mathbb E}\left[\sup_{n\in{\mathbb N}} M^{-\alpha n} W^{u}_n\right], \end{align}

where $\alpha\;:\!=\;d-u-\frac 12$ . Note that $\alpha\geq\frac 12$ , since $u\leq d-1$ . Now observe that the sequence of random variables $X_m\;:\!=\;\sup_{n\in\{1,\ldots,m\}} M^{-\alpha n} W^{u}_n$ , $m\in{\mathbb N}$ , is non-decreasing and converges as $m\to\infty$ to $\sup_{n\in{\mathbb N}} M^{-\alpha n} W^{u}_n$ . Moreover, for any $m\in{\mathbb N}$ , we have

\begin{align*} {\mathbb E}\left[X_m\right]&\leq {\mathbb E}\Bigg[\sum_{\ell=1}^n M^{-\alpha \ell} W^{u}_\ell \Bigg]=\sum_{\ell=1}^n M^{-\alpha \ell} {\mathbb E}\left[W^{u}_\ell\right]=\frac{1-M^{-\alpha n}}{M^{\alpha}-1}\leq \frac{1}{M^{\alpha}-1},\end{align*}

which, by monotone convergence, implies that the right-hand side of (5.7) is bounded (by $c_{u,k}$ times the latter constant, in which $\alpha\geq 1/2$ ). Finally, note that the chosen sequence $(h_n)$ is independent of the set T, and hence it can be used for all sets $T\subseteq\Sigma$ , $|T|\geq 2$ . Moreover, there are only finitely many choices for the constant $c_{u,k}$ (which depends on T via the dimension u of the resulting cube C). This shows the finiteness of the expectation in (5.5) and completes the proof.

First proof of Theorem 3.1. Observe that the volume $V_d(F_n)$ equals the number $N_n$ of cubes in the union $F_n$ times the volume of a single cube of level n, i.e. $V_d(F_n)=M^{-dn} N_n$ , $n\in{\mathbb N}_0$ . If we recall the expectation and variance of $N_n$ from (2.1), this implies ${\mathbb E}(V_d(F_n))= M^{-dn} {\mathbb E}(N_n) = p^n$ and

\begin{align*} \mathrm{Var}(V_d(F_n))&=M^{-2dn} \mathrm{Var}(N_n)=\begin{cases} \frac{1-p}{M^dp-1} (M^{-d}p)^n\left((M^dp)^n-1\right) & \text{ if } p\neq 1/M^d,\\[5pt] M^{-2dn}\cdot n\cdot (1-p) & \text { if } p=1/M^d, \end{cases}\end{align*}

from which the formula (3.1) follows at once.

Proof. On the one hand, by (6.1) and the invariance properties of the intrinsic volumes, we have for any $n\in{\mathbb N}$

(6.4) \begin{align} V_k\big(F_{2,n}^j\big)\overset{d}{=} V_k(\varphi_j(F_{n-1}))=M^{-k} V_k(F_{n-1}). \end{align}

On the other hand,

(6.5) \begin{align} V_k\big(F_{n}^j\big)=V_k\big(L_1^j\cap F_{2,n}^j\big)=V_k\big(F_{2,n}^j\big)\cdot Y_j, \end{align}

where $Y_j$ is the Bernoulli variable which determines whether $J_j$ is kept in the first step. Observe that $Y_j$ is independent of $F_{2,n}^j$ . Combining both equations proves the first assertion. The expressions for the expectation and variance can be deduced from the first equation, taking into account the independence of Y and $F_{2,n}^j$ and recalling that ${\mathbb E} Y={\mathbb E} Y^2=p$ and $\mathrm{Var}(Y)=p(1-p)$ .

Second proof of Theorem 3.1. Recall that $F_n=\bigcup_{j=1}^{M^d} F_n^j$ and that from the perspective of volume this union is disjoint. Hence $V_d(F_n)=\sum_j V_d(F_n^j)$ . Observe that this decomposes $V_d(F_n)$ into a sum of independent random variables. According to Lemma 6.1, the sets $F_n^j$ satisfy ${\mathbb E} V_d(F_n^j)= \frac p{M^{d}} {\mathbb E} V_d(F_{n-1})$ , which implies that

$${\mathbb E} V_d(F_n)=\sum_j p M^{-d} {\mathbb E} V_d(F_{n-1})= p \cdot{\mathbb E} V_d(F_{n-1}) $$

for any $n\in{\mathbb N}$ , where ${\mathbb E} V_d(F_0)=1$ . This is a recursion relation which implies immediately that ${\mathbb E} V_d(F_n)=p^n$ . Furthermore, Lemma 6.1 also yields a recursion equation for the variance of $V_d(F_n)$ ,

\begin{align*} \mathrm{Var}(V_d(F_n))&=\mathrm{Var}\Bigg(\sum_{j=1}^{M^d} V_d(F_n^j)\Bigg)=\sum_{j=1}^{M^d}\mathrm{Var}\left(V_d(F_n^j)\right)\\[5pt] &=\sum_{j=1}^{M^d} \frac{p}{M^{2d}}\left(\mathrm{Var}( V_d(F_{n-1}))+ (1-p)({\mathbb E} V_d(F_{n-1}))^2\right)\\[5pt] &= \frac{p}{M^{d}}\cdot \mathrm{Var}( V_d(F_{n-1})) + \frac{1-p}{M^{d}p}\cdot p^{2n}, \end{align*}

where we have used the independence of the $V_d(F_n^j)$ for the second equality. By induction, this leads to

\begin{align*} \mathrm{Var}(V_d(F_n))&=\frac{p}{M^{d}} \mathrm{Var}(V_d(F_0))+\frac{1-p}{M^{d}p}\cdot \sum_{\ell=0}^{n-1} \bigg(\frac{p}{M^{d}}\bigg)^{\ell} p^{2(n-\ell)}, \end{align*}

where the first summand vanishes, since $F_0=[0,1]^2$ is deterministic, and the second summand simplifies to the expression stated in Theorem 3.1.

Proof. Let $N'_{\!\!n}$ denote the number of basic cubes of level n contained in the intersection $I_n\;:\!=\;F_n^{\{1\}}\cap\ldots\cap F_n^{\{\ell\}}$ . Note that a basic cube $J_\sigma$ , $\sigma=\sigma_1\ldots\sigma_n\in\Sigma^n$ , of level n is contained in $I_n$ if and only if $J_{\sigma_1\ldots\sigma_{n-1}}$ is contained in $I_{n-1}$ and if, for each $i=1,\ldots,\ell$ , $Y_\sigma^{\{i\}}=1$ , i.e. $J_\sigma$ survives in $F_n^{\{i\}}$ . This implies that for each basic cube of level $n-1$ contained in $I_{n-1}$ , the random number of descendants (i.e. of subcubes of level n contained in $I_n$ ) is $\text{Bin}(M^d, p^\ell)$ -distributed, and hence $(N'_{\!\!n})$ forms a Galton–Watson process that is equivalent to the one associated to a fractal percolation F ^′ on $[0,1]^d$ with parameters M and $p^\ell$ . Now the assertion follows from the fact that $V_d(I_n)$ equals $N'_{\!\!n}$ times $M^{-dn}$ , the volume of a single basic cube of level n, which is also the volume $V_d(F'_n)$ . The formulas for the expectation and variance then follow directly from Theorem 3.1.

