Affiliation weighted networks with a differentially private degree sequence


Affiliation network is one kind of two-mode social network with two different sets of nodes (namely, a set of actors and a set of social events) and edges representing the affiliation of the actors with the social events. The asymptotic theorem of a differentially private estimator of the parameter in the private \(p_{0}\) model has been established. However, the \(p_{0}\) model only focuses on binary edges for one-mode network. In many case, the connections in many affiliation networks (two-mode) could be weighted, taking a set of finite discrete values. In this paper, we derive the consistency and asymptotic normality of the moment estimators of parameters in affiliation finite discrete weighted networks with a differentially private degree sequence. Simulation studies and a real data example demonstrate our theoretical results.

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We are very grateful to two anonymous referees and the Editor for their valuable comments that have greatly improved the manuscript. Luo’s research is partially supported by National Natural Science Foundation of China(No.11801576) and by the Fundamental Research Funds for the Central Universities(South-Central University for Nationalities(CZQ19010)).

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Correspondence to Tour Liu.

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In this appendix, we will present the proofs of Theorems 1 and 2. We start with some preliminaries. For a vector \({\mathbf {x}}=(x_1,\dots ,x_n)^{T}\in \mathbb {R}^n\), denote the \(\ell _{\infty }\) norm of \({\mathbf {x}}\) by \(\Vert {\mathbf {x}}\Vert _{\infty }=\max _{1\le i\le n}\mid x_i\mid \). For an \(n\times n\) matrix \(J=(J_{i,j}), \Vert J\Vert _{\infty }\) denotes the matrix norm induced by the \(\Vert \cdot \Vert _{\infty }\)-norm on vectors in \(\mathbb {R}^n\):

$$\begin{aligned} \Vert J\Vert _{\infty }=\max _{{\mathbf {x}}\ne 0}\frac{\Vert J{\mathbf {x}}\Vert _{\infty }}{\Vert {\mathbf {x}}\Vert _{\infty }}=\max _{1\le i\le n}\sum _{j=1}^{n}\mid J_{i,j}\mid . \end{aligned}$$

Let D be an open convex subset of \(\mathbb {R}^n\). We say an \(n\times n\) function matrix \(F({\mathbf {x}})\) whose elements \(F_{ij}({\mathbf {x}})\) are functions on vectors \({\mathbf {x}}\), is Lipschitz continuous on D if there exists a real number \(\lambda \) such that for any \({\mathbf {v}}\in R^n\) and any \(\mathbf {x,y}\in D,\)

$$\begin{aligned} \parallel F({\mathbf {x}})({\mathbf {v}})-F({\mathbf {y}})({\mathbf {v}})\parallel _{\infty }\le \lambda \parallel \mathbf {x-y}\parallel _{\infty }\parallel {\mathbf {v}}\parallel _{\infty }, \end{aligned}$$

where \(\lambda \) may depend on n but independent of \({\mathbf {x}}\) and \({\mathbf {y}}\). For fixed \(n,\lambda \) is a constant.

We present Lemma 1–3 in Yan et al. (2016) stated as three lemmas here, which will be used in the proofs.

Lemma 2

If \(V\in {{\mathcal {L}}}_{mn}(q,Q)\) with \(Q/q=o(n),\) then for large enough n,

$$\begin{aligned} \Vert V^{-1}-S \Vert \le \frac{c_1Q^2}{q^3mn}, \end{aligned}$$

where \(c_1\) is a constant that dose not depend on Mm and n, and \(\Vert A \Vert := \max _{i,j}|a_{i,j}|\) for a general matrix \(A=(a_{i,j})\).

Note that if Q and q are bounded constants, then the upper bound of the above approximation error is on the order of \((mn)^{-1},\) indicating that S is a high-accuracy approximation to \(V^{-1}.\) Further, based on the above proposition, we immediately have the following lemma.

Lemma 3

If \(V\in {{\mathcal {L}}}_{mn}(q,Q)\) with \(Q/q=o(n),\) then for a vector \({\mathbf {x}}\in R^{m+n-1},\)

$$\begin{aligned} \parallel V^{-1}{\mathbf {x}}\parallel _{\infty }\le \frac{2c_1Q^2}{q^3mn}+\frac{\mid x_{m+n}\mid }{v_{m+n,m+n}}+\max _{i=1,\dots ,m+n-1}\frac{\mid x_{i}\mid }{v_{i,i}}, \end{aligned}$$

where \(x_{m+n}:=\sum _{i=1}^{m}x_{i}-\sum _{i=m+1}^{m+n-1}x_{i}.\)

Lemma 4

Define a system of equations:

$$\begin{aligned} \begin{array}{rcl} F_i({\varvec{\theta }})&{}=&{}d_i-\sum _{k=1}^{n}f(\alpha _i+\beta _k),i=1,\dots ,m,\\ F_{m+j}({\varvec{\theta }})&{}=&{}b_j-\sum _{k=1}^{m}f(\alpha _k+\beta _j),j=1,\dots ,n-1,\\ F({\varvec{\theta }})&{}=&{}(F_1({\varvec{\theta }}),\dots ,F_m({\varvec{\theta }}),F_{m+1} ({\varvec{\theta }}),\dots ,F_{m+n-1}({\varvec{\theta }}))^T, \end{array} \end{aligned}$$

where \(f(\cdot )\) is a continuous function with the third derivative. Let \(D\subset \mathbb {R}^{m+n-1}\) be a convex set and assume for any \(\mathbf {x,y,v}\in D\), we have

$$\begin{aligned}&\parallel [F^{'}({\mathbf {x}})-F^{'}({\mathbf {y}})]{\mathbf {v}}\parallel _{\infty }\le K_1\parallel \mathbf {x-y}\parallel _{\infty }\parallel {\mathbf {v}}\parallel _{\infty },\\&\quad \max _{i=1,\dots ,m+n-1}\parallel F^{'}_i({\mathbf {x}})-F^{'}_i({\mathbf {y}})\parallel _{\infty }\le K_2\parallel \mathbf {x-y}\parallel _{\infty }, \end{aligned}$$

where \(F^{'}({\varvec{\theta }})\) is the Jacobin matrix of F on \({\varvec{\theta }}\) and \(F^{'}_{i}({\varvec{\theta }})\) is the gradient function of \(F_i\) on \({\varvec{\theta }}.\) Consider \({\varvec{\theta }}^{(0)}\in D\) with \(\Omega ({\varvec{\theta }}^{(0)},2\xi )\subset D\) where \(\xi =\parallel [F^{'}({\varvec{\theta }}^{(0)})]^{-1}F({\varvec{\theta }}^{(0)})\parallel _{\infty }\) for any \({\varvec{\theta }} \in \Omega ({\varvec{\theta }}^{(0)},2\xi ),\) we assume

$$\begin{aligned} F^{'}({{\varvec{\theta }}})\in {{\mathcal {L}}}_{mn}(q,Q)~~~~or~~-F^{'}({{\varvec{\theta }}})\in {{\mathcal {L}}}_{mn}(q,Q). \end{aligned}$$

