path: root/notes.tex

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\begin{document}

\section{Problem}

\paragraph{Data Model} 

There is a database $D$ of $N$ users. Each user $i\in\{1,\ldots, N\}$ is
represented by a vector of features $x_i\in\mathbf{R}^d$. To each user
$i$, is also associated a variable of interest $y_i$, which can be inferred
from $x_i$.

We will thus assume that there is a \emph{model} describing the relation
between $y_i$ and $x_i$, which in the most general form can be written as:
\begin{displaymath}
    y_i = h(x_i)
\end{displaymath}
where $h$ is an unknown function.

The database is valuable, because by observing the database, or a subset $S$ of
the database, it is possible to learn the unknown model function $h$ (or to
improve a prior knowledge of this function) and thus, to predict the variable
of interest associated with a given user.

\paragraph{Value} Consequently, the value $V(S)$ of a subset $S$ of the
database will be defined as the decrease of \emph{uncertainty} about $h$
induced by observing $S$. Depending on the definition of uncertainty used, it
is possible to distinguish two general approaches to define $V(S)$:
\begin{itemize}
    \item \emph{Information theoretic:} the knowledge of $h$ we have is
        represented by a probabilistic distribution, and the \emph{entropy} of
        $h$ is used as a measure of the uncertainty. The observation of the set
        $S$ allows us to update the distribution of $h$, and the uncertainty
        becomes the conditional entropy of $h$ given the observation. This
        leads to the definition of the value:
        \begin{displaymath}
            V(S) = H(h) - H(h\,|\,S) 
        \end{displaymath}
    \item \emph{Statistical:} in this case, we maintain a predictor $\tilde{y}$
        which given a user's feature vector $x$ gives a prediction
        $\tilde{y}(x)$ of his variable of interest. The uncertainty is then
        measured by the variance of the predictor at a given point $x$,
        $\sigma^2_x$. After observing a subset $S$ of the database, the
        predictor is updated, and its new variance is denoted by
        $\sigma_{x|S}^2$. Thus, we could define the value of $S$ by $V(S)
        = \sigma_x^2 - \sigma_{x|S}^2$. However, this definition depends on the
        point $x$. If we knew the distribution of $x$, one could simply take
        the expectation of the previous quantity as a definition of the value.
        Another way to solve the issue, is to average this quantity over the
        database, leading to the definition of the value:
        \begin{displaymath}
            V(S) = \frac{1}{N}\sum_{x\in D} \sigma_x^2 - \sigma_{x|A}^2
        \end{displaymath}
\end{itemize}

Once you have defined a notion of value, there are several problems you can
address:
\begin{enumerate}
    \item \emph{budget auctions:} the points in the database are associated
        with users. Each user $i$ has a cost $c_i$ for revealing his data. You
        are given a budget $B$ and you want to select a subset $S$ of users and
        a payment scheme which is truthful. The value of the subset $S$ should
        be as big as possible with the constraint that the payments should stay
        within budget.
    \item \emph{value sharing:} the goal is to find a value sharing method to
        split the whole value of a subset $S$ among its participants.
    \item \emph{k-best auctions:} this one is very similar to \emph{budget
        auctions}, but instead of having a budget constraint, the constraint is
        on the number of people in the subset $S$.
\end{enumerate}

\emph{[TODO: add references to the papers mentioning these problems, what we
know about the value functions defined above, etc.]}

These problems have already been studied, and optimal or near-optimal solution
are already known when the value function is submodular. 

\section{Data value in the Gaussian world}

In this section we will assume a multivariate Gaussian model:
\begin{displaymath}
    y = \beta^*x + \epsilon
\end{displaymath}

The prior distribution of $\beta$ is a multivariate normal distribution of mean
zero and covariance $\Sigma$. $\epsilon$ is the noise which is modeled with
a normal distribution of mean zero and variance $\sigma^2$.

Note that this model is the probabilistic model used in ridge regression. To
compute the value of a subset $S$ as defined above, we have to compute the
differential conditional entropy of the posterior distribution after observing
the set $S$, which is exactly the distribution which leads to the ridge
regression estimator. Thus the following computation can be seen as the study
of how the ridge regression estimator evolves as you observe more points.

Let us start by computing the entropy of the multivariate normal distribution:
\begin{displaymath}
    H(\beta) = -\frac{1}{C}
    \int_{b\in\mathbf{R}^d} \exp\left(-\frac{1}{2}b^*\Sigma^{-1}b\right)
    \log\left(-\frac{1}{C}\exp\left(-\frac{1}{2}b^*\Sigma^{-1}b\right)\right)db
\end{displaymath}
where:
\begin{displaymath}
    C = \big(\sqrt{2\pi}\big)^d\sqrt{\det\Sigma}
    = \int_{b\in\mathbf{R}^d}\exp\left(-\frac{1}{2}b^*\Sigma^{-1}b\right)db 
\end{displaymath}
\begin{displaymath}
    H(\beta) = \log(C) + \frac{1}{2C}\int_{b\in\mathbf{R}^d}
    b^*\Sigma^{-1}b\exp\left(-\frac{1}{2}b^*\Sigma^{-1}b\right)db
\end{displaymath}