Proof. First observe that, similarly as in the proof of Lemma 6.1,

(7.1) \begin{align} V_k\big(F_{n}^j\cap F_n^\ell\big)=V_k\big(L_1^j\cap F_{2,n}^j\cap L_1^\ell\cap F_{2,n}^\ell\big)=V_k\big(F_{2,n}^j\cap F_{2,n}^\ell\big)\cdot Y_j\cdot Y_\ell, \end{align}

where $Y_j$ and $Y_\ell$ are the Bernoulli variables which determine whether $J_j$ and $J_\ell$ , respectively, are kept in the first step. Note that $Y_j, Y_\ell, F_{2,n}^j, F_{2,n}^\ell$ are independent and that the product $Y_j\cdot Y_\ell$ is a Bernoulli variable with parameter $p^2$ . Now observe that, by (6.1), and with $K^{\{i\}}$ as in the statement, for any $n\in{\mathbb N}$ , $F_{2,n}^j\cap F_{2,n}^\ell\overset{d}{=} \varphi(K^{\{1\}}_{n-1}\cap K^{\{2\}}_{n-1})$ , where $\varphi\;:\;{\mathbb R}^{d-1}\to {\mathbb R}^d$ is the similarity with contraction ratio $1/M$ mapping $[0,1]^{d-1}$ to $J_j\cap J_\ell$ . This implies in particular that

(7.2) \begin{align} V_k\big(F_{2,n}^j\cap F_{2,n}^\ell\big)\overset{d}{=} V_k\big(\varphi\big(K^{\{1\}}_{n-1}\cap K^{\{2\}}_{n-1}\big)\big)=M^{-k} V_k\big(K^{\{1\}}_{n-1}\cap K^{\{2\}}_{n-1}\big). \end{align}

Combining both equations proves the first assertion. The expressions for the expectation and variance follow easily.

Proof of Theorem 3.2. Fix $n\in{\mathbb N}$ . For $j,\ell\in\Sigma$ , let $X_n\;:\!=\;V_{d-1}(F_n)$ , $X_n^{j}\;:\!=\;V_{d-1}(F_n^j)$ , and

\begin{align*} X_n^{j,\ell}\;:\!=\;V_{d-1}(F_n^j\cap F_n^\ell). \end{align*}

Note that all these random variables are nonnegative. The variables $X_n^j$ , $j\in\Sigma$ are i.i.d. Note also that $X_n^{j,\ell}=0$ whenever the corresponding cubes $J_j$ and $J_\ell$ are not direct neighbors. By the inclusion–exclusion principle, we have

(7.3) \begin{align} X_n&=\sum_{j\in\Sigma} X_n^j - \sum_{\{j,\ell\}\subset\Sigma} X_n^{j,\ell}. \end{align}

For the expectation this implies

\begin{align*} {\mathbb E}(X_n)&=\sum_{j\in\Sigma} {\mathbb E}\big(X_n^j\big) - \sum_{\{i,\ell\}\subset\Sigma} {\mathbb E}\big(X_n^{i,\ell}\big). \end{align*}

By Lemma 6.1, the first sum simplifies to $Mp {\mathbb E}(X_{n-1})$ . In the second term it is enough to sum over those pairs $\{i,\ell\}$ for which the corresponding cubes $J_i$ and $J_\ell$ are direct neighbors. There are $d (M-1) M^{d-1}$ such pairs in dimension d. All other summands are zero. Now observe that to direct neighbors Lemma 7.1 can be applied, which in combination with Corollary 6.1 yields

(7.4) \begin{align} {\mathbb E}\big(X_n^{i,\ell}\big)=\frac{p^2}{M^{d-1}} {\mathbb E} V_{d-1}\Big(K_{n-1}^{\{1\}}\cap K_{n-1}^{\{2\}}\Big)=\frac{p^2}{M^{d-1}} p^{2(n-1)}=\frac{p^{2n}}{M^{d-1}}. \end{align}

Therefore we conclude that for any $n\in{\mathbb N}$ ,

\begin{align*} {\mathbb E}(X_n)&=Mp {\mathbb E}(X_{n-1})-d (M-1) M^{d-1}\cdot M^{1-d} p^{2n}=Mp {\mathbb E}(X_{n-1})-d (M-1)p^{2n}, \end{align*}

where ${\mathbb E}(X_0)=V_{d-1}(J)=d$ . This is a recursion relation for the expectation and leads to

\begin{align*} {\mathbb E}(X_n)&=(Mp)^n {\mathbb E}(X_{0})-d (M-1) \sum_{i=1}^n (Mp)^{n-i} p^{2i}\\[5pt] &=d (Mp)^n \bigg(1-\frac{(M-1)p}{M-p}\bigg[1-\bigg(\frac pM\bigg)^n\bigg]\bigg), \end{align*}

which is equivalent to the expression stated in Equation (3.2), completing the proof of this equation.

For the computation of the variance of $X_n$ , the following statement is useful.

Lemma 7.2. For $i=1,2$ , let $Z_i,Z'_{\!\!i}$ be stochastically independent random variables. Then

\begin{align*} \mathrm{Cov}(Z_1\cdot Z'_{\!\!1}, Z_2\cdot Z'_{\!\!2})=\mathrm{Cov}(Z_1,Z_2)\cdot {\mathbb E}(Z'_{\!\!1}\cdot Z'_{\!\!2})+{\mathbb E} Z_1\cdot {\mathbb E} Z_2\cdot \mathrm{Cov}(Z'_{\!\!1}, Z'_{\!\!2}). \end{align*}

In particular, $\mathrm{Var}(Z_1\cdot Z'_{\!\!1})=\mathrm{Var}(Z_1)\cdot {\mathbb E}(Z'_{\!\!1}\cdot Z'_{\!\!1})+({\mathbb E} Z_1)^2\cdot \mathrm{Var}(Z'_{\!\!1}).$

Now we are ready to compute the variance of $X_n$ . Our starting point is again the central equation (7.3), which allows us to split the variance $\mathrm{Var}(X_n)$ into several terms which will then be treated separately. Equation (7.3) implies that

(7.5) \begin{align} \notag \mathrm{Var}(X_n)&=\mathrm{Var}\left(\sum_{j\in\Sigma} X_n^j\right) + \mathrm{Var}\left( \sum_{\{i,\ell\}\subset\Sigma} X_n^{i,\ell}\right)-2\cdot\mathrm{Cov}\left(\sum_{j\in\Sigma} X_n^j,\sum_{\{i,\ell\}\subset\Sigma} X_n^{i,\ell}\right)\\[5pt] &= \sum_{j\in\Sigma}\mathrm{Var}\big(X_n^j\big) + \sum_{\{i,\ell\}\subset\Sigma} \mathrm{Var}\big(X_n^{i,\ell}\big)+ \sum_{ \{i,\ell\}\neq\{i',\ell'\}} \mathrm{Cov}\Big(X_n^{i,\ell},X_n^{i',\ell'}\Big)\\[5pt] \notag &\qquad -2\sum_{j\in\Sigma}\sum_{\{i,\ell\}\subset\Sigma}\mathrm{Cov}\big(X_n^j,X_n^{i,\ell}\big), \end{align}

where we have used the independence of the $X_n^j$ in the first sum. Note that the sum in the third term extends over all pairs $(\{i,\ell\},\{i',\ell'\})$ of subsets of $\Sigma$ with two elements such that $\{i,\ell\}\neq\{i',\ell'\}$ . Since the $X_n^j$ are identically distributed, we get for the first term above

\begin{align*} \sum_{j\in\Sigma}\mathrm{Var}(X_n^j)&=M^d\cdot \mathrm{Var}\big(X_n^1\big)=M^d \frac{p}{M^{2(d-1)}}\left(\mathrm{Var}( X_{n-1})+ (1-p)({\mathbb E} X_{n-1})^2\right)\\[5pt] &=\frac{p}{M^{d-2}}\mathrm{Var}( X_{n-1})+ \frac{p(1-p)}{M^{d-2}}({\mathbb E} X_{n-1})^2, \end{align*}

where for the second equality we employed Equation (6.2) of Lemma 6.1. By Equation (3.2), we have

\begin{align*} ({\mathbb E} X_{n-1})^2&=(Mp)^{2n} \frac {d^2(1-p)^2}{p^2(M-p)^2} \left(1+\frac{M-1}{1-p}\left(\frac pM\right)^{n}\right)^2 \end{align*}

and thus

(7.6) \begin{align} \sum_{j\in\Sigma}\mathrm{Var}(X_n^j)&=\frac{p}{M^{d-2}}\mathrm{Var}( X_{n-1})+ \frac {d^2(1-p)^3}{M^{d-2}p(M-p)^2}\times\\[5pt] \notag &\quad\times(Mp)^{2n} \left(1+ \frac{M-1}{1-p}\left(\frac pM\right)^n\right)^2. \end{align}

Inserting this expression into Equation (7.5) yields a recursion relation for the variance $\mathrm{Var}(X_n)$ . In order for this to be of use, it remains to find explicit expressions for the other terms in (7.5). The discussion below of the three remaining terms will show that all of them are of some order lower than $(Mp)^{2n}$ and thus do not contribute to the first term in (3.3). Moreover, it will become clear that for certain parameter combinations the last term in (7.5) provides a contribution to the second term in (3.3).