For \(k=1,2,\dots ,\) define the Newton iterates \({\varvec{\theta }}^{(k+1)}={\varvec{\theta }}^{(k)}-[F^{'}({\varvec{\theta }}^{(k)})]^{-1}F({\varvec{\theta }}^{(k)}).\) Let

$$\begin{aligned} \rho =\frac{c_1(m+n-1)Q^2K_1}{2q^3mn}+\frac{K_2}{mq}. \end{aligned}$$

If \(\xi \rho <1/2,\) then \({\varvec{\theta }}^{(k)} \in \Omega ({\varvec{\theta }}^{(0)},2r),k=1,2,\dots ,\) are well defined and satisfy

$$\begin{aligned} \parallel {\varvec{\theta }}^{(k+1)}-{\varvec{\theta }}^{(0)}\parallel _{\infty }\le \xi /(1-\rho \xi ). \end{aligned}$$

Further, \(\lim _{k\rightarrow \infty }{\varvec{\theta }}^{(k)}\) exists and the limiting point is precisely the solution of \(F({\varvec{\theta }})=0\) in the rage of \({\varvec{\theta }} \in \Omega ({\varvec{\theta }}^{(0)},2\xi ).\)

Appendix A: Proof for Theorem 1

We define a system of functions:

$$\begin{aligned} \begin{array}{rcl} F_{i}({\varvec{\theta }})&{}=&{} {\tilde{d}}_i-\sum _{j=1}^{n}\frac{\sum _{k=0}^{r-1}ke^{k(\alpha _{i} + \beta _{j})}}{\sum _{k=0}^{r-1}e^{k(\alpha _{i} + \beta _{j})}} ,\quad i=1,\dots ,m,\\ F_{m+j}({\varvec{\theta }})&{}=&{} {\tilde{b}}_j-\sum _{i=1}^{m}\frac{\sum _{k=0}^{r-1}ke^{k(\alpha _{i} + \beta _{j})}}{\sum _{k=0}^{r-1}e^{k(\alpha _{i} + \beta _{j})}},\quad j=1,\dots ,n-1\\ F({\varvec{\theta }})&{}=&{} (F_{1}({{\varvec{\theta }}}),\dots ,F_{m+n-1}({{\varvec{\theta }}}))^T. \end{array} \end{aligned}$$

Note that the solution to the equation \(F({\varvec{\theta }})=0\) is precisely the private-parameter estimators. Then the Jacobin matrix \(F^{'}({\varvec{\theta }})\) of \(F({\varvec{\theta }})\) can be calculated as follows. For \(i=1,\dots ,m,\)

$$\begin{aligned}&\frac{\partial {F_i}}{\partial {{\alpha _l}}}=0,l=1,\dots ,m,l\ne i; \frac{\partial {F_i}}{\partial {{\alpha _i}}}=-\sum ^{n}_{j=1}\frac{\sum _{k=0}^{r-1}\sum _{l=k+1}^{r-1}(k-l)^{2}e^{(k+l)(\alpha _{i} + \beta _{j})}}{(\sum _{k=0}^{r-1}e^{k(\alpha _{i} + \beta _{j})})^2}, \\&\frac{\partial {F_i}}{\partial { {\beta _j}}}=-\frac{\sum _{k=0}^{r-1}\sum _{l=k+1}^{r-1}(k-l)^{2}e^{(k+l)(\alpha _{i} + \beta _{j})}}{(\sum _{k=0}^{r-1}e^{k(\alpha _{i} + \beta _{j})})^2},\quad j=1,\dots ,n-1 \end{aligned}$$

and for \(j=1,\dots ,n-1\)

$$\begin{aligned}&\frac{\partial {F_{m+j}}}{\partial { {\alpha _l}}}=-\frac{\sum _{k=0}^{r-1}\sum _{l=k+1}^{r-1}(k-l)^{2}e^{(k+l)(\alpha _{i} + \beta _{j})}}{(\sum _{k=0}^{r-1}e^{k(\alpha _{i} + \beta _{j})})^2},\quad l=1,\dots ,m,\\&\frac{\partial {F_{m+j}}}{\partial {{\beta _j}}}=-\sum ^{m}_{i=1}\frac{\sum _{k=0}^{r-1}\sum _{l=k+1}^{r-1}(k-l)^{2}e^{(k+l)(\alpha _{i} + \beta _{j})}}{(\sum _{k=0}^{r-1}e^{k(\alpha _{i} + \beta _{j})})^2};\frac{\partial {F_{m+j}}}{\partial {{\beta _k}}}=0,\\&\quad k=1,\dots ,n-1,k\ne j. \end{aligned}$$

Since \(2\Vert \varvec{\theta }\Vert _{\infty } \ge \alpha _{i} + \beta _{j}\), we have:

$$\begin{aligned} e^{2k(\alpha _{i} + \beta _{j})} \le {\left\{ \begin{array}{ll} e^{\alpha _{i} + \beta _{j}+2\Vert \varvec{\theta }\Vert _{\infty }}&{}\quad when ~~~ k =0,\\ e^{(k+(k-1))(\alpha _{i} + \beta _{j})+2\Vert \varvec{\theta }\Vert _{\infty }}&{}\quad when ~~~ k = 1,\ldots ,r-2, \\ e^{[(r-1)+(r-2)](\alpha _{i} + \beta _{j})+2\Vert \varvec{\theta }\Vert _{\infty }}&{}\quad when ~~~ k = r-1. \end{array}\right. } \end{aligned}$$

