\section{Introduction}

A trader on the stock market is usually assumed to make
his decisions relying on all the information which is generated
by the market events.
However it is registered that some people have more
detailed information than others, in the sense that they
act with the present time knowledge of some future event.
This is the so-called insider information and those dealers
taking advantage of it are the insiders.
The financial markets with economic agents possessing additional knowledge have been studied in
a number of papers (see e.g. \cite{A00, AIS98, GP99, Z07}).
We take approach originated in \cite{DuffieHuang} and \cite{Pikovsky}
assuming that the insider possesses some extra information stored in the random variable $G$
known at the beginning of the trading interval and not available to the regular trader.
The typical examples of $G$ are $G=S_{T+\delta}$, $G=\mathbf{1}_{[a,b]}(S_{T+\delta})$ or
$G=\sup_{t\in[0,T+\delta]}S_t$ ($\delta >0$),
where $S$ is a semimartingale representing the discounted stock price process and $T$ is fixed time
horizon till which the insider is allowed to trade.

In this paper we show how much better and with which strategies an insider
can perform on the market if he uses optimally the extra
information he has at his disposal.
The problem of pricing and perfect hedging of contingent claims is well understood in the
context of arbitrage-free models which are complete. In such models every contingent
claim can be replicated by a self-financing trading strategy. The cost
of replication equals the discounted expectation of
the claim under the unique equivalent martingale measure.
Moreover, this cost is the same for the insider and the regular trader.
Therefore instead of this strategy we will employ the quantile
hedging strategy of an insider for the replication, following idea of
F\"{o}llmer and Leukert \cite{FL99, FL00}.
That is, we will seek for the self-financing strategy that
\begin{itemize}
\item maximizes the probability of success of hedge under a given initial capital or
\item minimizes the initial capital under a given lower bound of the probability of the successful hedge.
\end{itemize}

This is the case when the insider is unwilling to put up the initial amount of capital required by a
perfect hedging.
This approach might be also seen as a
dynamic version of the VaR.

We use powerful technique of {\it grossissement de filtrations}
developed by Yor, Jeulin and Jacod \cite{J80, JY85}
and utilize the results of Amendinger
\cite{A00} and Amendinger et al. \cite{AIS98}.

The paper is organized as follows. In Section \ref{market} we present the main results. In Section \ref{numerical}
we analyze in detail some examples. Finally, in Section \ref{proofs} we give the proofs of the main results.

\section{Main results}\label{market}

Let $(\Omega,\mathcal{F},\mathbb{F},\mathbb{P})$ be a complete probability space
and $S=(S_t)_{t\geq 0}$ be an $(\mathbb{F},\mathbb{P})$-semimartingale representing the discounted stock price process.
Assume that the filtration $\mathbb{F}=(\mathcal{F}_t)_{t\geq 0}$ is the natural filtration of $S$ satisfying usual conditions with the trivial $\sigma$-algebra $\mathcal{F}_0$. Thus, the regular trader makes his portfolio decisions according to the information flow
$\mathbb{F}$. In addition to the regular trader we will consider the insider, whose knowledge will be modelled by the \textit{initial enlargement} of $\mathbb{F}$, that is filtration $\mathbb{G}=(\mathcal{G}_t)_{t\geq 0}$ given by:
$$\mathcal{G}_t=\mathcal{F}_t \vee
\sigma (G),
$$
where $G$ is an $\mathcal{F}$-measurable random variable. In particular, $G$ can be an
$\mathcal{F}_{T+\delta}$-measurable random variable ($\delta >0$) for $T$ being a fixed time horizon representing
the expiry date of the hedged contingent claim.