Analysis of the second term in (7.5). First recall that $X_n^{j,\ell}=V_{d-1}(F_n^j\cap F_n^\ell)=0$ whenever the corresponding cubes $J_j$ and $J_\ell$ are not direct neighbors, which implies $\mathrm{Var}(X_n^{j,\ell})=0$ in this case. Recall also that there are $d(M-1)M^{d-1}$ pairs $\{j,\ell\}$ such that $J_j$ and $J_\ell$ are direct neighbors. By Lemma 7.1, we get for all such pairs $\{j,\ell\}$ that $\mathrm{Var}(X_n^{j,\ell})$ equals

\begin{align*} &\frac{p^2}{M^{2(d-1)}}\Big(\mathrm{Var}\Big( V_{d-1}\Big(K_{n-1}^{\{1\}}\cap K_{n-1}^{\{2\}}\Big)\Big)+ (1-p^2)\Big({\mathbb E} V_{d-1}\Big(K_{n-1}^{\{1\}}\cap K_{n-1}^{\{2\}}\Big)\Big)^2\Big), \end{align*}

where $K^{\{1\}}, K^{\{2\}}$ are independent $(d-1)$ -dimensional fractal percolations. Hence, by Corollary 6.1, we have $({\mathbb E} V_{d-1}(K_{n-1}^{\{1\}}\cap K_{n-1}^{\{2\}}))^2=p^{4(n-1)}$ and, for $p^2\neq M^{1-d}$ ,

\begin{align*} \mathrm{Var}\Big( V_{d-1}\Big(K_{n-1}^{\{1\}}\cap K_{n-1}^{\{2\}}\Big)\Big)&=p^{4(n-1)} \frac{1-p^2}{M^{d-1}p^2-1}\Bigg(1-\bigg(\frac 1{M^{d-1}p^2}\bigg)^{n-1}\Bigg).\end{align*}

For $p^2= M^{1-d}$ , a different expression for the variance is provided by Corollary 6.1. Inserting these expressions into the above formula yields

(7.7) \begin{align} \mathrm{Var}(X_n^{j,\ell})&=\frac{1-p^2}{M^{d-1}}\cdot p^{4n} \begin{cases} \frac{1}{M^{d-1}p^2-1} \left(1 - \left(M^{d-1}p^{2}\right)^{-n}\right), &M^{d-1}p^2\neq 1,\\[5pt] n, &M^{d-1}p^2= 1. \end{cases} \end{align}

This provides explicit expressions for the second term in (7.5).

Analysis of the third term in (7.5). First note that the covariances of the form $\mathrm{Cov}\Big(X_n^{i,\ell},X_n^{i',\ell'}\Big)$ appearing in the third term vanish whenever $\{i,\ell\}\cap\{i',\ell'\}=\emptyset$ , simply because in this case the variables $X_n^{i,\ell}$ and $X_n^{i',\ell'}$ are determined by disjoint subfamilies of the basic random variables $Y_\sigma$ , $\sigma\in\Sigma^*$ , and are therefore independent. Since also $\{i,\ell\}\neq\{i',\ell'\}$ , all relevant terms in the third sum refer to pairs of the form $(\{i,\ell\},\{i,\ell'\})$ with $\ell\neq\ell'$ , where $J_\ell$ and $J_{\ell'}$ are both direct neighbors of $J_i$ . Now observe that the covariance $\mathrm{Cov}\Big(X_n^{i,\ell},X_n^{i,\ell'}\Big)$ depends on the relative position of $J_\ell$ and $J_{\ell'}$ , since this determines which of the basic variables $Y_\sigma$ determining $F_n^i$ are relevant for both quantities $X_n^{i,\ell}$ and $X_n^{i,\ell'}$ . Since $J_i\cap J_\ell$ and $J_i\cap J_{\ell'}$ are facets of $J_i$ , and any two facets of a cube either are disjoint (if they lie on opposite sides, i.e. in parallel hyperplanes) or intersect in a $(d-2)$ -face (i.e. in a cube of dimension $d-2$ ), we are left with exactly these two cases. The remaining task is now to determine how many covariance terms there are for each of these two types, and to provide an explicit expression for each of them. We start with the easier one, for which the cubes $J_\ell$ , $J_i$ , and $J_{\ell'}$ have to be arranged in a row (in this order) in one of the 2d directions. (Note that because of the independent summation through all pairs $\{i,\ell\}$ and all pairs $\{i,\ell'\}$ , it is a different contribution if $\ell$ and $\ell'$ are interchanged.) There are $2d(M-2)M^{d-1}$ such terms. (For $M=2$ this situation is not possible, and accordingly the number is zero.) The only basic variable which affects both quantities $X_n^{i,\ell}$ and $X_n^{i,\ell'}$ is $Y_i$ ; i.e. only the first construction step affects this covariance term. To compute it, we separate the influence of the first construction step from the rest (just as in Lemma 6.1) and use the independence. We have, for any $n\in{\mathbb N}$ ,

\begin{align*} \mathrm{Cov}\Big(X_n^{i,\ell},X_n^{i,\ell'}\Big)&= \mathrm{Cov}\left(V_{d-1}\Big(F_{2,n}^i\cap L_1^i\cap F_n^\ell\Big),V_{d-1}\Big(F_{2,n}^i\cap L_1^{i}\cap F_n^{\ell'}\Big)\right)\\[5pt] &=\mathrm{Cov}\left(V_{d-1}\Big(F_{2,n}^i\cap F_n^\ell\Big)\cdot Y_i,V_{d-1}\Big(F_{2,n}^i\cap F_n^{\ell'}\Big)\cdot Y_i\right). \end{align*}

Setting $\widetilde{X}_n\;:\!=\;V_{d-1}(F_{2,n}^i\cap F_n^\ell)$ and $\widetilde{X}_n'\;:\!=\;V_{d-1}\Big(F_{2,n}^i\cap F_n^{\ell'}\Big)$ and noting that these two variables are independent, identically distributed, and independent of $Y_i$ , with mean given by

(7.8) \begin{align} {\mathbb E} X_n^{i,\ell}= {\mathbb E}\Big(V_{d-1}\Big(F_{2,n}^i\cap F_n^\ell\Big)\cdot Y_i\Big)={\mathbb E} \widetilde{X}_n\cdot {\mathbb E} Y_i={\mathbb E} \widetilde{X}_n\cdot p, \end{align}

we obtain, using the first formula in Lemma 7.2, that

\begin{align*} \mathrm{Cov}\Big(X_n^{i,\ell},X_n^{i,\ell'}\Big)&= \mathrm{Cov}\left(\widetilde{X}_n,\widetilde{X}_n'\right) {\mathbb E}(Y_i\cdot Y_i)+ {\mathbb E} \widetilde{X}_n\cdot {\mathbb E} \widetilde{X}_n'\cdot \mathrm{Cov}(Y_i,Y_i)\\[5pt] &=\left({\mathbb E} \widetilde{X}_n\right)^2 \mathrm{Var}(Y_i) =\frac {1-p}{p} \big({\mathbb E} X_n^{i,\ell}\big)^2. \end{align*}