Therefore, we have

$$\begin{aligned} \begin{aligned} e^{0\times (\alpha _{i} + \beta _{j})}&\le \sum _{k=0}^{0}\sum _{l=1}^{r-1}e^{(k+l)(\alpha _{i} + \beta _{j})}e^{2\Vert \varvec{\theta }\Vert _{\infty }} \\ \sum _{k=1}^{r-2}e^{2k(\alpha _{i} + \beta _{j})}&\le \sum _{k=1}^{r-1}\sum _{l=0}^{k-1}e^{(k+l)(\alpha _{i} + \beta _{j})}e^{2\Vert \varvec{\theta }\Vert _{\infty }} \\ e^{2(r-1)(\alpha _{i} + \beta _{j})}&\le \sum _{k=r-2}^{r-2}\sum _{l=k+1}^{r-1}e^{(k+l)(\alpha _{i} + \beta _{j})}e^{2\Vert \varvec{\theta }\Vert _{\infty }} \end{aligned} \end{aligned}$$

Combing the above three inequalities, it yields such that

$$\begin{aligned} \begin{aligned} \sum _{k=0}^{r-1}e^{2k(\alpha _{i} + \beta _{j})}&\le \sum _{k=0}^{0}\sum _{l=1}^{r-1}e^{(k+l)(\alpha _{i} + \beta _{j})}e^{2\Vert \varvec{\theta }\Vert _{\infty }} + \sum _{k=1}^{r-1}\sum _{l=0}^{k-1}e^{(k+l)(\alpha _{i} + \beta _{j})}e^{2\Vert \varvec{\theta }\Vert _{\infty }} \\&\quad + \sum _{k=r-2}^{r-2}\sum _{l=k+1}^{r-1}e^{(k+l)(\alpha _{i} + \beta _{j})}e^{2\Vert \varvec{\theta }\Vert _{\infty }}\\&\le \sum _{0\le k \ne l\le r-1} e^{(k+l)(\alpha _{i} + \beta _{j})}e^{2\Vert \varvec{\theta }\Vert _{\infty }} \end{aligned} \end{aligned}$$

Here, the notation “\(\sum _{0\le k \ne l\le r-1}\)” is a shorthand for the double summation “\(\sum _{k=0}^{r-1}\sum _{l=0, l\ne k}^{r-1}\)”. Therefore,

$$\begin{aligned} \begin{aligned}&\frac{\frac{1}{2}\sum _{k\ne l}(k-l)^{2}e^{(k+l)(\alpha _{i} + \beta _{j})}}{(\sum _{k=0}^{r-1}e^{k(\alpha _{i} + \beta _{j})})^2} \ge \frac{\frac{1}{2}\sum _{k\ne l}e^{(k+l)(\alpha _{i} + \beta _{j})}}{(\sum _{k=0}^{r-1}e^{k(\alpha _{i} + \beta _{j})})^2}\\&= \frac{\frac{1}{2}\sum _{k\ne l}e^{(k+l)(\alpha _{i} + \beta _{j})}}{\sum _{k \ne l}e^{(k+l)(\alpha _{i} + \beta _{j})}+\sum _{k=0}^{r-1}e^{2k(\alpha _{i} + \beta _{j})}} \ge \frac{\sum _{k\ne l}e^{(k+l)(\alpha _{i} + \beta _{j})}}{2(1+e^{\Vert \varvec{\theta }\Vert _{\infty }})\sum _{k\ne l}e^{(k+l)(\alpha _{i} + \beta _{j})}}\\&=\frac{1}{2(1+e^{\Vert \varvec{\theta }\Vert _{\infty }})} \end{aligned} \end{aligned}$$

On the other hand, it is easy to verify that

$$\begin{aligned} \frac{\frac{1}{2}\sum _{k \ne l}(k-l)^{2}e^{(k+l)(\alpha _{i} + \beta _{j})} }{{(\sum _{k=0}^{r-1}e^{k(\alpha _{i} + \beta _{j})})^2}} \le \frac{1}{2}\max _{k \ne l}(k-l)^{2} \le \frac{(r-1)^{2}}{2} \end{aligned}$$

Consequently, when \(\varvec{\theta } \in \Omega (\varvec{\theta }^{*},2\xi )\), for any \(i \ne j\) , we have

$$\begin{aligned} \frac{1}{2(1+e^{2(\Vert \varvec{\theta }\Vert _{\infty }+2\xi )})} \le - F_{i,j}^{'} \le \frac{(r-1)^{2}}{2} \end{aligned}$$

According to definition of \(L_{mn}(q,Q)\), we have that \(-F^{'}_{i,j} \in L_{mn}(q,Q)\), where

$$\begin{aligned} m = \frac{1}{2(1+e^{2(\Vert \varvec{\theta }\Vert _{\infty }+2\xi )})},\quad M = \frac{(r-1)^{2}}{2} \end{aligned}$$

The constants \(K_{1}, K_{2}\) and r in the upper bounds of Lemma 4 are given below.

Lemma 5

Take \(D = R^{m+n-1}\) and \(\varvec{\theta }^{0} = \varvec{\theta }^{*}\) in Lemma 4. Assume

$$\begin{aligned} \max \{\max _{i=1,\dots ,m}|{\tilde{d}}_i-\mathbb {E}(d_i)|,\max _{j=1,\dots ,n}|{\tilde{b}}_j-\mathbb {E}(b_j)|\}= O_p( \sqrt{n\log n } + \kappa \sqrt{\log n} ). \end{aligned}$$

Then we can choose the constants K1, K2 and r in Lemma 4 as

$$\begin{aligned} K_1=n,K_2=n/2,\xi \le \frac{(\log n)^{1/2}}{n^{1/2}}\left( c_{11}e^{6\parallel {\varvec{\theta }}^{*}\parallel _{\infty }}+c_{12} e^{2\parallel {\varvec{\theta }}^{*}\parallel _{\infty }}\right) , \end{aligned}$$

where \(c_{11},c_{12}\) are constants.


