Appendix: Miscellaneous Results
Standard Deviation Estimates
We use the following standard deviation estimates for sums of independent Poisson and Bernoulli random variables.
Lemma 13
Suppose \(W_i, 1 \le i \le m\) are independent Bernoulli random variables satisfying \(\mu _1 \le {\mathbb {P}}(W_1=1) = 1-{\mathbb {P}}(W_1~=~0) \le \mu _2.\) For any \(0< \epsilon < \frac{1}{2},\)
$$\begin{aligned} {\mathbb {P}}\left( \sum _{i=1}^{m} W_i > m\mu _2(1+\epsilon ) \right) \le \exp \left( -\frac{\epsilon ^2}{4}m\mu _2\right) \end{aligned}$$
(7.1)
and
$$\begin{aligned} {\mathbb {P}}\left( \sum _{i=1}^{m} W_i < m\mu _1(1-\epsilon ) \right) \le \exp \left( -\frac{\epsilon ^2}{4}m\mu _1\right) \end{aligned}$$
(7.2)
Estimates (7.1) and (7.2) also hold if \(\{W_i\}\) are independent Poisson random variables with \(\mu _1 \le {\mathbb {E}}W_1 \le \mu _2.\)
For completeness, we give a quick proof.
Proof of Lemma 13
First suppose that \(\{W_i\}\) are independent Poisson with
\(\mu _1 \le {\mathbb {E}}W_i \le \mu _2\) so that \({\mathbb {E}}e^{sW_i} = \exp \left( {\mathbb {E}}W_i (e^{s}-1)\right) \le \exp \left( \mu _2(e^{s}-1)\right) \) for \(s~>~0.\) By Chernoff bound, we then have
$$\begin{aligned} {\mathbb {P}}\left( \sum _{i=1}^{m} W_i > m\mu _2(1+\epsilon )\right) \le e^{-sm\mu _2(1+\epsilon )} \exp \left( m\mu _2(e^{s}-1)\right) = e^{m\mu _2 \varDelta _1}, \end{aligned}$$
where \(\varDelta _1 = e^{s} -1 -s-s\epsilon .\) For \(s \le 1,\) we have the bound
$$\begin{aligned} e^{s}-1-s = \sum _{k\ge 2} \frac{s^{k}}{k!} \le s^2\sum _{k\ge 2} \frac{1}{k!} = s^2(e-2) \le s^2 \end{aligned}$$
and so we set \(s = \frac{\epsilon }{2}\) to get that \(\varDelta _1 \le s^2-s\epsilon =\frac{-\epsilon ^2}{4}.\)
Similarly, for \(s > 0,\) we have
$$\begin{aligned} {\mathbb {E}}e^{-sW_i} = \exp \left( {\mathbb {E}}W_i(e^{-s}-1)\right) \le \exp \left( \mu _1(e^{-s}-1)\right) \end{aligned}$$
and so
$$\begin{aligned} {\mathbb {P}}\left( \sum _{i=1}^{m} W_i < m\mu _1(1-\epsilon )\right) \le e^{sm\mu _1(1-\epsilon )} \exp \left( m\mu _1(e^{-s}-1)\right) = e^{-m\mu _1\varDelta _2}, \end{aligned}$$
where \(\varDelta _2 = 1 -s-e^{-s} + s\epsilon .\) For \(s \le 1,\) the term \(e^{-s}\le 1-s + \frac{s^2}{2}\) and so we get \(\varDelta _2 \ge -\frac{s^2}{2}+s\epsilon =\frac{\epsilon ^2}{2}\) for \(s = \epsilon .\)
The proof for the Binomial distribution follows from the fact that if \(\{W_i\}_{1 \le i \le m}\) are independent Bernoulli distributed with \(\mu _1 \le {\mathbb {E}}W_i \le \mu _2,\) then for \(s > 0\) we have \({\mathbb {E}}e^{s W_i} = 1-{\mathbb {E}}W_i + e^{s} {\mathbb {E}}W_i \le \exp \left( {\mathbb {E}}W_i (e^{s}-1)\right) \le \exp \left( \mu _2(e^{s}-1)\right) \) and similarly \({\mathbb {E}}e^{-s W_i} \le \exp \left( \mu _1(e^{-s}-1)\right) .\) The rest of the proof is then as above. \(\square \)
Proof of the Monotonicity Property (2.14)
For \(\alpha \le 1\), we couple the original Poisson process \(\mathcal{P}\) and the homogenous process \(\mathcal{P}_{\delta }\) in the following way. Let \(V_{i}, i \ge 1\) be i.i.d. random variables each with density \(f(\cdot )\) and let \(N_V\) be a Poisson random variable with mean \(n,\) independent of \(\{V_i\}.\) The nodes \(\{V_i\}_{1 \le i \le N_V}\) form a Poisson process with intensity \(nf(\cdot )\) which we denote as \(\mathcal{P}\) and colour green.
