path: root/proof.tex

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\begin{document}

\section{Budget feasible mechanism}

\subsection{Data model}
\begin{itemize}
    \item set of $n$ users $\mathcal{N} = \{1,\ldots, n\}$
    \item each user has a public vector of features $x_i\in\mathbf{R}^d$ and some
        undisclosed variable $y_i\in\mathbf{R}$
    \item \textbf{Ridge regression model:}
        \begin{itemize}
            \item $y_i = \beta^*x_i + \varepsilon_i$
            \item $\varepsilon_i \sim \mathcal{N}(0,\sigma^2)$, 
                $(\varepsilon_i)_{i\in \mathcal{N}}$ are mutually independent.
            \item prior knowledge of $\beta$: $\beta\sim\mathcal{N}(0,\kappa I_d)$
        \end{itemize}
\end{itemize}

\subsection{Economics}

\begin{itemize}
    \item Value function:
        \begin{align*}
            \forall S\subset\mathcal{N},\; V(S)
            & = \frac{1}{2}\log\det\left(I_d
                    + \frac{\kappa}{\sigma^2}\sum_{i\in S} x_ix_i^*\right)\\
            & = \frac{1}{2}\log\det X(S)
        \end{align*}
    \item each user $i$ has a cost $c_i$
    \item the auctioneer has a budget constraint $B$
    \item optimisation problem:
        \begin{displaymath}
            OPT(V,\mathcal{N}, B) = \max_{S\subset\mathcal{N}} \left\{ V(S)\,|\,
            \sum_{i\in S}c_i\leq B\right\}
        \end{displaymath}
\end{itemize}

\subsection{Relaxations of the value function}

We say that $R_\mathcal{N}:[0,1]^n\rightarrow\mathbf{R}$ is a relaxation of the
value function $V$ over $\mathcal{N}$ if it coincides with $V$ at binary
points. Formally, for any $S\subset\mathcal{N}$, let $\mathbf{1}_S$ denote the
indicator vector of $S$. $R_\mathcal{N}$ is a relaxation of $V$ over
$\mathcal{N}$ iff:
\begin{displaymath}
    \forall S\subset\mathcal{N},\; R_\mathcal{N}(\mathbf{1}_S) = V(S)
\end{displaymath}

We can extend the optimisation problem defined above to a relaxation by
extending the cost function:
\begin{displaymath}
    \forall \lambda\in[0,1]^n,\; c(\lambda)
    = \sum_{i\in\mathcal{N}}\lambda_ic_i
\end{displaymath}
The optimisation problem becomes:
\begin{displaymath}
    OPT(R_\mathcal{N}, B) =
    \max_{\lambda\in[0,1]^n}\left\{R_\mathcal{N}(\lambda)\,|\, c(\lambda)\leq B\right\}
\end{displaymath}
The relaxations we will consider here rely on defining a probability
distribution over subsets of $\mathcal{N}$.

Let $\lambda\in[0,1]^n$, let us define:
\begin{displaymath}
    P_\mathcal{N}(S,\lambda) = \prod_{i\in S}\lambda_i
    \prod_{i\in\mathcal{N}\setminus S}(1-\lambda_i)
\end{displaymath}
$P_{\mathcal{N}}(S,\lambda)$ is the probability of picking the set $S$ if we select
a subset of $\mathcal{N}$ at random by deciding independently for each point to
include it in the set with probability $\lambda_i$ (and to exclude it with
probability $1-\lambda_i$).

We will consider two relaxations of the value function $V$ over $\mathcal{N}$:
\begin{itemize}
    \item the \emph{multi-linear extension} of $V$:
        \begin{align*}
            F_\mathcal{N}(\lambda) 
            & = \mathbb{E}_{S\sim P_\mathcal{N}(S,\lambda)}\big[\log\det X(S)\big]\\
            & = \sum_{S\subset\mathcal{N}} P_\mathcal{N}(S,\lambda) V(S)\\
            & = \sum_{S\subset\mathcal{N}} P_{\mathcal{N}}(S,\lambda) \log\det X(S)\\
        \end{align*}
    \item the \emph{concave relaxation} of $V$:
        \begin{align*}
            L_{\mathcal{N}}(\lambda) 
            & = \log\det \mathbb{E}_{S\sim P_\mathcal{N}(S,\lambda)}\big[X(S)\big]\\
            & = \log\det\left(\sum_{S\subset N}
                P_\mathcal{N}(S,\lambda)X(S)\right)\\
            & = \log\det\left(I_d
                + \frac{\kappa}{\sigma^2}\sum_{i\in\mathcal{N}}
                    \lambda_ix_ix_i^*\right)
        \end{align*}
\end{itemize}

\begin{lemma}
    The \emph{concave relaxation} $L_\mathcal{N}$ is concave\footnote{Hence
    this relaxation is well-named!}.
\end{lemma}

\begin{proof}
    This follows almost immediately from the concavity of the $\log\det$
    function over symmetric positive semi-definite matrices. More precisely, if
    $A$ and $B$ are two symmetric positive semi-definite matrices, then:
    \begin{multline*}
        \forall\alpha\in [0, 1],\; \log\det\big(\alpha A + (1-\alpha) B\big)\\
        \geq \alpha\log\det A + (1-\alpha)\log\det B
    \end{multline*}
\end{proof}