For fixed mn, we first derive \(K_1\) and \(K_2\) in the inequalities of Lemma 4. Let \({\mathbf {x}},{\mathbf {y}}\in R^{m+n-1}\) and

$$\begin{aligned} {F_i}^{'}({\varvec{\theta }})=({F^{'}_{i,1}}({\varvec{\theta }}),\dots ,{F^{'}_{i,m+n-1}}({\varvec{\theta }}):= (\frac{\partial {F_i}}{\partial {\alpha _1}},\dots , \frac{\partial {F_i}}{\partial {\alpha _m}},\frac{\partial {F_i}}{\partial {\beta _1}},\dots , \frac{\partial {F_i}}{\partial {\beta _{n-1}}}). \end{aligned}$$

Then, for \(i=1,\dots ,m\), we have

$$\begin{aligned} \frac{\partial ^2{F_i}}{\partial {{\alpha _l}}\partial { {\alpha _s}}}&=0,\quad s\ne l ,\\ \frac{\partial ^2{F_i}}{\partial {{\alpha _i}}^2}&= - \sum ^{n}_{j=1}\frac{1}{2}\frac{\sum _{k=0}^{r-1}\sum _{l=0}^{r-1}\sum _{s=0}^{r-1}(k-l)^{2}(k+l-2s)e^{(k+l+s)(\alpha _{i}+\beta _{j})}}{\left( \sum _{k=0}^{r-1}e^{k(\alpha _{i} + \beta _{j})}\right) ^3},\\ \frac{\partial ^2{F_i}}{\partial { {\alpha _i}}\partial { {\beta _s}}}&=-\frac{1}{2}\frac{\sum _{k=0}^{r-1}\sum _{l=0}^{r-1}\sum _{s=0}^{r-1}(k-l)^{2}(k+l-2s)e^{(k+l+s)(\alpha _{i}+\beta _{j})}}{\left( \sum _{k=0}^{r-1}e^{k(\alpha _{i} + \beta _{j})}\right) ^3},\\&\quad s=1,\dots ,n-1;\\ \frac{\partial ^2{F_{i}}}{\partial {{\beta _l}}\partial { {\beta _s}}}&=0,s\ne l ,\\ \frac{\partial ^2{F_i}}{\partial {{\beta _j}}^2}&=-\frac{1}{2}\frac{\sum _{k=0}^{r-1}\sum _{l=0}^{r-1}\sum _{s=0}^{r-1}(k-l)^{2}(k+l-2s)e^{(k+l+s)(\alpha _{i}+\beta _{j})}}{\left( \sum _{k=0}^{r-1}e^{k(\alpha _{i} + \beta _{j})}\right) ^3},\\&\quad j=1,\dots ,n-1. \end{aligned}$$

Since \(k+l-2s \le 2(r-1)\) and

$$\begin{aligned} \sum _{k=0}^{r-1}\sum _{k=0}^{r-1}\sum _{k=0}^{r-1}e^{(k+l+s)(\alpha _{i}+\beta _{j})} = \left( \sum _{k=0}^{r-1}e^{k(\alpha _{i}+\beta _{j})}\right) ^{3} \end{aligned}$$

We have

$$\begin{aligned} \begin{aligned}&\frac{1}{2}\frac{\sum _{k=0}^{r-1}\sum _{l=0}^{r-1}\sum _{s=0}^{r-1}(k-l)^{2}(k+l-2s)e^{(k+l+s)(\alpha _{i}+\beta _{j})}}{\left( \sum _{k=0}^{r-1}e^{k(\alpha _{i} + \beta _{j})}\right) ^3}\\&\quad \le \frac{(r-1)^{3}\sum _{k=0}^{r-1}\sum _{l=0}^{r-1}\sum _{s=0}^{r-1}e^{(k+l+s)(\alpha _{i}+\beta _{j})}}{\left( \sum _{k=0}^{r-1}e^{k(\alpha _{i} + \beta _{j})}\right) ^3}\\&\quad =(r-1)^{3} \end{aligned} \end{aligned}$$

By the mean value theorem for vector-valued functions (Lang 1993, p.341), we have

$$\begin{aligned} F^{'}_i({\mathbf {x}})-F^{'}_i({\mathbf {y}})=J^{(i)}({\mathbf {x}}-{\mathbf {y}}), \end{aligned}$$


$$\begin{aligned} J^{(i)}_{s,l}=\int ^1_0\frac{\partial F^{'}_{i,s}}{\partial {{\varvec{\theta }}}_l}(t{\mathbf {x}}+(1-t){\mathbf {y}})dt,~~s,l=1,\dots ,m+n-1. \end{aligned}$$


$$\begin{aligned} \max _s\sum _{l}^{m+n-1}\mid J^{(i)}\mid \le 2(n-1)(r-1)^{3}~~, ~~ \sum ^{}_{s,l}\mid J^{(i)}_{(s,l)}\mid \le 4(n-1)(r-1)^{3} \end{aligned}$$

Similarly, for \(i=m+1,\dots ,m+n-1,\) we also have \(F^{'}_i({\mathbf {x}})-F^{'}_i({\mathbf {y}})=J^{(i)}({\mathbf {x}}-{\mathbf {y}})\) and \(\sum ^{}_{s,l}\mid J^{(i)}_{(s,l)}\mid \le 4(m-1)(r-1)^{3}.\)


$$\begin{aligned} \parallel F^{'}_i({\mathbf {x}})-F^{'}_i({\mathbf {y}})\parallel _{\infty }\le & {} \parallel J^{(i)}\parallel _{\infty }\parallel \mathbf {x-y}\parallel _{\infty }\le 2(n-1)(r-1)^{3}\parallel \mathbf {x-y}\parallel _{\infty },\\&i=1,\dots ,m+n-1, \end{aligned}$$

and for \(\forall ~{\mathbf {v}}\in R^{m+n-1}\),

$$\begin{aligned} \begin{array}{rcl} \parallel [F^{'}_i({\mathbf {x}})-F^{'}_i({\mathbf {y}})]{\mathbf {v}}\parallel _{\infty }&{}=&{}\displaystyle \max _{i}\mid \sum _{j=1}^{m+n-1}(F_{i,j}^{'}({\mathbf {x}})-F_{i,j}^{'}({\mathbf {y}}))v_j\mid \\ &{}=&{}\displaystyle \max _{i}\mid (\mathbf {x-y})J^{(i)}{\mathbf {v}}\mid \\ &{}\le &{}\parallel \mathbf {x-y}\parallel _{\infty }\parallel {\mathbf {v}}\parallel _{\infty }\sum ^{}_{k,j}\mid J^{(i)}_{(s,l)}\mid \\ &{}\le &{} 4(n-1)(r-1)^{3} \parallel \mathbf {x-y}\parallel _{\infty }\parallel {\mathbf {v}}\parallel _{\infty } \end{array} \end{aligned}$$

so we choose \(K_1=4(n-1)(r-1)^{3}\) and \(K_2=2(n-1)(r-1)^{3}\) in the inequalities of Lemma 4.

