+ Stego-security states that the knowledge of $K$ does not help to make the
+ difference between $p(X)$ and $p(Y)$. This definition implies the following
+ property:
+ $$p(Y|K_1)= \cdots = p(Y|K_{N_k})=p(Y)=p(X)$$
+ This property is equivalent to a zero Kullback-Leibler divergence, which is the
+ accepted definition of the "perfect secrecy" in steganography~\cite{Cachin2004}.
+
+
+\subsection{The negation mode is stego-secure}
+To make this article self-contained, this section recalls theorems and proofs of stego-security for negation mode published in~\cite{gfb10:ip}.
+
+\begin{proposition} \emph{dhCI dissimulation} of Definition \ref{def:dhCI} with
+negation mode and CIIS strategy-adapter is stego-secure, whereas it is not the
+case when using CIDS strategy-adapter.
+\end{proposition}
+
+
+\begin{proof} On the one hand, let us suppose that $X \sim
+\mathbf{U}\left(\mathbb{B}^n\right)$ when using \linebreak CIIS$(K,\_,\_,l)$.
+We prove by a
+mathematical induction that $\forall t \in \mathds{N}, X^t \sim
+\mathbf{U}\left(\mathbb{B}^n\right)$.
+
+The base case is immediate, as $X^0 = X \sim
+\mathbf{U}\left(\mathbb{B}^n\right)$. Let us now suppose that the statement $X^t
+\sim \mathbf{U}\left(\mathbb{B}^n\right)$ holds until for some $t$.
+Let $e \in
+\mathbb{B}^n$ and \linebreak $\mathbf{B}_k=(0,\cdots,0,1,0,\cdots,0) \in
+\mathbb{B}^n$ (the digit $1$ is in position $k$).
+
+So
+$P\left(X^{t+1}=e\right)=\sum_{k=1}^n
+P\left(X^t=e\oplus\mathbf{B}_k,S^t=k\right)$ where
+$\oplus$ is again the bitwise exclusive or.
+These two events are independent when
+using CIIS strategy-adapter
+(contrary to CIDS, CIIS is not built by using $X$),
+ thus:
+$$P\left(X^{t+1}=e\right)=\sum_{k=1}^n
+P\left(X^t=e\oplus\mathbf{B}_k\right) \times P\left(S^t=k\right).$$
+
+According to the
+inductive hypothesis: $P\left(X^{n+1}=e\right)=\frac{1}{2^n} \sum_{k=1}^n
+P\left(S^t=k\right)$. The set of events $\left \{ S^t=k \right \}$ for $k \in
+\llbracket 1;n \rrbracket$ is a partition of the universe of possible, so
+$\sum_{k=1}^n P\left(S^t=k\right)=1$. Finally,
+$P\left(X^{t+1}=e\right)=\frac{1}{2^n}$, which leads to $X^{t+1} \sim
+\mathbf{U}\left(\mathbb{B}^n\right)$. This result is true for all $t \in
+\mathds{N}$ and then for $t=l$.
+
+Since $P(Y|K)$ is $P(X^l)$ that is proven to be equal to $P(X)$,
+we thus have established that,
+$$\forall K \in [0;1], P(Y|K)=P(X^{l})=P(X).$$
+So dhCI dissimulation with CIIS
+strategy-adapter is stego-secure.
+
+On the other hand, due to the definition of CIDS, we have \linebreak
+$P(Y=(1,1,\cdots,1)|K)=0$.
+%\JFC{Pourquoi? Justifier davantage là ou dans la def de CIDS}
+So there is no uniform repartition for the stego-contents $Y|K$.
+\end{proof}
+
+
+
+To sum up, Alice and Bob can counteract Eve's attacks in WOA setup, when using
+dhCI dissimulation with CIIS strategy-adapter. To our best knowledge, this is
+the second time an information hiding scheme has been proven to be stego-secure:
+the former was the spread-spectrum technique in natural marking
+configuration with $\eta$ parameter equal to 1 \cite{Cayre2008}.
+
+
+
+
+
+\subsection{A new class of $\varepsilon$-stego-secure schemes}
+
+Let us prove that,
+\begin{theorem}\label{th:stego}
+Let $\epsilon$ be positive,
+$l$ be any size of LSCs,
+$X \sim \mathbf{U}\left(\mathbb{B}^l\right)$,
+$f_l$ be an image mode s.t.
+$\Gamma(f_l)$ is strongly connected and
+the Markov matrix associated to $f_l$
+is doubly stochastic.
+In the instantiated \emph{dhCI dissimulation} algorithm
+with any uniformly distributed (u.d.) strategy-adapter
+that is independent from $X$,
+there exists some positive natural number $q$ s.t.
+$|p(X^q)- p(X)| < \epsilon$.
+\end{theorem}
+
+
+\begin{proof}
+Let $\textit{deci}$ be the bijection between $\Bool^{l}$ and
+$\llbracket 0, 2^l-1 \rrbracket$ that associates the decimal value
+of any binary number in $\Bool^{l}$.
+The probability $p(X^t) = (p(X^t= e_0),\dots,p(X^t= e_{2^l-1}))$ for $e_j \in \Bool^{l}$ is thus equal to
+$(p(\textit{deci}(X^t)= 0,\dots,p(\textit{deci}(X^t)= 2^l-1))$ further denoted by $\pi^t$.
+Let $i \in \llbracket 0, 2^l -1 \rrbracket$,
+the probability $p(\textit{deci}(X^{t+1})= i)$ is
+\[
+ \sum\limits^{2^l-1}_{j=0}
+\sum\limits^{l}_{k=1}
+p(\textit{deci}(X^{t}) = j , S^t = k , i =_k j , f_k(j) = i_k )
+\]
+\noindent
+where $ i =_k j $ is true iff the binary representations of
+$i$ and $j$ may only differ for the $k$-th element,
+and where
+$i_k$ abusively denotes, in this proof, the $k$-th element of the binary representation of
+$i$.
+
+Next, due to the proposition's hypotheses on the strategy,
+$p(\textit{deci}(X^t) = j , S^t = k , i =_k j, f_k(j) = i_k )$ is equal to
+$\frac{1}{l}.p(\textit{deci}(X^t) = j , i =_k j, f_k(j) = i_k)$.
+Finally, since $i =_k j$ and $f_k(j) = i_k$ are constant during the
+iterative process and thus does not depend on $X^t$, we have
+\[
+\pi^{t+1}_i = \sum\limits^{2^l-1}_{j=0}
+\pi^t_j.\frac{1}{l}
+\sum\limits^{l}_{k=1}
+p(i =_k j, f_k(j) = i_k ).
+\]