By Equation (7.4), this yields for the $2d(M-2)M^{d-1}$ terms for which the corresponding cubes $J_\ell$ , $J_i$ , and $J_{\ell'}$ are arranged in a row, and for any $n\in{\mathbb N}$ ,

(7.9) \begin{align} \mathrm{Cov}\Big(X_n^{i,\ell},X_n^{i,\ell'}\Big)&= \frac {1-p}{p} \left(\frac{p^{2n}}{M^{d-1}}\right)^2=\frac {1-p}{M^{2(d-1)}}p^{4n-1}. \end{align}

Now let us look at the second type of covariance term occurring in the third sum in (7.5), for which the cubes involved, namely $J_i$ , $J_\ell$ , and $J_{\ell'}$ , intersect in a $(d-2)$ -face of $J_i$ . First let us determine the number of such terms. It is convenient to determine first the number of $(d-2)$ -dimensional cubes at which this intersection can happen. There are $\binom{d}{2}(M-1)^2M^{d-2}$ such cubes. (Indeed, choose $d-2$ out of d directions to span the direction space L of (the affine hull of) the cube (or equivalently, choose 2 out of d to span the orthogonal complement of L). For each direction space there are $(M-1)^2M^{d-2}$ such cubes: there are $(M-1)^2$ in each 2-dimensional ‘layer’, and there are $M^{d-2}$ such layers.) Each such $(d-2)$ -cube C is surrounded by four d-dimensional cubes; hence there are four choices for $J_i$ . Then the two cubes which are direct neighbors of $J_i$ must be $J_\ell$ and $J_{\ell'}$ , and the only choice we can make is which one is $J_\ell$ . Hence for each $(d-2)$ -cube there are eight possible choices, and so the total number of terms of the second type in the third sum of (7.5) is $8\binom{d}{2}(M-1)^2M^{d-2}$ . It remains to compute the covariance for this type.

Setting (in analogy with the first type)

$$ \widehat{X}_n\;:\!=\;V_{d-1}\big(F_{2,n}^i\cap F_{2,n}^\ell\big) \text{ and } \widehat{X}'_{\!\!n}\;:\!=\;V_{d-1}\Big(F_{2,n}^i\cap F_{2,n}^{\ell'}\Big) $$

and noting that these two variables are independent, identically distributed, and independent of $Y_i$ , $Y_\ell$ , and $Y_{\ell'}$ , we obtain for the mean

(7.10) \begin{align} {\mathbb E} X_n^{i,\ell}= {\mathbb E}(V_{d-1}(F_{2,n}^i\cap F_{2,n}^\ell)\cdot Y_i\cdot Y_\ell)={\mathbb E} \widehat{X}_n\cdot {\mathbb E} Y_i\cdot {\mathbb E} Y_\ell={\mathbb E} \widehat{X}_n\cdot p^2, \end{align}

and for the covariance, using the first formula in Lemma 7.2 and (7.4),

(7.11) \begin{align} \mathrm{Cov}\Big(X_n^{i,\ell},X_n^{i,\ell'}\Big)&= \mathrm{Cov}\left(\widehat{X}_n\cdot Y_i Y_\ell,\widehat{X}'_{\!\!n}\cdot Y_iY_{\ell'}\right) \notag\\[5pt] &=\mathrm{Cov}\left(\widehat{X}_n,\widehat{X}'_{\!\!n}\right) {\mathbb E}(Y_i^2 Y_\ell Y_{\ell'})+ {\mathbb E} \widehat{X}_n\cdot {\mathbb E} \widehat{X}'_{\!\!n}\cdot \mathrm{Cov}(Y_i Y_\ell,Y_i Y_{\ell'})\notag\\[5pt] &=\mathrm{Cov}\left(\widehat{X}_n,\widehat{X}'_{\!\!n}\right) \cdot p^3+\left({\mathbb E} \widehat{X}_n\right)^2 p^3(1-p) \notag\\[5pt] &=\mathrm{Cov}\left(\widehat{X}_n,\widehat{X}'_{\!\!n}\right) \cdot p^3+\frac{1-p}{M^{2(d-1)}}\cdot p^{4n-1}. \end{align}

But this time $\widehat{X}_n$ and $\widehat{X}'_{\!\!n}$ are not independent, so the covariance term does not vanish. In order to derive a recursion for the left-hand side, notice that the intersection $J_i\cap J_\ell$ is $(d-1)$ -dimensional and intersects $M^{d-1}$ of the second-level cubes $J_{ij}$ contained in $J_i$ . The intersection $J_i\cap J_\ell\cap J_{\ell'}$ is $(d-2)$ -dimensional and intersects $M^{d-2}$ of the second-level cubes $J_{ij}$ contained in $J_i$ . Denote the latter cubes by $J_{i1}, J_{i2},\ldots, J_{iM^{d-2}}$ , and let R be the union of all the remaining second-level cubes contained in $J_i$ ; i.e., set $R\;:\!=\;\cup\{J_{ij}\;:\;j\in\Sigma\setminus\{1,\ldots,M^{d-2}\}\}$ . Observe that

\begin{align*} \widehat{X}_n= V_{d-1}(F_{2,n}^i\cap F_{2,n}^\ell)=\underbrace{V_{d-1}(F_{2,n}^i\cap R\cap F_{2,n}^\ell)}_{\;=:\;\widetilde{R}}+\sum_{k=1}^{M^{d-2}} \underbrace{V_{d-1}(F_{2,n}^i\cap J_{ik}\cap F_{2,n}^\ell)}_{\;=:\;\widetilde{Q}_k}, \end{align*}

where all the terms on the right are independent. The same holds with $\ell$ replaced by $\ell'$ , for which we denote the corresponding terms by $\widetilde{R}'$ and $\widetilde{Q}'_{\!\!k}$ , respectively. What happens in $J_{ik}$ is now a scaled version of what happens in $J_i$ . More precisely, we have, for any $k\in\{1,\ldots, M^{d-2}\}$ ,

(7.12) \begin{align} \widetilde{Q}_k & \overset{d}{=} M^{-d+1}V_{d-1}\big(F_{n-1}^i\cap F_{n-1}^\ell\big)=M^{-d+1} X_{n-1}^{i,\ell}. \end{align}

(Note that the corresponding sets are not equal in distribution, owing to intersections with neighboring cubes, which are fortunately of lower dimension and thus do not contribute to the intrinsic volume of order $d-1$ .) Taking into account the dependence structure of $\widetilde{R},\widetilde{R}',\widetilde{Q}_k,\widetilde{Q}_k$ and (7.12), we conclude that

\begin{align*} \mathrm{Cov}\left(\widehat{X}_n,\widehat{X}'_{\!\!n} \right)&=\mathrm{Cov}\left(\widetilde{R}+\sum_k \widetilde{Q}_k,\widetilde{R}'+\sum_k \widetilde{Q}'_{\!\!k}\right)\\[5pt] &=\sum_k \mathrm{Cov}\left(\widetilde{Q}_k,\widetilde{Q}'_{\!\!k}\right)\\[5pt] &= M^{d-2} \cdot \mathrm{Cov}\left(M^{1-d}\cdot X_{n-1}^{i,\ell},M^{1-d}\cdot X_{n-1}^{i,\ell'}\right)\\[5pt] &= M^{-d} \mathrm{Cov}\left(X_{n-1}^{i,\ell},X_{n-1}^{i,\ell'}\right). \end{align*}