It’s obvious that \(-F^{'}({ {\varvec{\theta }}}^{*})\in \mathcal{L}_{mn}(q_{*},Q_{*})\) where,

$$\begin{aligned} q_{*}=\frac{1}{2(1+e^{2\parallel {{\varvec{\theta }}^{*}}\parallel _{\infty }})},\quad Q_{*}=\frac{(r-1)^{2}}{2}. \end{aligned}$$

Note that

$$\begin{aligned} F( {\varvec{\theta }}^*)=(d_1-\mathbb {E}(d_1),\dots ,d_m-\mathbb {E}(d_m),b_1-\mathbb {E}(b_1),\dots ,b_{n-1}-\mathbb {E}(b_{n-1})). \end{aligned}$$

By Lemmas 2 and 3, we have

$$\begin{aligned} \xi =\parallel [F^{'}( {\varvec{\theta }}^*)]^{-1}F( {\varvec{\theta }}^*)\parallel _{\infty }&\le \frac{2c_1(m+n-1)Q_{*}^2\parallel F( {\varvec{\theta }}^*)\parallel _{\infty }}{q_*^3mn}\\&\quad +\displaystyle \max _{i=1,\dots ,m+n-1}\frac{\mid F_{i}( {\varvec{\theta }}^*)\mid }{v_{i,i}}+\frac{\mid F_{m+n}( {\varvec{\theta }}^*)\mid }{v_{m+n,m+n}}\\&\le \frac{(\log n)^{1/2}}{n^{1/2}}\left( c_{11}e^{6\parallel {\varvec{\theta }}^{*}\parallel _{\infty }}+c_{12} e^{2\parallel {\varvec{\theta }}^{*}\parallel _{\infty }}\right) \end{aligned}$$

where \(c_{11},c_{12}\) are constants. \(\square \)

We present several results that we will use in the following lemmas. Recall that a random variable X is sub-exponential with parameter \(\kappa > 0\) (Vershynin R 2012) if

$$\begin{aligned}{}[\mathbb {E}|X|^p]^{1/p} \le \kappa p \quad \text { for all } p \ge 1. \end{aligned}$$

and sub-exponential random variables satisfy the concentration inequality.

Theorem 3

(Corollary 5.17 in Vershynin R (2012)). Let \(X_{1},\ldots ,X_{n}\) be independent centered variables, and suppose each \(X_{i}\) is sub-exponential with parameter \(\kappa \). Then for every \(\epsilon >0\),

$$\begin{aligned} P\left( |\frac{1}{n}\sum _{i}^{n}|\geqslant \epsilon \right) \leqslant \exp [-\gamma n\cdot \min \left( \frac{\epsilon ^{2}}{\gamma ^{2}},\frac{\epsilon }{\kappa }\right) ], \end{aligned}$$

where \(\gamma >0\) is an absolute constant.

Note that if X is a sub-exponential random variable with parameter X, then the centered random variable \(X-\mathbb {E}[X]\) is also sub-exponential with parameter \(2 \kappa \). This follows from the triangle inequality applied to the p-norm, followed by Jensen’s inequality for \(p \ge 1\):

$$\begin{aligned} \begin{aligned} \big [\mathbb {E}\big |X-\mathbb {E}[X]\big |^p\big ]^{1/p}&\le [\mathbb {E}|X|^p]^{1/p} + \big |\mathbb {E}[X]\big | \le 2[\mathbb {E}|X|^p]^{1/p}. \end{aligned} \end{aligned}$$

Lemma 6

Let X be a discrete Laplace random variable with the probability distribution

$$\begin{aligned} \mathbb {P}(X=x)= \frac{1-\lambda }{1+\lambda } \lambda ^{|x|},~~x=0, \pm 1, \ldots , \lambda \in (0,1). \end{aligned}$$

Then X is sub-exponential with parameter \(2( \log \frac{1}{\lambda } )^{-1}\).


Note that

$$\begin{aligned} \mathbb {E}|X|^p= & {} \frac{ 2(1-\lambda )}{1+\lambda } \sum _{x=0}^\infty \lambda ^x x^p \le \frac{ 2(1-\lambda )}{1+\lambda } \int _0^\infty t^p e^{-t\log \frac{1}{\lambda }} dt \\\le & {} \frac{ 2(1-\lambda )}{1+\lambda }(\frac{1}{ \log \frac{1}{\lambda } })^{p+1} \Gamma (p). \end{aligned}$$

It follows that

$$\begin{aligned}{}[\mathbb {E}|X|^p]^{1/p}< 2^{1/p} (\frac{1}{ \log \frac{1}{\lambda } })^{1+1/p} p < 2p \frac{1}{ \log \frac{1}{\lambda } } \end{aligned}$$

\(\square \)

The following lemma assures that condition (A5) hold with a large probability.

Lemma 7

Let \(\kappa _{mn}=2(r-1)(-\log \lambda _{mn})^{-1}=4(r-1)^{2}/\epsilon _{mn}\), where \(\lambda _{mn} \in (0,1)\). We have

$$\begin{aligned} \max \left\{ \max _{i=1,\dots ,m}|{\tilde{d}}_i-\mathbb {E}(d_i)|,\max _{j=1,\dots ,n}|{\tilde{b}}_j-\mathbb {E}(b_j)|\right\} = O_p\left( \sqrt{n\log n } + \kappa _{mn}\sqrt{\log n}\right) .\nonumber \\ \end{aligned}$$