We will assume that the market is complete and arbitrage-free for the regular trader, hence there exists a unique equivalent martingale measure $\mathbb{Q}_{\mathbb{F}}$ such that $S$ is an $(\mathbb{F},\mathbb{Q}_{\mathbb{F}})$-martingale on $[0,T]$.
Denote by $(Z_{t}^{\mathbb{F}})_{t\in[0,T]}$ the density process of $\mathbb{Q}_\mathbb{F}$ with respect to $\mathbb{P}$, i.e.:
$$Z_{t}^{\mathbb{F}}=\left.\frac{{\rm d}\mathbb{Q}_\mathbb{F}}{{\rm d}\mathbb{P}}\right|_{\mathcal{F}_t}.$$

We will consider the contingent claim $H$ being $\mathcal{F}_T$-measurable, nonnegative random variable and the replicating
investment strategies for insider, which are expressed in terms of the integrals with respect to $S$.
To define them properly we assume that $S$ is a $(\mathbb{G},\mathbb{P})$-semimartingale
which follows from the requirement:
\begin{equation}
\mathbb{P}(G\in\cdot|\mathcal{F}_t)\ll \mathbb{P}(G\in\cdot)\qquad \mathbb{P}-{\rm a.s.} \label{abs_cont}
\end{equation}
for all $t\in [0,T]$ (see e.g. \cite{A00} and \cite{JY85}).
In fact, we assume from now on more, that is that the measure $\mathbb{P}(G\in\cdot|\mathcal{F}_t)$ and the law of $G$ are equivalent
for all $t\in [0,T]$:
\begin{equation}
\mathbb{P}(G\in\cdot|\mathcal{F}_t)\sim \mathbb{P}(G\in\cdot)\qquad \mathbb{P}-{\rm a.s.} \label{equiv}
\end{equation}
Under the condition (\ref{equiv}) there exists equivalent
$\mathbb{G}$-martinagle measure $\mathbb{Q}_{\mathbb{G}}$ defined by:
\begin{equation}\label{QG}
\mathbb{Q}_\mathbb{G}(A)=\int_{A}\frac{Z_{T}^{\mathbb{F}}}{p_{T}^{G}}(\omega){\rm d}\mathbb{P}(\omega),\qquad A\in\mathcal{G}_T,\end{equation}
where
$$p_{t}^{x}\;\mathbb{P}(G\in {\rm d}x)$$
is a version of $\mathbb{P}(G\in {\rm d}x|\mathcal{F}_t);$
see \cite{A00} and \cite{AIS98} (and also Theorems \ref{ther1} and \ref{ther2}).

For $\mathbb{H}\in\{\mathbb{F},\mathbb{G}\}$ we will consider only self-financing admissible trading strategies $(V_0,\xi)$ on $[0,T]$
for which the value process
$$V_{t}=V_{0}+\int_{0}^{t}\xi_u\;{\rm d}S_u,\qquad t\in[0,T],$$
is well defined, where an initial capital $V_{0}\geq 0$ is $\mathcal{H}_0$-measurable, a process $\xi$ is $\mathbb{H}$-predictable
and
$$V_t\geq 0,\qquad \mathbb{P}-{\rm a.s.}$$
for all $t\in [0,T]$.
Denote all admissible strategies associated to the filtration $\mathbb{H}\in\{\mathbb{F},\mathbb{G}\}$ by $\mathcal{A}^{\mathbb{H}}$.

Under assumption (\ref{equiv}) the insider can perfectly replicate the contingent
claim $H\in L^{1}(\mathbb{Q}_\mathbb{F})\cap L^{1}(\mathbb{Q}_\mathbb{G})$:
$$\mathbb{E}_{\mathbb{Q}_\mathbb{G}}(H|\mathcal{G}_t)=H_{0}+\int_{0}^{t}\xi_{u}\;{\rm d}S_{u},\qquad\mathbb{P}-{\rm a.s.}$$
where $H_{0}=\mathbb{E}_{\mathbb{Q}_\mathbb{G}}(H|\mathcal{G}_0)$. Moreover, from \cite{A00} and \cite{AIS98} it follows that $H_{0}=\mathbb{E}_{\mathbb{Q}_\mathbb{G}}H$ (see Theorem \ref{ther1}).
In this paper we will analyze the case when the insider is unwilling to pay the initial capital
$H_0$ required by a perfect hedge.
We will consider the following pair of dual problems.