Plugging this into (7.11) yields the desired recursion for the covariance of the second type:

(7.13) \begin{align} \mathrm{Cov}\Big(X_n^{i,\ell},X_n^{i,\ell'}\Big)&= \frac {p^3}{M^d}\cdot \mathrm{Cov}\Big(X_{n-1}^{i,\ell},X_{n-1}^{i,\ell'}\Big) + \frac{1-p}{M^{2(d-1)}}\cdot p^{4n-1},\qquad n\in{\mathbb N}, \end{align}

where $\mathrm{Cov}\Big(X_0^{i,\ell},X_0^{i,\ell'}\Big)\;:\!=\;0$ . We conclude that, for $n\in{\mathbb N}$ ,

(7.14) \begin{align} \mathrm{Cov}\Big(X_n^{i,\ell},X_n^{i,\ell'}\Big) &= \frac{1-p}{M^{2(d-1)}}\cdot p^{4n-1} \sum_{k=0}^{n-1} \left(\frac 1{M^dp}\right)^k \notag\\[5pt] &=\begin{cases} \frac{1-p}{M^{d-2}(M^d p-1)}\cdot p^{4n}\left(1-(M^{d}p)^{-n}\right), & M^dp\neq 1, \\[5pt] \frac{1-p}{M^{2(d-1)}}\cdot p^{4n-1} \cdot n, & M^dp= 1. \end{cases} \end{align}

Analysis of the last term in (7.5). First note that the covariances $\mathrm{Cov}(X_n^{j},X_n^{i,\ell})$ in the last term vanish whenever $j\notin\{i,\ell\}$ , because the variables $X_n^{j}$ and $X_n^{i,\ell}$ are determined by disjoint subfamilies of the basic random variables $Y_\sigma$ , $\sigma\in\Sigma^*$ , in this case and are therefore independent. Hence, all relevant terms in the last sum refer to pairs $(j, \{i,\ell\})$ such that $J_i$ and $J_\ell$ are direct neighbors and $j\in\{i,\ell\}$ . Let us first determine the number of such terms. For each pair $\{i,l\}$ there are two choices for j (namely, $j=i$ and $j=\ell$ ), and there are $d(M-1)M^{d-1}$ pairs such that $J_i$ and $J_\ell$ are direct neighbors. Hence there are $2d(M-1)M^{d-1}$ such terms. To find an expression for this covariance, we choose $j=i$ without loss of generality. Again we start by separating the influence of the first construction step from the rest. Setting $\widehat{X}_n\;:\!=\;V_{d-1}(F_{2,n}^i\cap F_{2,n}^\ell)$ and $\check{X}_n\;:\!=\;V_{d-1}(F_{2,n}^i)$ we infer, using the first formula in Lemma 7.2,

\begin{align*} \mathrm{Cov}(X_n^{i},X_n^{i,\ell}) &=\mathrm{Cov}\left(V_{d-1}(F_{2,n}^i)\cdot Y_i,V_{d-1}(F_{2,n}^i\cap F_{2,n}^{\ell})\cdot Y_iY_\ell\right)\\[5pt] &=\mathrm{Cov}\left(\check{X}_n,\widehat{X}_n\right) {\mathbb E}\left(Y_i^2Y_\ell\right)+ {\mathbb E} V_{d-1}(F_{2,n}^i)\cdot {\mathbb E} \widehat{X}_n\cdot \mathrm{Cov}(Y_i, Y_i Y_\ell)\\[5pt] &=\mathrm{Cov}\left(\check{X}_n,\widehat{X}_n\right)\cdot p^2 + {\mathbb E} V_{d-1}(F_{2,n}^i) \cdot {\mathbb E} \widehat{X}_n\cdot p^2(1-p). \end{align*}

Now recall from (7.10) that ${\mathbb E} \widehat{X}_n\cdot p^2={\mathbb E} X_n^{i,\ell}$ , for which an explicit expression is provided in (7.4). Moreover, by (6.4), we have ${\mathbb E} \check{X}_n=M^{1-d} {\mathbb E} V_{d-1}(F_{n-1})$ , and for the latter an expression is given by (3.2). Hence, setting $\gamma_n\;:\!=\;\mathrm{Cov} (\check{X}_{n},\widehat{X}_{n})$ , $n\in{\mathbb N}$ (and noting that $\gamma_1=0$ ), we infer that for any $n\in{\mathbb N}$ ,

(7.15) \begin{align} \mathrm{Cov}\big(X_n^{i},X_n^{i,\ell}\big) &= \gamma_n \cdot p^2 + \big(Mp^3\big)^{n} \frac {d(1-p)^2}{M^{2(d-1)}p(M-p)} \left(1+\frac{M-1}{1-p}\left(\frac pM\right)^{n}\right). \end{align}

This time again, the covariance term $\gamma_n$ on the right does not vanish. But we can derive a recursion for $\gamma_n$ using an approach similar to the one used for the second type in the third term. The intersection $J_i\cap J_\ell$ is $(d-1)$ -dimensional and intersects $M^{d-1}$ of the second-level cubes $J_{ij}$ contained in $J_i$ . Denote the latter cubes by $Q_1, Q_2,\ldots, Q_{M^{d-1}}$ and let R be the union of all the remaining second-level cubes contained in $J_i$ , i.e. $R=\cup\{J_{ij}\;:\;j\in\Sigma, J_{ij}\neq Q_m $ for all $m\}$ . Observe that $V_{d-1}(F_{2,n}^i\cap R\cap F_{2,n}^\ell)=0$ and therefore

\begin{align*} \widehat{X}_n= V_{d-1}(F_{2,n}^i\cap F_{2,n}^\ell)=\sum_{k=1}^{M^{d-1}} \underbrace{V_{d-1}(F_{2,n}^i\cap Q_k\cap F_{2,n}^\ell)}_{\;=:\;\widehat{Q}_k}, \end{align*}

where the terms on the right are independent. What happens in $Q_k$ is now again a scaled version of what happens in $J_i$ . More precisely, we have, for $k\in\{1,\ldots, M^{d-1}\}$ ,

\begin{align*} \widehat{Q}_k & \overset{d}{=} M^{1-d}V_{d-1}(F_{n-1}^i\cap F_{n-1}^\ell)=M^{1-d} X_{n-1}^{i,\ell}. \end{align*}

Observe that for any $n\in{\mathbb N}$ ,

(7.16) \begin{align} \gamma_n&= \mathrm{Cov}\left(\check{X}_{n},\sum_{k=1}^{M^{d-1}} \widehat{Q}_k\right) =\sum_{k=1}^{M^{d-1}} \mathrm{Cov}\left(\check{X}_{n},\widehat{Q}_k\right). \end{align}