Let \(U_i, i \ge 1\) be i.i.d. random variables each with density \(\epsilon _2-f(\cdot )\) where \(\epsilon _2 \ge 1\) is as in (1.1) and let \(N_U\) be a Poisson random variable with mean \(n(\epsilon _2-1).\) The random variables \((\{U_i\},N_U)\) are independent of \((\{V_i\},N_V)\) and the nodes \(\{U_i\}_{1 \le i \le N_U}\) form a Poisson process with intensity \(n(\epsilon _2-f(\cdot ))\) which we denote as \(\mathcal{P}_{ext}\) and colour red. The nodes of \(\mathcal{P}\) and \(\mathcal{P}_{ext}\) together form a homogenous Poisson process with intensity \(n\epsilon _2,\) which we denote as \(\mathcal{P}_{\delta }\) and define it on the probability space \((\varOmega _{\delta },\mathcal{F}_{\delta }, {\mathbb {P}}_{\delta }).\)
Let \(\omega _{\delta } \in \varOmega _{\delta }\) be any configuration and as above let \(\{i^{(\delta )}_{j}\}_{1 \le j \le Q_{\delta }}\) be the indices of the squares in \(\{R_j\}\) containing at least one node of \(\mathcal{P}_{\delta }\) and let \(\{i_{j}\}_{1 \le j \le Q}\) be the indices of the squares in \(\{R_j\}\) containing at least one node of \(\mathcal{P}.\) The indices in \(\{i^{(\delta )}_j\}\) and \(\{i_j\}\) depend on \(\omega _{\delta }.\) Defining \(S_{\alpha } = S_{\alpha }(\omega _{\delta })\) and \(S^{(\delta )}_{\alpha } = S_{\alpha }^{(\delta )}(\omega _{\delta })\) as before, we have that \(S_{\alpha }\) is determined only by the green nodes of \(\omega _{\delta }\) while \(S^{(\delta )}_{\alpha }\) is determined by both green and red nodes of \(\omega _{\delta }.\)
From the monotonicity property, we therefore have that \(S_{\alpha }(\omega _{\delta }) \le S^{(\delta )}_{\alpha }(\omega _{\delta })\) and so for any \(x > 0\) we have
$$\begin{aligned} {\mathbb {P}}_{\delta }(S^{(\delta )}_{\alpha }< x) \le {\mathbb {P}}_{\delta }(S_{\alpha }< x) = {\mathbb {P}}_0(S_{\alpha } < x), \end{aligned}$$
(7.3)
proving (2.14).
If \(\alpha > 1,\) we perform a slightly different analysis. Letting \(\epsilon _1 \le 1\) be as in (1.1), we construct a Poisson process \(\mathcal{P}_{ext}\) with intensity \(n(f(\cdot )-\epsilon _1)\) and colour nodes of \(\mathcal{P}_{ext}\) red. Letting \(\mathcal{P}_{\delta }\) be another independent Poisson process with intensity \(n\epsilon _1,\), we colour nodes of \(\mathcal{P}_{\delta }\) green. The superposition of \(\mathcal{P}_{ext}\) and \(\mathcal{P}_{\delta }\) is a Poisson process with intensity \(nf(\cdot ),\) which we define on the probability space \((\varOmega _{\delta },\mathcal{F}_{\delta },{\mathbb {P}}_{\delta }).\) In this case, the sum \(S_{\alpha }\) is determined by both green and red nodes while \(S^{(\delta )}_{\alpha }\) is determined only by the green nodes. Again using the monotonicity property of \(S_{\alpha },\) we get (7.3). \(\square \)
Additive Relations
If \(\mathrm{TSP}(x_1,\ldots ,x_j), j \ge 1\) denotes the length of the TSP cycle with vertex set \(\{x_1,\ldots ,x_j\},\) then for any \(k \ge 1,\) we have
$$\begin{aligned}&\mathrm{TSP}(x_1,\ldots ,x_{j+k}) \le \mathrm{TSP}(x_1,\ldots ,x_j) \nonumber \\&\quad + \mathrm{TSP}(x_{j+1},\ldots ,x_{j+k}) + (c_2\sqrt{2})^{\alpha } \end{aligned}$$
(7.4)
and if \( \alpha \le 1\) and the edge weight function \(h\) is a metric, then
$$\begin{aligned} \mathrm{TSP}(x_1,\ldots ,x_j) \le \mathrm{TSP}(x_1,\ldots ,x_{j+k}). \end{aligned}$$
(7.5)
Proof of (7.4) and (7.5)
To prove (7.4), suppose \(\mathcal{C}_{1}\) is the minimum weight spanning cycle formed by the nodes \(\{x_l\}_{1 \le l \le j}\) and \(\mathcal{C}_2\) is the minimum weight spanning cycle formed by \(\{x_l\}_{j+1 \le l \le j+k}.\) Let \(e_1 = (u_1,v_1) \in \mathcal{C}_1\) and \(e_2 = (u_2,v_2) \in \mathcal{C}_2\) be any two edges. The cycle
$$\begin{aligned} \mathcal{C}_{tot} = \left( \mathcal{C}_1 \setminus \{e_1\}\right) \cup \left( \mathcal{C}_2 \setminus \{e_2\}\right) \cup \{(u_1,u_2), (v_1,v_2)\} \end{aligned}$$
obtained by removing the edges \(e_1,e_2\) and adding the “cross”-edges \((u_1,u_2)\) and \((v_1,v_2)\) is a spanning cycle containing all the nodes \(\{x_l\}_{1 \le l \le j+k}.\) The edges \((u_1,u_2)\) and \((v_1,v_2)\) have a Euclidean length of at most \(\sqrt{2}\) and so a weight of at most \((c_2\sqrt{2})^{\alpha }\) using the bounds for the metric \(h\) in (1.2). This proves (7.4).