\begin{problem}\label{P1}
Let $\alpha$ be a given $\mathcal{G}_0$-measurable random variable taking values in $[0,1]$. We are looking for a strategy $(\alpha\mathbb{E}_{\mathbb{Q}_\mathbb{G}}H,\xi)\in\mathcal{A}^\mathbb{G}$ which maximizes for any realization of $G$ the insider's probability of a successful hedge
$$\mathbb{P}\left(\left.\alpha\mathbb{E}_{\mathbb{Q}_\mathbb{G}}H+\int_0^T\xi_t \;{\rm d}S_t\geq H\right|\mathcal{G}_0\right),\qquad \mathbb{P}-{\rm a.s.}$$
\end{problem}
\begin{problem}\label{P2}
Let $\epsilon$ be a given $\mathcal{G}_0$-measurable random variable taking values in $[0,1]$. We are looking for a minimal $\mathcal{G}_0$-measurable random variable $\alpha$ for which there exists $\xi$ such that $(\alpha\mathbb{E}_{\mathbb{Q}_\mathbb{G}}H,\xi)\in\mathcal{A}^\mathbb{G}$ and
\begin{equation}\label{P2inequality}\mathbb{P}\left(\left.\alpha\mathbb{E}_{\mathbb{Q}_\mathbb{G}}H+\int_0^T\xi_t \;{\rm d}S_t\geq H \right|\mathcal{G}_0\right)\geq 1-\epsilon,\qquad \mathbb{P}-{\rm a.s.}\end{equation}
\end{problem}
\begin{remark}
Recall that in the quantile hedging problem for the usual trader we maximize the objective probability $\mathbb{P}(\alpha\mathbb{E}_{\mathbb{Q}_\mathbb{F}}H+\int_0^T\xi_t \;{\rm d}S_t\geq H)$, where $\alpha$ is number from $[0,1]$.
In the Problems \ref{P1}-\ref{P2} we use conditional probability, since
now the insider's perception of the market at time $t=0$ depends on the knowledge described by $\mathcal{G}_0$.
\end{remark}
The set
\begin{equation}\label{succset}\left\{\alpha\mathbb{E}_{\mathbb{Q}_\mathbb{G}}H+\int_0^T\xi_t \;{\rm d}S_t\geq H\right\}\end{equation}
we will call the \textit{success set}.
Denote
$$\mathbb{Q}^*(A)=\frac{\mathbb{E}_{\mathbb{Q}_\mathbb{G}}(H\mathbf{1}_A)}{\mathbb{E}_{\mathbb{Q}_\mathbb{G}}(H)},\qquad A\in\mathcal{G}_T.$$

The following theorems solve Problems \ref{P1} and \ref{P2}.
\begin{theorem}\label{maxprobab}
Suppose that there exists a $\mathcal{G}_0$-measurable random variable $k$ such that
\begin{equation}\label{eqk}\mathbb{Q}^*\left(\left.\left.\frac{\;{\rm d}\mathbb{Q}^*}{\;{\rm d}\mathbb{P}}\right|_{\mathcal{G}_T}\leq k\right|\mathcal{G}_0\right)=\alpha.\end{equation}
Then the maximal probability of a success set solving Problem \ref{P1} equals:
$$\mathbb{P}\left(\left.\left.\frac{\;{\rm d}\mathbb{Q}^*}{\;{\rm d}\mathbb{P}}\right|_{\mathcal{G}_T}\leq k\right|\mathcal{G}_0\right)$$
and it is realized by the strategy
$$\left(\mathbb{E}_{\mathbb{Q}_\mathbb{G}}\left[\left.H\mathbf{1}_{\left\{\left.\frac{\;{\rm d}\mathbb{Q}^*}{\;{\rm d}\mathbb{P}}\right|_{\mathcal{G}_T}\leq k\right\}}\right|\mathcal{G}_0\right],\tilde{\xi}\right),$$
which replicates the payoff $H\mathbf{1}_{\left\{\left.\frac{\;{\rm d}\mathbb{Q}^*}{\;{\rm d}\mathbb{P}}\right|_{\mathcal{G}_T}\leq k\right\}}$.
\end{theorem}