Now we split $\check{X}_{n}=V_{d-1}(F_{2,n}^i)$ in a similar way using the decomposition $J_i=R\cup\bigcup_k Q_k$ . It is clear that the main contribution of the kth term above will come from $\mathrm{Cov}(V_{d-1}(F_{2,n}^i\cap Q_k),\widehat{Q}_k)$ , but there are further contributions this time which are due to the fact that the sets $F_{2,n}^i\cap Q_j$ are full-dimensional, so that their intersections cannot be neglected. In fact, in order to separate the contributions of the cubes $Q_k$ we need more refined notation. Without loss of generality, we can assume that the basic cubes $J_{ij}$ contained in $J_i$ are enumerated in such a way that $J_{ik}=Q_k$ for $k=1,\ldots,M^{d-1}$ . Let $F_{2,n}^{ij}$ be the union of those cubes $J_\sigma$ of level n (i.e. $\sigma=\sigma_1\ldots\sigma_n\in\Sigma^n$ ) contained in $F_{2,n}$ such that $\sigma_1=i$ and $\sigma_2=j$ , $j\in\Sigma$ , and let $F_{2,n}^{i,R}\;:\!=\;\bigcup_{j>M^{d-1}} F_{2,n}^{ij}$ . By the inclusion–exclusion principle, and ignoring lower-dimensional intersections, we have

\begin{align*} \check{X}_{n}=V_{d-1}\Big(F_{2,n}^{i,R}\Big)+ \sum_{j=1}^{M^{d-1}} \bigg[ &V_{d-1}\Big(F_{2,n}^{ij}\Big)- V_{d-1}\Big(F_{2,n}^{ij}\cap F_{2,n}^{i,R}\Big)\bigg]\\[5pt] &-\sum_{\{j,m\}:Q_j\sim Q_m} V_{d-1}\Big(F_{2,n}^{ij}\cap F_{2,n}^{im}\Big), \end{align*}

where $Q_j\sim Q_m$ in the last sum means that $Q_j$ and $Q_m$ are direct neighbors. Now we insert this representation into the kth covariance term in (7.16) and sort out which of the terms still have some of the basic random variables in common with $\widehat{Q}_k$ and which are independent. First observe that for any k, $\mathrm{Cov}(V_{d-1}(F_{2,n}^{i,R}),\widehat{Q}_k)=0$ , since the variables are determined by disjoint families of the basic random variables of the process. For a similar reason, the covariances $\mathrm{Cov}(V_{d-1}(F_{2,n}^{ij}),\widehat{Q}_k)$ and $\mathrm{Cov}(V_{d-1}(F_{2,n}^{ij}\cap F_{2,n}^{i,R}),\widehat{Q}_k)$ vanish whenever $j\neq k$ . Finally, $\mathrm{Cov}(V_{d-1}(F_{2,n}^{ij}\cap F_{2,n}^{im}),\widehat{Q}_k)=0$ whenever $k\notin\{j,m\}$ . Hence we get

(7.17) \begin{align} \mathrm{Cov}\left(\check{X}_{n},\widehat{Q}_k\right) &=\mathrm{Cov}\Big(V_{d-1}\Big(F_{2,n}^{ik}\Big),\widehat{Q}_k\Big)-\mathrm{Cov}\Big(V_{d-1}\Big(F_{2,n}^{ik}\cap F_{2,n}^{i,R}\Big),\widehat{Q}_k\Big)\notag\\[5pt] &\qquad - \sum_{m:Q_m\sim Q_k} \mathrm{Cov}\Big(V_{d-1}\Big(F_{2,n}^{ik}\cap F_{2,n}^{im}\Big),\widehat{Q}_k\Big). \end{align}

In order to derive a recursion, note that analogously to Lemma 6.1, we have

\begin{align*} &\left(V_{d-1}\Big(F_{3,n}^{ik}\Big),V_{d-1}\Big(F_{3,n}^i\cap Q_k\cap F_{2,n}^\ell\Big)\right)\\[5pt] &\qquad\overset{d}{=} M^{1-d} \left(V_{d-1}\Big(F_{2,n-1}^i\Big), V_{d-1}\Big(F_{2,n-1}^i\cap F_{2,n-1}^\ell\Big)\right)=M^{1-d} \left(\check{X}_{n-1},\widehat{X}_{n-1}\right),\end{align*}

which implies for the first term in the right-hand side of (7.17)

\begin{align*} \mathrm{Cov}&\Big(V_{d-1}\Big(F_{2,n}^{ik}\Big),\widehat{Q}_k\Big) = \mathrm{Cov}\Big(V_{d-1}\Big(F_{3,n}^{ik}\Big)\cdot Y_{ik},V_{d-1}\Big(F_{3,n}^{i}\cap Q_k\cap F_{2,n}^\ell\Big)\cdot Y_{ik}\Big)\\[5pt] &=\mathrm{Cov}\Big(V_{d-1}\Big(F_{3,n}^{ik}\Big),V_{d-1}\Big(F_{3,n}^{i}\cap Q_k\cap F_{2,n}^\ell\Big)\Big)\cdot {\mathbb E}\big(Y_{ik}^2\big)\\[5pt] &\qquad +{\mathbb E} V_{d-1}(F_{3,n}^{ik}) {\mathbb E} V_{d-1}(F_{3,n}^{i}\cap Q_k\cap F_{2,n}^\ell)\mathrm{Var}(Y_{ik})\\[5pt] &= M^{2(1-d)} \mathrm{Cov}(\check{X}_{n-1},\widehat{X}_{n-1})\cdot p +M^{2(1-d)} {\mathbb E}\check{X}_{n-1}\cdot {\mathbb E} \widehat{X}_{n-1}\cdot p(1-p)\\[5pt] &=\frac p{M^{2(d-1)}} \cdot \gamma_{n-1} +\frac{1-p}{M^{3(d-1)}} p^{2n-3} \cdot {\mathbb E}\check{X}_{n-1}.\end{align*}

Here we have used (7.10) and (7.4) to replace ${\mathbb E} \widehat{X}_{n-1}$ in the last equality. Similarly, employing (6.2) and (3.2) of Theorem 3.2, we can replace ${\mathbb E}\check{X}_{n-1}$ with some explicit expression. We conclude that

(7.18) \begin{align} \mathrm{Cov}(V_{d-1}&(F_{2,n}^{ik}),\widehat{Q}_k) =\frac p{M^{2(d-1)}} \cdot \gamma_{n-1} +\frac{1-p}{M^{3(d-1)}} p^{2n-2} \cdot {\mathbb E} V_{d-1}(F_{n-2}) \notag \\[5pt] &=\frac p{M^{2(d-1)}} \gamma_{n-1} + (Mp^3)^{n} \frac {d(1-p)^2}{M^{4d-3}p^4(M-p)} \left(1+\frac{M-1}{1-p}\left(\frac pM\right)^{n-1}\right). \end{align}

Inserting this into (7.17) and combining it with (7.16) yields the desired recursion relation for $\gamma_n$ , provided we can compute the two missing terms in (7.17). For the second term in (7.17), $\mathrm{Cov}(V_{d-1}(F_{2,n}^{ik}\cap F_{2,n}^{i,R}),\widehat{Q}_k)$ , observe that we can replace $F_{2,n}^{i,R}$ by $F_{2,n}^{ik'}$ without changing the random variable $V_{d-1}(F_{2,n}^{ik}\cap F_{2,n}^{i,R})$ , where k ^′ is the unique index such that $J_{ik'}$ is the unique direct neighbor of $J_{ik}$ contained in R. Indeed, all the other cubes in R have lower-dimensional intersections with $F_{2,n}^{ik}$ . Similarly, in $\widehat{Q}_k=V_{d-1}(F_{2,n}^i\cap Q_k\cap F_{2,n}^\ell)$ we can replace the set $F_{2,n}^i\cap Q_k$ by $F_{2,n}^{ik}$ and the set $F_{2,n}^\ell$ by $F_{2,n}^{\ell \ell'}$ , where $\ell'$ is the unique index such that the cubes $J_{ik}$ and $J_{\ell \ell'}$ are direct neighbors. Hence,

\begin{align*} \mathrm{Cov}\Big(V_{d-1}\Big(F_{2,n}^{ik}\cap F_{2,n}^{i,R}\Big),\widehat{Q}_k\Big)&=\mathrm{Cov}\Big(V_{d-1}\Big(F_{2,n}^{ik}\cap F_{2,n}^{ik'}\Big),V_{d-1}\Big(F_{2,n}^{ik}\cap F_{2,n}^{\ell \ell'}\Big)\Big).\end{align*}