By Hoeffding’s inequality (Hoeffding 1963), and \(m<n\) we have

$$\begin{aligned} \begin{aligned}&P\left( |d_i-\mathbb {E}(d_i)|\ge (r-1)\sqrt{n\log {n}}\right) \\&\le 2\exp {\left\{ -\frac{2(r-1)^{2}(n-1)\log (n-1)}{(m-1)(r-1)^{2}}\right\} }=\frac{2}{n^{2n/m}} \le \frac{2}{n^2} \end{aligned} \end{aligned}$$


$$\begin{aligned} \begin{array}{rcl} P\left( \displaystyle \max _{i}|d_i-\mathbb {E}(d_i)|\ge (r-1)\sqrt{n\log {n}}\right) &{}\le &{}P\left( \displaystyle \bigcup _{i}|d_i-\mathbb {E}(d_i)|\ge (r-1)\sqrt{n\log {n}}\right) \\ &{}\le &{}\displaystyle \sum _{i=1}^{m}P\left( |d_i-\mathbb {E}(d_i)|\ge (r-1)\sqrt{n\log {n}}\right) \\ &{}\le &{}m\times \frac{2}{n^2}\le \frac{2}{n}. \end{array}\end{aligned}$$

Similarly, we have

$$\begin{aligned} P\left( \displaystyle \max _{j}|b_j-\mathbb {E}(b_j)|\ge (r-1)\sqrt{n\log {n}}\right) \le \frac{2}{n}. \end{aligned}$$


$$\begin{aligned} \begin{array}{rcl} &{}~&{}P\left( \displaystyle \max \left\{ \max _{i=1,\dots ,m}|d_i-\mathbb {E}(d_i)|,\max _{j=1,\dots ,n}|b_j-\mathbb {E}(b_j)|\right\} \ge (r-1)\sqrt{n\log {n}}\right) \\ &{}\le &{}P\left( \displaystyle \max _{i}|d_i-\mathbb {E}(d_i)|\ge (r-1)\sqrt{n\log {n}}\right) +P\left( \displaystyle \max _{j}|b_j-\mathbb {E}(b_j)|\ge (r-1)\sqrt{n\log {n}}\right) \\ &{}\le &{}\frac{4}{n}. \end{array} \end{aligned}$$

Then, we have

$$\begin{aligned} \max \left\{ \max _{i=1,\dots ,m}|{\tilde{d}}_i-\mathbb {E}(d_i)|,\max _{j=1,\dots ,n}|{\tilde{b}}_j-\mathbb {E}(b_j)|\right\} \le (r-1)\sqrt{n\log (n)}. \end{aligned}$$

Note that \(\{e_i^+\}_{i=1}^n\) and \(\{ e_i^- \}_{i=1}^n\) are independently discrete Laplace random variables and sub-exponential with the same parameter \(\kappa _{mn}\) by Lemma 6. By the concentration inequality in Theorem 3, we have

$$\begin{aligned} \mathbb {P}\left( \max _{i=1, \ldots , m} |e_i^+| \ge 2\kappa _{mn} \sqrt{\frac{\log n}{\gamma }}\right) \le \sum _i \mathbb {P}\left( |e_i^+| \ge 2\kappa _{mn} \sqrt{\frac{\log n}{\gamma }}\right) \le n\times e^{-2\log n} = \frac{1}{n}\nonumber \\ \end{aligned}$$


$$\begin{aligned} \mathbb {P}\left( |\sum _{i=1}^m e_i^+| \ge 2\kappa _{mn} \sqrt{\frac{n\log n}{\gamma }}\right) \le 2 \exp \left( - \frac{\gamma }{n} \times \frac{n\log n}{\gamma }\right) = \frac{2}{n}, \end{aligned}$$

where \(\gamma \) is an absolute constant appearing in the concentration inequality. So, with probability at least \(1-4n/(n-1)^2-2/n\), we have

$$\begin{aligned} \max _{i=1, \ldots , m} |{\tilde{d}}_{i} - \mathbb {E}(d_i) | \le \max _i|d_i- \mathbb {E}(d_i) | + \max _i |e_i^+| \le (r-1)\sqrt{n\log n } + 2\kappa _{mn} \sqrt{\frac{\log n}{\gamma }} . \end{aligned}$$

Similarly, with probability at least \(1-4n/(n-1)^2-2/n\), we have

$$\begin{aligned} \max _{i=1, \ldots , n} |{\tilde{b}}_i- \mathbb {E}(b_i) | \le (r-1)\sqrt{n\log n } + 2\kappa _{mn} \sqrt{\frac{\log n}{\gamma }} . \end{aligned}$$

Let A and B be the events:

$$\begin{aligned} \begin{array}{rcl} A &{} = &{} \left\{ \max _{i=1, \ldots , m} |{\tilde{d}}_i - \mathbb {E}(d_i) | \le (r-1)\sqrt{n\log n } + 2\kappa _{mn} \sqrt{\frac{\log n}{\gamma }} \right\} , \\ B &{} = &{} \left\{ \max _{i=1, \ldots , n} |{\tilde{b}}_i - \mathbb {E}(b_i) | \le (r-1)\sqrt{n\log n } + 2\kappa _{mn} \sqrt{\frac{\log n}{\gamma }}\right\} . \end{array} \end{aligned}$$

Consequently, as n goes to infinity, we have

$$\begin{aligned} \mathbb {P}(A\bigcap B) \ge 1 - \mathbb {P}(A^c) - \mathbb {P}(B^c) \ge 1- 8n/(n-1)^2-4/n \rightarrow 1. \end{aligned}$$

This completes the proof. \(\square \)

It can be easily checked that \(-F'(\theta )\in \mathcal {L}_n(m, M)\), (Zhang et al. 2016) where \(M=(r-1)^2/2\) and \(m= 1/2( 1+ e^{ 2\Vert \theta \Vert _\infty })^2\). We are now ready to present the proof of Theorem 1.