It suffices to prove (7.5) for \(k =1.\) Let \(\mathcal{C} = (y_1,\ldots ,y_{j+1},y_1)\) be any cycle with vertex set \(\{y_i\}_{1 \le i \le j+1} = \{x_i\}_{1 \le i \le j+1}\) and without loss of generality suppose that \(y_{j+1} = x_{j+1}.\) Removing the edges \((y_j,y_{j+1})\) and \((y_{j+1},y_1),\) and adding the edge \((y_1,y_j)\) we get a new cycle \(\mathcal{C}'\) with vertex set \(\{x_i\}_{1 \le i \le j}.\)
Since the edge weight function \(h\) is a metric, we have by triangle inequality that \(h(y_j,y_1) \le h(y_j,y_{j+1}) + h(y_{j+1},y_1).\) Using \((a+b)^{\alpha } \le a^{\alpha } + b^{\alpha }\) for \(a,b > 0\) and \(0 < \alpha \le 1\), we get that \(h^{\alpha }(y_j,y_1) \le h^{\alpha }(y_j,y_{j+1}) + h^{\alpha }(y_{j+1},y_1).\) Therefore, the weight \(W(\mathcal{C}')\) of \(\mathcal{C}'\)
$$\begin{aligned} W(\mathcal{C'})= & {} \sum _{i=1}^{j-1}h^{\alpha }(y_i,y_{i+1}) + h^{\alpha }(y_j,y_1) \\&\le \sum _{i=1}^{j} h^{\alpha }(y_i,y_{i+1}) + h^{\alpha }(y_{j+1},y_1) = W(\mathcal{C}). \end{aligned}$$
Therefore, \(\mathrm{TSP}(x_1,\ldots ,x_j) \le W(\mathcal{C}') \le W(\mathcal{C}).\) Taking minimum over all cycles \(\mathcal{C}\) with vertex set \(\{x_i\}_{1 \le i \le j+1},\) we get (7.5) for \(k=1.\) \(\square \)
Moments of Random Variables
Let \(X \ge 1\) be any integer valued random variable such that
$$\begin{aligned} {\mathbb {P}}(X \ge l) \le e^{-\theta (l-1)} \end{aligned}$$
(7.6)
for all integers \(l \ge 1\) and some constant \(\theta > 0\) not depending on \(l.\) For every integer \(r \ge 1,\)
$$\begin{aligned} {\mathbb {E}}X^{r} \le r\sum _{l\ge 1} l^{r-1} {\mathbb {P}}(X \ge l) \le r\sum _{l \ge 1} l^{r-1} e^{-\theta (l-1)} \le \frac{r!}{(1-e^{-\theta })^{r}} \end{aligned}$$
(7.7)
Proof of (7.7)
For \(r \ge 1\), we have
$$\begin{aligned} {\mathbb {E}}X^{r} = \sum _{l \ge 1} l^{r} {\mathbb {P}}(X= l) = \sum _{l \ge 1}l^{r} {\mathbb {P}}(X \ge l) - l^{r}{\mathbb {P}}(X \ge l+1) \end{aligned}$$
(7.8)
and substituting the \(l^{r}\) in the final term of (7.8) with \( (l+1)^{r} - ((l+1)^{r}-l^{r})\), we get
$$\begin{aligned} {\mathbb {E}}X^r= & {} \sum _{l \ge 1} \left( l^{r} {\mathbb {P}}(X \ge l) - (l+1)^{r} {\mathbb {P}}(X \ge l+1)\right) \nonumber \\&+ \;\;\;\sum _{l \ge 1} ((l+1)^{r}-l^{r}) {\mathbb {P}}(X \ge l+1) \nonumber \\&= 1 + \sum _{l \ge 1}((l+1)^{r}-l^{r}) {\mathbb {P}}(X \ge l+1) \nonumber \\&= \sum _{l \ge 0}((l+1)^{r}-l^{r}) {\mathbb {P}}(X \ge l+1) \end{aligned}$$
(7.9)
where the second equality is true since \(l^{r}{\mathbb {P}}(X \ge l) \le l^{r}e^{-\theta (l-1)} \longrightarrow 0\) as \(l~\rightarrow ~\infty .\) Using \((l+1)^{r} - l^{r} \le r\cdot (l+1)^{r-1}\) in (7.9), we get the first relation in (7.7).