Note that the cubes $J_{ik'}$ , $J_{ik}$ , and $J_{\ell \ell'}$ lie in a row, and therefore we are in the situation of the first type of covariance studied for the third term (cf. (7.9) and the paragraph before this equation), but one level further down. More precisely, we have

\begin{align*} \left(V_{d-1}\Big(F_{2,n}^{ik}\cap F_{2,n}^{ik'}\Big),\right.&\left.V_{d-1}\Big(F_{2,n}^{ik}\cap F_{2,n}^{\ell \ell'}\Big)\right)\\[5pt] &\overset{d}{=} M^{1-d} \left(V_{d-1}\Big(F_{n-1}^{i}\cap F_{n-1}^{i'}\Big),V_{d-1}\Big(F_{n-1}^{i}\cap F_{n-1}^{\ell}\Big)\right),\end{align*}

where i ^′ is the (unique) index such that $J_{i'}$ , $J_i$ , and $J_\ell$ lie in a row. (Note that i ^′ might not exist if $J_i$ touches the boundary of J, but then one can shift the situation or think of $F_{n-1}^{i'}$ as being part of a second fractal percolation in the corresponding unit-size square neighboring J. This will not change the stated distributional relation.) Employing (7.9), we conclude that for any $n\in{\mathbb N}$ , $n\geq 2$ ,

(7.19) \begin{align} \mathrm{Cov}\Big(V_{d-1}\Big(F_{2,n}^{ik}\cap F_{2,n}^{i,R}\Big),\widehat{Q}_k\Big) &=M^{2(1-d)}\mathrm{Cov}\Big(X_{n-1}^{i,i'},X_{n-1}^{i,\ell}\Big)=\frac {1-p}{M^{4(d-1)}}p^{4n-5}, \end{align}

while this covariance vanishes for $n\leq 1$ .

The last sum in (7.17) consists of $2(d-1)$ terms of the form $\mathrm{Cov}(V_{d-1}(F_{2,n}^{ik}\cap F_{2,n}^{im}),\widehat{Q}_k)$ , for which the corresponding cubes $J_{ik}$ and $J_{im}$ are direct neighbors. As for the previous term, we can use the relation $\widehat{Q}_k=V_{d-1}\Big(F_{2,n}^{ik}\cap F_{2,n}^{\ell\ell'}\Big)$ , where $\ell'$ was the unique index such that the cubes $J_{ik}$ and $J_{\ell \ell'}$ are direct neighbors. Hence,

\begin{align*} \mathrm{Cov}\big(V_{d-1}\big(F_{2,n}^{ik}\cap F_{2,n}^{im}\big),\widehat{Q}_k\big)&=\mathrm{Cov}\Big(V_{d-1}\Big(F_{2,n}^{ik}\cap F_{2,n}^{im}\Big),V_{d-1}\Big(F_{2,n}^{ik}\cap F_{2,n}^{\ell \ell'}\Big)\Big),\end{align*}

where now the cubes $J_{im}$ and $J_{\ell\ell'}$ have a non-empty intersection. Thus we are in the situation of the second type of covariance studied for the second term (cf. (7.12) and the paragraph before this equation), just one level further down. More precisely, we have

\begin{align*} \left(V_{d-1}\big(F_{2,n}^{ik}\cap F_{2,n}^{im}\big),\right.&\left.V_{d-1}\Big(F_{2,n}^{ik}\cap F_{2,n}^{\ell \ell'}\Big)\right)\\[5pt] &\overset{d}{=} M^{1-d} \left(V_{d-1}\Big(F_{n-1}^{i}\cap F_{n-1}^{i'}\Big),V_{d-1}\big(F_{n-1}^{i}\cap F_{n-1}^{\ell}\big)\right),\end{align*}

where now i ^′ is one of the indices such that $J_{i'}$ and $J_i$ are direct neighbors and $J_{i'}$ and $J_\ell$ are distinct and have a non-empty intersection. Employing (7.12), we conclude that for $p>M^{-d}$ and any $n\in{\mathbb N}$ , $n\geq 2$ ,

(7.20) \begin{align} \sum_{m: Q_k\sim Q_m}\mathrm{Cov}(&V_{d-1}\big(F_{2,n}^{ik}\cap F_{2,n}^{im}\big),\widehat{Q}_k) =2(d-1)M^{2(1-d)}\mathrm{Cov}\Big(X_{n-1}^{i,i'},X_{n-1}^{i,\ell}\Big) \notag\\[5pt] &= \frac{2(d-1)(1-p)}{M^{3d-4}(M^d p-1)}\cdot p^{4(n-1)}\left(1-(M^{d}p)^{-n+1}\right), \end{align}

while this covariance vanishes for $n\leq 1$ .

Now, plugging the expressions derived in (7.18), (7.19), and (7.20) into (7.17) and summing over all $k=1,...,M^{d-1}$ , we obtain the desired recursion for the covariance $\gamma_n$ in (7.16). For $n\geq 2$ and $M^dp\neq 1$ , we have

\begin{align*} \gamma_n &=\frac p{M^{d-1}} \cdot \gamma_{n-1} + (Mp^3)^{n} \frac {d(1-p)^2}{M^{3d-2}(M-p)p^4} \left(1+\frac{M-1}{1-p}\left(\frac pM\right)^{n-1}\right)\\[5pt] &\qquad -\frac {1-p}{M^{3(d-1)}}p^{4n-5}-\frac{2(d-1)(1-p)}{M^{2d-3}(M^d p-1)}\cdot p^{4(n-1)}\left(1-(M^{d}p)^{-n+1}\right)\\[5pt] &=\frac p{M^{d-1}} \cdot \gamma_{n-1} + \tilde c_1\cdot (Mp^3)^{n} +\tilde c_2\cdot p^{4n} + \tilde c_3 \cdot \left(M^{-d}p^3\right)^n, \end{align*}

where $\tilde c_1\;:\!=\;\frac {d(1-p)^2}{M^{3d-2}p^4(M-p)} $ , $\tilde c_3\;:\!=\;\frac{2(d-1)(1-p)}{M^{d-3}p^3(M^d p-1)}$ and

\begin{align*} \tilde c_2&\;:\!=\;\frac{d(M-1)(1-p)}{M^{3(d-1)}p^5(M-p)}-\frac {1-p}{M^{3(d-1)}p^5}-\frac{2(d-1)(1-p)}{M^{2d-3}(M^d p-1)p^4}\\[5pt] &=\frac{1-p}{M^{3(d-1)}p^5}\left(d\frac{M-1}{M-p}-1-2(d-1)\frac{M^dp}{M^dp-1}\right). \end{align*}

By induction, we get from the above recursion that for $n\geq 2$ ,

\begin{align*} \gamma_n =& (Mp^3)^n \tilde c_1\frac{M^dp^2}{M^dp^2-1} +p^{4n} \tilde c_2 \frac{M^{d-1}p^3}{M^{d-1}p^3-1} +\left(\frac{p^{3}}{M^d}\right)^{n} \tilde c_3 \frac{p^2}{p^2-M}\\[5pt] &-\left(\frac{p}{M^{d-1}}\right)^{n}\underbrace{\left(\tilde c_1 \frac{(M^{d}p^2)^2}{M^dp^2-1}+ \tilde c_2 \frac{(M^{d-1}p^3)^2}{M^{d-1}p^3-1}+ \tilde c_3\frac{(M^{-1}p^2)^2}{M^{-1}p^2-1}\right)}_{\;=:\;\tilde c_4}, \end{align*}

where we have used that $\gamma_1=0$ , and where we have excluded the case $p^2=M^{-d}$ , in which the constant in the first term has to be replaced by $\tilde c_1(n-1)$ and the first summand in $\tilde c_4$ by 0. In the sequel, we will only need an asymptotic expression for $\gamma_n$ as $n\to\infty$ . It turns out that which term is the leading one depends on the parameter combination. For $p>M^{-d}$ we obtain, as $n\to\infty$ ,

\begin{align*} \gamma_n &=\begin{cases} \tilde c_1\cdot \frac{M^{d}p^2}{M^{d}p^2-1} (Mp^3)^{n}+ O(\max\{p^{4}, p/M^{d-1}\}^n) & \text{ if } p>M^{-d/2},\\[5pt] \tilde c_1 \cdot (Mp^3)^{n} \cdot (n-1)+ O((Mp^3)^{n}) & \text{ if } p^2= M^{-d},\\[5pt] \tilde c_4 \cdot (p/M^{d-1})^n + O((Mp^3)^n) & \text{ if } p< M^{-d/2}. \end{cases} \end{align*}