Proof of Theorem 1

Assume that condition (A5) holds. Recall that the Newton’s iterates in Lemma 4, \( {\varvec{\theta }}^{(k+1)}={\varvec{\theta }}^{(k)}-[F^{'}( {\varvec{\theta }}^{(k)})]^{-1}F( {\varvec{\theta }}^{(k)})\) with \( {\varvec{\theta }}^{(0)}= {\varvec{\theta }}^{*}.\) If \( {\varvec{\theta }}\in \Omega ( {\varvec{\theta }}^{*},2\xi )\), then \(-F^{'}({ {\varvec{\theta }}}^{*})\in {{\mathcal {L}}}_{mn}(q,Q)\) with

$$\begin{aligned} q=\frac{1}{2(1+e^{2(\parallel {{\varvec{\theta }}^{*}}\parallel _{\infty }+2\xi )})},\quad Q=\frac{(r-1)^{2}}{2}. \end{aligned}$$

To apply Lemma 4, we need to calculate r and \(\rho r\) in this theorem. Let

$$\begin{aligned} {\tilde{F}}_{m+n}(\theta ^*)= \sum _{i=1}^{m}F_{i}(\theta ^*)-\sum _{i=1}^{n-1}F_{n+i}(\theta ^*)={\tilde{b}}_n-\sum _{i=1}^{n-1}\frac{\displaystyle \sum \nolimits ^{q-1}_{k=0}k e^{k({\alpha }_{i}+{\beta }_{n})}}{\displaystyle \sum \nolimits ^{q-1}_{k=0}e^{k({\alpha }_{i}+{\beta }_{n})}}+\sum _{i=1}^{m}e_{i}^{+} - \sum _{i=1}^{n-1}e_{i}^{-}. \end{aligned}$$

By Lemma 7, we have

$$\begin{aligned} |{\tilde{F}}_{m+n}(\theta ^*)| = O_p( (1+\kappa _{mn} )\sqrt{n\log n} ). \end{aligned}$$

By Lemma 2,

$$\begin{aligned} \xi&= \parallel [F^{'} \theta ^*]^{-1}F( \theta ^*)\parallel _{\infty }\nonumber \\&\leqslant (m+n-1)\Vert V^{-1} - S \Vert \Vert F(\theta ^*) \Vert _\infty + \max _{i=1,\dots ,m+n-1}\frac{\mid F_{i}( \theta ^*)\mid }{v_{i,i}} + \frac{\mid F_{m+n}(\theta ^*)\mid }{v_{m+n,m+n}}\nonumber \\&\leqslant \frac{2c_{1}(r-1)^{2}( 1 + e^{2(\Vert \theta ^*\Vert _\infty ) } )^6}{mn} \left( (r-1)\sqrt{(n-1)\log (n-1)}+\kappa _{mn}\sqrt{\log n } \right) \nonumber \\&\quad +\frac{2(1+e^{2\theta ^*\parallel _{\infty }})}{n-1} \left( (1+\kappa _{mn})\sqrt{n\log n}+(r-1)\sqrt{(n-1)\log (n-1)}+\kappa _{mn}\sqrt{\log n} \right) \nonumber \\&= O_p \left( \frac{(\log n)^{1/2}}{n^{1/2}}(1+\kappa _{mn}) e^{6\parallel \theta ^{*}\parallel _{\infty }} \right) . \end{aligned}$$

By Lemma (5), we have

$$\begin{aligned} \rho = \frac{ c_1(2n-1)Q^24(n-1)(r-1)^3}{2q^3n^2} + \frac{ 2(n-1) }{ q(n-1)(r-1)^3 } = O( e^{6\Vert \theta ^*\Vert _\infty } ) \end{aligned}$$

By Lemma (5) and condition (A5), for sufficient small \(\xi \),

$$\begin{aligned} \begin{array}{rcl} \rho \xi &{}\le &{}\left[ \frac{ c_1(2n-1)Q^24(n-1)(r-1)^3}{2q^3n^2} + \frac{ 2(n-1) }{ q(n-1)(r-1)^3 }\right] \times O_p \left( \frac{(\log n)^{1/2}}{n^{1/2}}(1+\kappa _{mn}) e^{6\parallel \theta ^{*}\parallel _{\infty }} \right) \\ &{}\le &{}O_p \left( \frac{(\log n)^{1/2}}{n^{1/2}}(1+\kappa _{mn}) e^{12\parallel \theta ^{*}\parallel _{\infty }} \right) \end{array} \end{aligned}$$

Note that if \((1+\kappa ) e^{12\Vert \theta ^*\Vert _\infty } = o( (n/\log n)^{1/2} )\), then \(\xi =o(1)\), and \(\rho \xi \rightarrow 0\) as \(n\rightarrow \infty \). Consequently, there exists N, when \(n\ge N\), \(\rho \xi <\frac{1}{2}\), by Lemma 4, \(\lim _{n\rightarrow \infty }{\widehat{{\varvec{\theta }}}}^{(n)}\) exists. Denote the limit as \(\widehat{\varvec{\theta }}\), then it satisfies

$$\begin{aligned} \parallel {\widehat{{\varvec{\theta }}}}- {\varvec{\theta }}^*\parallel _{\infty }\le 2r=O( \frac{(\log n)^{1/2}}{n^{1/2}}(1+\kappa _{mn}) e^{12\parallel \theta ^{*}\parallel _{\infty }})=o(1) \end{aligned}$$

By Lemma 7, condition (A5) holds with probability one, thus the above inequality also holds with probability one. The uniqueness of the parameter estimator comes from Proposition 5 in Yan et al. (2016) of Sect. 5. \(\square \)

Appendix B: Proof for Theorem 2

We first present one proposition. Since \(d_i=\sum ^{n}_{j= 1}a_{i,j}\) and \(b_j=\sum ^{m}_{ j= 1}a_{i,j}\) are sums of n and m independent random variables, by the central limit theorem for the bounded case in Loeve (1977), p. 289, we know that \({v_{i,i}}^{-1/2}(d_i-\mathbb {E}(d_i))\) and \({v_{m+j,m+j}}^{-1/2}(b_j-\mathbb {E}(b_j))\) are asymptotically standard normal if \(v_{i,i}\), \(v_{m+j,m+j}\) diverges. Note that

$$\begin{aligned}&\frac{m}{2(1+e^{2\parallel {{\varvec{\theta }}^{*}}\parallel _{\infty }})}\le v_{i,i}\le \frac{m(r-1)^{2}}{2}.\\&\quad \frac{n}{2(1+e^{2\parallel {{\varvec{\theta }}^{*}}\parallel _{\infty }})}\le v_{m+j,m+j}\le \frac{n(r-1)^{2}}{2}. \end{aligned}$$

Following Yan (2020), we have the following proposition.

Proposition 1

Let \(\kappa _{mn}=2(r-1)(-\log \lambda _{mn})^{-1}\), where \(\lambda _{mn}=\exp (-\epsilon _{mn}/2(r-1))\).