\begin{theorem}\label{minCostQH}
Suppose that there exists a $\mathcal{G}_0$-measurable random variable $k$ such that
\begin{equation}\label{eqk2}\mathbb{P}\left(\left.\left.\frac{\;{\rm d}\mathbb{Q}^*}{\;{\rm d}\mathbb{P}}\right|_{\mathcal{G}_T}\leq k\right|\mathcal{G}_0\right)=1-\epsilon.\end{equation}
Then the minimal $\mathcal{G}_0$-measurable random variable $\alpha$ solving Problem \ref{P2} equals:
$$\mathbb{Q}^*\left(\left.\left.\frac{\;{\rm d}\mathbb{Q}^*}{\;{\rm d}\mathbb{P}}\right|_{\mathcal{G}_T}\leq k\right|\mathcal{G}_0\right)$$
and it is realized by the strategy
$$\left(
\mathbb{E}_{\mathbb{Q}_\mathbb{G}}\left[\left.H\mathbf{1}_{\left\{\left.\frac{\;{\rm d}\mathbb{Q}^*}{\;{\rm d}\mathbb{P}}\right|_{\mathcal{G}_T}\leq k\right\}}\right|\mathcal{G}_0\right],\tilde{\xi}\right)$$ being the perfect hedge of $H\mathbf{1}_{\left\{\left.\frac{\;{\rm d}\mathbb{Q}^*}{\;{\rm d}\mathbb{P}}\right|_{\mathcal{G}_T}\leq k\right\}}.$
\end{theorem}

\begin{remark}
The assumptions that there exists $k$ satisfying (\ref{eqk}) and (\ref{eqk2}) are satisfied if
$\mathbb{P}(Z_T^{\mathbb{F}}H=0|\mathcal{G}_0)<\alpha$ and $\mathbb{P}(Z_T^{\mathbb{F}}H=0|\mathcal{G}_0)<1-\epsilon$ respectively, and
$Z_T^{\mathbb{F}}H$ has the conditional density on $\mathbb{R}_+$ given $G=g$ - see Section \ref{numerical} for the examples.
\end{remark}

The proofs of these theorems are given in Section \ref{proofs}.

\section{Numerical examples}\label{numerical}

In this section we consider the standard Black-Scholes model
in which the price evolution is described by the equation
$${\rm d}S_{t}=\sigma S_{t}{\rm d}W_{t}+\mu S_{t}{\rm d}t,$$
where $W$ is a Brownian motion, $\sigma,\mu>0$.
For simplicity we assume that interest rate is zero.
We analyze the Problem \ref{P2} for two examples of the insider information
and provide numerical results for pricing the vanilla call option, where
$$H=(S_T-K)^+$$
and $K$ is a strike price.