\begin{lemma}[Rounding]
    For any feasible $\lambda\in[0,1]^n$, there exists a feasible
    $\bar{\lambda}\in[0,1]^n$ such that at most one of its component is
    fractional, that is, lies in $(0,1)$ and:
    \begin{displaymath}
        F_{\mathcal{N}}(\lambda)\leq F_{\mathcal{N}}(\bar{\lambda})
    \end{displaymath}
\end{lemma}

\begin{proof}
    We give a rounding procedure which given a feasible $\lambda$ with at least
    two fractional components, returns some $\lambda'$ with one less fractional
    component, feasible such that:
    \begin{displaymath}
        F(\lambda) \leq F(\lambda')
    \end{displaymath}
    Applying this procedure recursively yields the lemma's result.

    Let us consider such a feasible $\lambda$. Let $i$ and $j$ be two
    fractional components of $\lambda$ and let us define the following
    function:
    \begin{displaymath}
        F_\lambda(\varepsilon) = F(\lambda_\varepsilon)
        \quad\textrm{where} \quad
        \lambda_\varepsilon = \lambda + \varepsilon\left(e_i-\frac{c_i}{c_j}e_j\right)
    \end{displaymath}

    It is easy to see that if $\lambda$ is feasible, then:
    \begin{multline}\label{eq:convex-interval}
        \forall\varepsilon\in\Big[\max\Big(-\lambda_i,(\lambda_j-1)\frac{c_j}{c_i}\Big), \min\Big(1-\lambda_i, \lambda_j
        \frac{c_j}{c_i}\Big)\Big],\;\\
            \lambda_\varepsilon\;\;\textrm{is feasible}
    \end{multline}

    Furthermore, the function $F_\lambda$ is convex, indeed:
    \begin{align*}
        F_\lambda(\varepsilon)
        & = \mathbb{E}_{S'\sim P_{\mathcal{N}\setminus\{i,j\}}(S',\lambda)}\Big[
        (\lambda_i+\varepsilon)\Big(\lambda_j-\varepsilon\frac{c_i}{c_j}\Big)V(S'\cup\{i,j\})\\
        & + (\lambda_i+\varepsilon)\Big(1-\lambda_j+\varepsilon\frac{c_i}{c_j}\Big)V(S'\cup\{i\})\\
        & + (1-\lambda_i-\varepsilon)\Big(\lambda_j-\varepsilon\frac{c_i}{c_j}\Big)V(S'\cup\{j\})\\
        & + (1-\lambda_i-\varepsilon)\Big(1-\lambda_j+\varepsilon\frac{c_i}{c_j}\Big)V(S')\Big]\\
    \end{align*}
    Thus, $F_\lambda$ is a degree 2 polynomial whose dominant coefficient is:
    \begin{multline*}
        \frac{c_i}{c_j}\mathbb{E}_{S'\sim P_{\mathcal{N}\setminus\{i,j\}}(S',\lambda)}\Big[
            V(S'\cup\{i\})+f(S'\cup\{i\})\\
        -V(S'\cup\{i,j\})-f(S')\Big]
    \end{multline*}
    which is positive by submodularity of $V$. Hence, the maximum of
    $F_\lambda$ over the interval given in \eqref{eq:convex-interval} is
    attained at one of its limit, at which either the $i$-th or $j$-th component of
    $\lambda_\varepsilon$ becomes integral.
\end{proof}

\begin{lemma}
    There exists $C>0$ such that:
    \begin{displaymath}
        \forall\lambda\in[0,1]^n,\; C\,L_\mathcal{N}(\lambda)\leq
        F_\mathcal{N}(\lambda)\leq L_{\mathcal{N}}(\lambda)
    \end{displaymath}
\end{lemma}

\begin{lemma}
    \begin{displaymath}
        OPT(L_\mathcal{N}, B) \leq \frac{1}{C}OPT(V,\mathcal{N},B)
        + \max_{i\in\mathcal{N}}V(i)
    \end{displaymath}
\end{lemma}

\begin{algorithm}
    \caption{Budget feasible mechanism for ridge regression}
    \begin{algorithmic}[1]
    \State $i^* \gets \argmax_{j\in\mathcal{N}}V(j)$
    \State $x^* \gets \argmax_{x\in[0,1]^n} \{L_{\mathcal{N}\setminus\{i^*\}}(x)
                                    \,|\, c(x)\leq\frac{B}{2}\}$
        \Statex
        \If{$L(x^*) < CV(\{i^*\})$}
            \State \textbf{return} $\{i^*\}$
        \Else
            \State $i \gets \argmax_{1\leq j\leq n}\frac{V(\{j\})}{c_j}$
            \State $S \gets \emptyset$
            \While{$c_i\leq \frac{B}{2}\frac{V(S\cup\{i\})-V(S)}{V(S\cup\{i\})}$}
                \State $S \gets S\cup\{i\}$
                \State $i \gets \argmax_{j\in\{1,\ldots,n\}\setminus S}
                \frac{V(S\cup\{j\})-V(S)}{c_j}$
            \EndWhile
            \State \textbf{return} $S$
        \EndIf
    \end{algorithmic}
\end{algorithm}
\end{document}