  1. (i)

    If \(\kappa _{mn} (\log n)^{1/2} e^{2\Vert \theta ^*\Vert _\infty }=o(1)\) and \(e^{\Vert \theta ^*\Vert _\infty }=o( n^{1/2} )\), then for any fixed \(k \ge 1\), as \(n\rightarrow \infty \), the vector consisting of the first k elements of \(S({\tilde{g}} - \mathbb {E}g )\) is asymptotically multivariate normal with mean zero and covariance matrix given by the upper left \(k \times k\) block of S.

  2. (ii)


    $$\begin{aligned} s_{mn}^2=\mathrm {Var}\left( \sum _{i=1}^m e_i^+ - \sum _{i=1}^{n-1} e_i^-\right) = (m+n-1)\frac{ 2\lambda _{mn}}{ (1-\lambda _{mn})^2}. \end{aligned}$$

    Assume that \(s_{mn}/v_{m+n,m+n}^{1/2} \rightarrow c\) for some constant c. For any fixed \(k \ge 1\), the vector consisting of the first k elements of \(S({\tilde{g}} - \mathbb {E}g )\) is asymptotically k-dimensional multivariate normal distribution with mean \({\mathbf {0}}\) and covariance matrix

    $$\begin{aligned} \mathrm {diag}\left( \frac{1}{v_{1,1}}, \ldots , \frac{1}{v_{k,k}}\right) + \left( \frac{1}{v_{m+n,m+n}} + \frac{s_{mn}^2}{v_{m+n,m+n}^2}\right) {\mathbf {1}}_k {\mathbf {1}}_k^\top , \end{aligned}$$

    where \({\mathbf {1}}_k\) is a k-dimensional column vector with all entries 1.

To complete the proof of Theorem 2, we need two lemmas as follows.

Lemma 8

Let \(R=V^{-1}-S\) and \(U=Cov[R\{{\mathbf {g}}-\mathbb {E}{\mathbf {g}}\}]\). Then

$$\begin{aligned} \Vert U \Vert \le \Vert V^{-1}-S\Vert +\frac{6(q-1)^2(1+e^{2\parallel {{\varvec{\theta }}^{*}}\parallel _{\infty }})^{2}}{4mn}. \end{aligned}$$


Note that

$$\begin{aligned} U=RVR^T=(V^{-1}-S)V(V^{-1}-S)^T=(V^{-1}-S)-S(I-VS), \end{aligned}$$

where I is a \((m+n-1)\times (m+n-1)\) identity matrix, and by Yan et al. (2016), we have

$$\begin{aligned} \mid \{S(I-VS)\}_{i,j}\mid =\mid w_{i,j}\mid \le \frac{6(r-1)^2(1+e^{2\parallel {{\varvec{\theta }}^{*}}\parallel _{\infty }})^{2}}{4mn}. \end{aligned}$$


$$\begin{aligned} \parallel U \parallel \le \parallel V^{-1}-S\parallel +\parallel \{S(I_{m+n-1}-VS)\}\parallel \le \parallel V^{-1}-S\parallel +\frac{6(r-1)^2(1+e^{2\parallel {{\varvec{\theta }}^{*}}\parallel _{\infty }})^{2}}{4mn}. \end{aligned}$$

\(\square \)

Lemma 9

Let \(\kappa _{mn}=2(r-1)(-\log \lambda _{mn})^{-1}=4(r-1)^2\epsilon _{mn}^{-1}\). If \((1+\kappa _{mn})^2 e^{ 18\Vert \theta ^*\Vert _\infty } = o( (n/\log n)^{1/2} )\), then for any i,

$$\begin{aligned} {\widehat{\theta }}_i- \theta _i^* = [V^{-1} ({\tilde{g}} - \mathbb {E}g ) ]_i + o_p( n^{-1/2} ). \end{aligned}$$


The proof is very similar to the proof of Lemma 9 in Yan et al. (2016). It only requires verification of the fact that all the steps hold by replacing g with \({\tilde{g}}\). \(\square \)


By Lemma 9 and noting that \(V^{-1}=S+R\), we have

$$\begin{aligned} ({\widehat{\theta }}-\theta )_i= [S({\tilde{g}} - \mathbb {E}g) ]_i+ [R \{ {\tilde{g}} - \mathbb {E}g \}]_i + o_p( n^{-1/2} ). \end{aligned}$$

By (A6) and \(\Vert {\tilde{g}} - g \Vert _\infty = O_p( \kappa _{mn} \sqrt{\log n})\), we have

$$\begin{aligned}{}[R ( {\tilde{g}} - g )]_i = O_p\left( n \frac{ M^2}{m^3n^2} \kappa _{mn} \sqrt{\log n}\right) = O_p\left( \frac{ \kappa _{mn} (\log n)^{1/2}e^{6\Vert \theta ^*\Vert _\infty } }{n}\right) , \end{aligned}$$


$$\begin{aligned} m=\frac{1}{2(1+e^{2\Vert \theta ^*\Vert _\infty })^2}, ~~M=\frac{(r-1)^2}{2}. \end{aligned}$$

If \(\kappa _{mn} e^{6\Vert \theta ^*\Vert _\infty } = o( (n/\log n)^{1/2})\), then \([R \{ {\widetilde{g}} - g \}]_i=o_p(n^{-1/2})\). Combing Lemma 8, it yields

$$\begin{aligned}{}[R ( {\tilde{g}} - \mathbb {E}g )]_i= [R({\tilde{g}} - g)]_i + [R(g - \mathbb {E}g)]_i = o_p(n^{-1/2}). \end{aligned}$$


$$\begin{aligned} ({\widehat{\theta }}-\theta )_i = [S({\tilde{g}} - \mathbb {E}g) ]_i + o_p( n^{-1/2} ). \end{aligned}$$

Theorem 2 immediately follows from Proposition 1. \(\square \)

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Luo, J., Liu, T. & Wang, Q. Affiliation weighted networks with a differentially private degree sequence. Stat Papers (2021).

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  • Affiliation networks
  • Asymptotic normality
  • Consistency
  • Finite discrete weight
  • Differential privacy

Mathematics Subject Classification

  • 62E20
  • 62F12