One can notice that:
\begin{displaymath}
    \frac{1}{C}\int_{b\in\mathbf{R}^d}
    b^*\Sigma^{-1}b\exp\left(-\frac{1}{2}b^*\Sigma^{-1}b\right)db
    = \expt{b^*\Sigma^{-1}b}
\end{displaymath}
where $b$ follows a multivariate normal distribution of variance $\Sigma$
and mean zero.
\begin{displaymath}
    \expt{b^*\Sigma^{-1}b} = \expt{tr(\Sigma^{-1}bb^*)} = tr(\Sigma^{-1}\Sigma)
    = 1
\end{displaymath}
\begin{displaymath}
    H(\beta) = \frac{1}{2}\log\big((2\pi)^d\det\Sigma\big) + \frac{1}{2}
    = \frac{1}{2}\log\big((2\pi e)^d\det\Sigma\big)
\end{displaymath}
\section{Submodularity}

In this section, we will consider that we are given a \emph{universe} set $U$.
A set function $f$ is a function defined on the power set of $U$, $\mathfrak{P}(U)$.

A set function $f$ defined on $\mathfrak{P}(U)$ will be said \emph{increasing} if
it is increasing with regards to inclusion, that is:

\begin{displaymath}
  \forall\,S\subseteq T,\quad f(S)\leq f(T)
\end{displaymath}

A \emph{decreasing} function on $\mathfrak{P}(U)$ is defined similarly.

A set function $f$ defined on $\mathfrak{P}(U)$ is said to be
\emph{submodular} if it verifies the diminishing returns property, that is,
the marginal increments when adding a point to a set, is a set decreasing
function. More formally, for any point $x$ in $U$, we can define the marginal
increment of $f$ regarding $x$, it is the set function defined as:
\begin{displaymath}
    \Delta_x f(S) = f(S\cup\{x\}) - f(S)
\end{displaymath}
Then, $f$ is \emph{submodular} iff. for all $x$ in $U$, $\Delta_x f$ is a set
decreasing function.

Similarly, a \emph{supermodular} is a function whose marginal increments are set
increasing functions.
\begin{prop}
  Let $R:\mathbf{R}\rightarrow \mathbf{R}$ be a decreasing concave function and
  $f:\mathfrak{P}(U)\rightarrow\mathbf{R}$ be a decreasing submodular function,
  then the composed function $R\circ f$ is increasing and supermodular.
\end{prop}

\begin{proof}
  The increasingness of $R\circ f$ follows immediately from the decreasingness
  of $R$ and $f$.
  
  For the supermodularity, let $S$ and $T$ be two sets such that $S\subseteq
  T$. By decreasingness of $f$, we have:
  \begin{displaymath}
    \forall\,V,\quad f(T)\leq f(S)\quad\mathrm{and}\quad f(T\cup V)\leq f(S\cup V)
  \end{displaymath}
  
  Thus, by concavity of $R$:
  \begin{displaymath}\label{eq:base}
    \forall\,V,\quad\frac{R\big(f(S)\big)-R\big(f(S\cup V)\big)}{f(S)-f(S\cup V)}
    \leq\frac{R\big(f(T)\big)-R\big(f(T\cup V)\big)}{f(T)-f(T\cup V)}
  \end{displaymath}
  
  $f$ is decreasing, so multiplying this last inequality by
  $f(S)-f(S\cup V)$ and $f(T)-f(T\cup V)$ yields:
  \begin{multline}
    \forall V,\quad\Big(R\big(f(S)\big)-R\big(f(S\cup V)\big)\Big)\big(f(T)-f(T\cup V)\big)\\
    \leq \Big(R\big(f(T)\big)-R\big(f(T\cup V)\big)\Big)\big(f(S)-f(S\cup V)\big)
  \end{multline}

  $f$ is submodular, so:
  \begin{displaymath}
    f(T\cup V)-f(T)\leq f(S\cup V) - f(S)
  \end{displaymath}
  
  $R\circ f$ is increasing, so:
  \begin{displaymath}
    R\big(f(S)\big)-R\big(f(S\cup V)\big)\leq 0
  \end{displaymath}

  By combining the two previous inequalities, we get:
    \begin{multline*}
    \forall V,\quad\Big(R\big(f(S)\big)-R\big(f(S\cup V)\big)\Big)\big(f(S)-f(S\cup V)\big)\\
    \leq \Big(R\big(f(S)\big)-R\big(f(S\cup V)\big)\Big)\big(f(T)-f(T\cup V)\big)
  \end{multline*}

  Injecting this last inequality into \eqref{eq:base} gives:
  \begin{multline*}
    \forall V,\quad\Big(R\big(f(S)\big)-R\big(f(S\cup V)\big)\Big)\big(f(S)-f(S\cup V)\big)\\
    \leq \Big(R\big(f(T)\big)-R\big(f(T\cup V)\big)\Big)\big(f(S)-f(S\cup V)\big)
  \end{multline*}

  Dividing left and right by $f(S)-f(S\cup V)$ yields:
  \begin{displaymath}
    \forall V,\quad R\big(f(S)\big)-R\big(f(S\cup V)\big)
    \leq R\big(f(T)\big)-R\big(f(T\cup V)\big)
  \end{displaymath}
  which is exactly the supermodularity of $R\circ f$.
\end{proof}

\end{document}