Now we can complete the analysis of the fourth term. Inserting the last expression into Equation (7.15) and noting that the second term in (7.15) is always of order $O((Mp^{3})^n)$ , we conclude that for any $p>M^{-d}$ , as $n\to\infty$ ,

(7.21) \begin{align} \mathrm{Cov}(X_n^{i},X_n^{i,\ell})&= \begin{cases} \hat c_1\cdot(Mp^3)^{n}+ O(\max\{p^{4}, p/M^{d-1}\}^n) & \text{if } p>M^{-d/2},\\[5pt] \tilde c_1 \cdot (Mp^3)^{n} \cdot (n-1)+ O((Mp^3)^{n})& \text{if } p^2= M^{-d},\\[5pt] \tilde c_4 \cdot (p/M^{d-1})^n + O((Mp^3)^n)& \text{if } p< M^{-d/2}, \end{cases} \end{align}

where $ \hat c_1\;:\!=\;\frac{d(1-p)}{M^{2(d-1)}p(M-p)} (1+\frac{1-p}p\frac{1}{M^dp^2-1})$ .

Computation of the variance of $X_n=V_{d-1}(F_n)$ . Now that we have determined all the terms on the right-hand side of (7.5), we can go back to this equation and put all the pieces together. Setting $\beta_n\;:\!=\;\mathrm{Var}(X_n)$ and combining (7.6) with (7.5), we get

(7.22) \begin{align} \notag \beta_n &= \frac{p}{M^{d-2}}\beta_{n-1} + \frac {d^2(1-p)^3}{M^{d-2}p(M-p)^2} (Mp)^{2n} \left(1+ \frac{M-1}{1-p}\left(\frac pM\right)^n\right)^2\\[5pt] & \quad + \sum_{\{i,\ell\}\subset\Sigma}\hspace{-2mm} \mathrm{Var}(X_n^{i,\ell})+ \sum_{ \{i,\ell\}\neq\{i',\ell'\}}\hspace{-4mm} \mathrm{Cov}\Big(X_n^{i,\ell},X_n^{i',\ell'}\Big) -2\sum_{j\in\Sigma}\sum_{\{i,\ell\}\subset\Sigma}\mathrm{Cov}\left(X_n^j,X_n^{i,\ell}\right). \end{align}

Observe from (7.7) that the first sum in the second line above satisfies

\begin{align*} \sum_{\{i,\ell\}\subset\Sigma} \mathrm{Var}(X_n^{i,\ell})&= \begin{cases} O(p^{4n}), &M^{d-1}p^2>1,\\[5pt] O(p^{4n}\cdot n), &M^{d-1}p^2= 1,\\[5pt] O((p^2/M^{d-1})^n), &M^{d-1}p^2< 1, \end{cases} \end{align*}

as $n\to\infty$ . So the order of the leading term depends on whether p is below, at, or above the value $M^{-\frac{d-1}2}$ . In any case this term is certainly dominated by $p^{2n}$ as $n\to\infty$ (i.e., $\sum_{\{i,\ell\}\subset\Sigma} \mathrm{Var}(X_n^{i,\ell})=o(p^{2n})$ ) and thus vanishes rapidly for $n\to\infty$ . Equations (7.9) and (7.14) show that the situation for the second sum in (7.22) is easier. For $p>M^{-d}$ and any $n\in{\mathbb N}$ , we have

\begin{align*} \sum_{ \{i,\ell\}\neq\{i',\ell'\}} \mathrm{Cov}\Big(X_n^{i,\ell},X_n^{i',\ell'}\Big) &=\frac {2d(M-2)(1-p)}{M^{d-1}p}p^{4n} + 8\binom{d}{2}(M-1)^2\times \\[5pt] &\quad \frac{1-p}{M^d p-1}\cdot p^{4n}\left(1-(M^{d}p)^{-n}\right), \end{align*}

and thus, as $n\to\infty$ ,

\begin{align*} \sum_{ \{i,\ell\}\neq\{i',\ell'\}} \mathrm{Cov}\Big(X_n^{i,\ell},X_n^{i',\ell'}) &= O(p^{4n}\Big). \end{align*}

For the last sum in (7.22), we infer from (7.21) that in all three cases this term is certainly dominated by $(Mp)^{2n}$ as $n\to\infty$ . Thus, the second term in (7.22) (of order $(Mp)^{2n}$ ) is always the leading one. But we also need the correct second-order term in order to say something about the speed of convergence of $\mathrm{Var}(Z_n^{d-1})$ . The recursion for $\beta_n$ will produce a second term of order $(p/M^{d-2})^n$ , but it turns out that for some parameters there are others of higher order which we will therefore have to take into account in the recursion. Note that for $p^2\geq1/M^{d-1}$ we have $Mp^3\geq p/M^{d-2}$ , and thus we include in the recursion the corresponding term of order $(Mp^3)^n$ , which is the term of highest order among the remaining ones for these parameters. This is not necessary for $p^2<1/M^{d-1}$ . More precisely, we have for any $p>M^{-d}$ , as $n\to\infty$ ,

\begin{align*} \beta_n=\frac{p}{M^{d-2}}\cdot \beta_{n-1} + \tilde{c}\cdot(Mp)^{2n}+\begin{cases} \hat c_1\cdot(Mp^3)^{n}+O(q^n) &\text{if } p^2\geq M^{-(d-1)},\\[5pt] O((Mp^3)^n) & \text{if } p^2< M^{-(d-1)}, \end{cases}\end{align*}

where $\tilde{c}\;:\!=\;\frac {d^2(1-p)^3}{M^{d-2}p(M-p)^2}$ , $\hat c_1$ is the same as in (7.21), and q is some number strictly less than $Mp^3$ (depending on the parameter combination). By induction, this yields

\begin{align*} \beta_n=\tilde{c} \frac{M^{d}p}{M^{d}p-1}(Mp)^{2n} + \begin{cases} \hat c_1 \frac{M^{d-1}p^2}{M^{d-1}p^2-1}(Mp^3)^n+ O(s^n) & \text{if } p^2>1/M^{d-1},\\[5pt] \hat c_1 (Mp^3)^n \cdot n+O((Mp^3)^n) & \text{if } p^2=1/M^{d-1},\\[5pt] \bar c \left(\frac{p}{M^{d-2}}\right)^n +O((Mp^3)^n) & \text{if } p^2<1/M^{d-1},\\[5pt] \end{cases} \end{align*}

as $n\to\infty$ , where $\tilde c$ and $\hat c_1$ are as above, $s<Mp^3$ is some number, and $\bar c\in{\mathbb R}$ is some constant which we do not give explicitly here, as doing so would require us to determine all the terms in the above recursion exactly. Here we have used that $\beta_0=\mathrm{Var}(V_{d-1}(F_0))=0$ and that $p>M^{-d}$ is equivalent to $M^2p^2>p/M^{d-2}$ . Note that $\tilde{c} \frac{M^{d}p}{M^{d}p-1}=\bar c_2$ , i.e. the constant in the first-order term of $\mathrm{Var}(V_{d-1}(F_n))$ is as asserted in (3.3), and also the second-order terms given in (3.3) are transparent from the above equation. This completes the proof of the formula for the variance in Theorem 3.2.