\subsection{The case of $G=W_{T+\delta}$}
It means that insider knows the stock price $G=S_{T+\delta}$
after the expiry date $T$.
In this case we have:
\begin{eqnarray*}
\lefteqn{\mathbb{P}(W_{T+\delta}\in {\rm d}z|\mathcal{F}_t)=\mathbb{P}(W_{T+\delta}-W_{t}+W_{t}\in {\rm d}z|\mathcal{F}_t)}\\
&&=  \frac{1}{\sqrt{2\pi(T+\delta-t)}}\exp\Big(-\frac{(z-W_{t})^{2}}{2(T+\delta-t)}\Big){\rm d}z\\
&& =  p_{t}^{z}\mathbb{P}(W_{T+\delta}\in {\rm d}z),
\end{eqnarray*}
where
$$p_{t}^{z}=\sqrt{\frac{T+\delta}{T+\delta-t}}
\exp\Big(-\frac{(z-W_{t})^{2}}{2(T+\delta-t)}+\frac{z^{2}}{2(T+\delta)}\Big).$$
Therefore:
\begin{eqnarray*}
\lefteqn{ \left.\frac{{\rm d}\mathbb{Q}_\mathbb{G}}{{\rm d}\mathbb{P}}\right|_{\mathcal{G}_T}=\frac{Z_T^{\mathbb{F}}}{p_T^G}=
\frac{\left.\frac{{\rm d}\mathbb{Q}_\mathbb{F}}{{\rm d}\mathbb{P}}\right|_{\mathcal{F}_T}}{p_T^G}=
\frac{\exp\Big(-\frac{\mu}{\sigma}W_T-\frac{1}{2}\big(\frac{\mu}{\sigma}\big)^{2}T\Big)
}{\sqrt{\frac{T+\delta}{\delta}}\exp\Big(-\frac{(W_{T+\delta}-W_T)^{2}}{2\delta}
+\frac{W_{T+\delta}^{2}}{2(T+\delta)}\Big)}}\\
&& =  \sqrt{\frac{\delta}{T+\delta}}\exp\Big(-\frac{\mu}{\sigma}W_T-\frac{1}{2}
\Big(\frac{\mu}{\sigma}\Big)^{2}T+\frac{(W_{T+\delta}-W_T)^{2}}{2\delta}-\frac{W_{T+\delta}^{2}}{2(T+\delta)}\Big)
\end{eqnarray*}
and
\begin{equation}\label{eq1ex1}\left.\frac{{\rm d}\mathbb{Q}^*}{{\rm d}\mathbb{P}}\right|_{\mathcal{G}_T}
=\frac{H}{\mathbb{E}_{\mathbb{Q}_\mathbb{G}}H}\left.\frac{{\rm d}\mathbb{Q}_\mathbb{G}}{{\rm d}\mathbb{P}}\right|_{\mathcal{G}_T}.\end{equation}
Note that
$\left.\frac{{\rm d}\mathbb{Q}^*}{{\rm d}\mathbb{P}}\right|_{\mathcal{G}_T}$ is a random variable with the conditional density
on $\mathbb{R}_+$ given $G=g$ and for given $\epsilon\in[0,1]$ we can find a $\mathcal{G}_0$-measurable random variable $k$ such that
$\mathbb{P}\left(\left.\left.\frac{{\rm d}\mathbb{Q}^*}{{\rm d}\mathbb{P}}\right|_{\mathcal{G}_T}\leq k\right|\mathcal{G}_0\right)=1-\epsilon$
if $\mathbb{P}(H=0|\mathcal{G}_0)<1-\epsilon$. Therefore, by Theorem \ref{minCostQH} the cost of the quantile hedging for the insider can be reduced in this case by the factor:
\begin{equation}\label{eq2ex1}\alpha=\mathbb{Q}^*\left(\left.\left.\frac{\;{\rm d}\mathbb{Q}^*}{\;{\rm d}\mathbb{P}}\right|_{\mathcal{G}_T}\leq k\right|\mathcal{G}_0\right).\end{equation}
Below, we provide the values of $\alpha$ for $\mu=0.08$, $\sigma=0.25$, $S_0=100$, $K=110$, $T=0.25$, $\delta=0.02$, and different values of $G$ and $\epsilon$. In the programme we use simple fact that $\mathbb{E}[f(W_T)|G=g]=f(g-W(\delta))$ for a measurable function $f$.\newline
\begin{center}
\begin{tabular}{|c|c|c|c|c|c|c|c|c|c|c|c|c|}
\hline
~ & \multicolumn{12}{|c|}{$G$}\\
\hline
~ & ~ & 105 & 106 & 107 & 108 & 109 & 110 & 111 & 112 & 113 & 114 & 115\\
\hline
\multirow{6}{*}{$\epsilon$} & 0.01 & 0.05 & 0.09 & 0.13 & 0.17 & 0.22 & 0.27 & 0.32 & 0.37 & 0.42 & 0.46 & 0.51\\
& 0.05 & $<0.01$ & 0.01 & 0.04 & 0.07 & 0.10 & 0.14 & 0.18 & 0.23 & 0.28 & 0.33 & 0.38\\
& 0.10 & $<0.01$ & $<0.01$ & 0.01 & 0.03 & 0.05 & 0.08 & 0.12 & 0.16 & 0.21 & 0.25 & 0.30\\
& 0.15 & $<0.01$ & $<0.01$ & $<0.01$ & 0.01 & 0.03 & 0.05 & 0.08 & 0.12 & 0.16 & 0.21 & 0.25\\
& 0.20 & $<0.01$ & $<0.01$ & $<0.01$ & $<0.01$ & 0.01 & 0.03 & 0.06 & 0.09 & 0.13 & 0.17 & 0.21\\
& 0.25 & $<0.01$ & $<0.01$ & $<0.01$ & $<0.01$ & $<0.01$ & 0.02 & 0.04 & 0.07 & 0.10 & 0.14 & 0.18\\
\hline
\end{tabular}
\end{center}
~\newline
\subsection{The case of $G=\mathbf{1}_{\left\{W_{T+\delta}\in[a,b]\right\}}$}