We prove the second relation in (7.7) by induction as follows. Let \(\gamma = e^{-\theta } < 1\) and \(J_r := \sum _{l \ge 1} l^{r-1} \gamma ^{l-1}\) so that
$$\begin{aligned} J_{r+1}(1 - \gamma ) = \sum _{l \ge 1} l^{r} \gamma ^{l-1} - \sum _{l \ge 1}l^{r} \gamma ^{l} = \sum _{l \ge 1} \left( l^{r}-(l-1)^{r} \right) \gamma ^{l-1}. \end{aligned}$$
Using \(l^{r}-(l-1)^r \le r\cdot l^{r-1}\) for \(l \ge 1\) we therefore get that
$$\begin{aligned} J_{r+1}(1-\gamma ) \le r\sum _{l \ge 1}l^{r-1} \gamma ^{l-1} = rJ_r \end{aligned}$$
and so the second relation in (7.7) follows from induction. \(\square \)
Scaling and Translation Relations
For a set of nodes \(\{x_1,\ldots ,x_n\}\) in the unit square \(S,\) recall from Sect. 1 that \(K_n(x_1,\ldots ,x_n)\) is the complete graph formed by joining all the nodes by straight line segments and the edge \((x_i,x_j)\) is assigned a weight of \(d^{\alpha }(x_i,x_j),\) where \(d(x_i,x_j)\) is the Euclidean length of the edge \((x_i,x_j).\) We denote \(\mathrm{TSP}(x_1,\ldots ,x_n)\) to be the length of the minimum spanning cycle of \(K_n(x_1,\ldots ,x_n)\) with edge weights obtained as in (1.4),
\(\underline{Scaling }\): For any \(a > 0,\) consider the graph \(K_n(ax_1,\ldots ,ax_n)\) where the length of the edge between the vertices \(ax_1\) and \(ax_2\) is simply \(a\) times the length of the edge between \(x_1\) and \(x_2\) in the graph \(K_n(x_1,\ldots ,x_n)\) Using the definition of TSP in (1.4), we then have \(\mathrm{TSP}(ax_1,\ldots ,ax_n) = a^{\alpha } \mathrm{TSP}(x_1,\ldots ,x_n)\) and so if \(Y_1,\ldots ,Y_n\) are \(n\) nodes uniformly distributed in the square \(aS\) of side length \(a,\) then
$$\begin{aligned} \mathrm{TSP}(n;a) := \mathrm{TSP}(Y_1,\ldots ,Y_n) = a^{\alpha } \mathrm{TSP}(X_1,\ldots ,X_n), \end{aligned}$$
where \(X_i = \frac{Y_i}{a}, 1 \le i \le n\) are i.i.d. uniformly distributed in \(S.\) Recalling the notation \(\mathrm{TSP}_n = \mathrm{TSP}(X_1,\ldots ,X_n)\) from (1.4), we therefore get
$$\begin{aligned} {\mathbb {E}} \mathrm{TSP}(n;a) = a^{\alpha } {\mathbb {E}}\mathrm{TSP}_n. \end{aligned}$$
(7.10)
\(\underline{Translation }\): For \(b \in {\mathbb {R}}^2\), consider the graph \(K_n(x_1+b,\ldots ,x_n+b).\) Using the translation property \((b2),\) the weight \(h(x_1+b,x_2+b) \le h_0 \cdot h(x_1,x_2),\) the weight of the edge between \(x_1\) and \(x_2.\) Using the definition of TSP in (1.4), we therefore have
$$\begin{aligned} \mathrm{TSP}(x_1+b,\ldots ,x_n+b) \le h_0^{\alpha } \cdot \mathrm{TSP}(x_1,\ldots ,x_n), \end{aligned}$$
(7.11)
obtaining the desired bound. \(\square \)