In this example the insider knows the range of the stock price $S_{T+\delta}$
after the expiry date $T$.
The straightforward calculation yields:
\begin{eqnarray*}
\mathbb{P}(G=1|\mathcal{F}_t) & = & \frac{1}{\sqrt{2\pi(T+\delta-t)}}\int_{a-W_t}^{b-W_t}\exp\Big(-\frac{u^2}{2(T+\delta-t)}\Big)\;{\rm d}u\\
~ & = & \Phi\left(\frac{b-W_t}{\sqrt{T+\delta-t}}\right)-\Phi\left(\frac{a-W_t}{\sqrt{T+\delta-t}}\right),\\
\end{eqnarray*}
where $\Phi$ is c.d.f. of the standard normal distribution. Thus,
$$p_t^1=\frac{\mathbb{P}(G=1|\mathcal{F}_t)}{\mathbb{P}(G=1)}=\frac{\Phi\left(\frac{b-W_t}{\sqrt{T+\delta-t}}\right)-\Phi\left(\frac{a-W_t}{\sqrt{T+\delta-t}}\right)}{\Phi\left(\frac{b}{\sqrt{T+\delta}}\right)-\Phi\left(\frac{a}{\sqrt{T+\delta}}\right)},$$
and similarly
$$p_t^0=\frac{1+\Phi\left(\frac{a-W_t}{\sqrt{T+\delta-t}}\right)-\Phi\left(\frac{b-W_t}{\sqrt{T+\delta-t}}\right)}{1+\Phi\left(\frac{a}{\sqrt{T+\delta}}\right)-\Phi\left(\frac{b}{\sqrt{T+\delta}}\right)}.$$
Hence
\begin{eqnarray*}
\lefteqn{\left.\frac{{\rm d}\mathbb{Q}_\mathbb{G}}{{\rm d}\mathbb{P}}\right|_{\mathcal{G}_T} =  \frac{\exp\left(-\frac{\mu}{\sigma}W_T-\frac{1}{2}\left(\frac{\mu}{\sigma}\right)^2T\right)}{
p_T^0\mathbf{1}_{\{G=0\}}+p_T^1\mathbf{1}_{\{G=1\}}}}\\
&& =  \left.\exp\left(-\frac{\mu}{\sigma}W_T-\frac{1}{2}\left(\frac{\mu}{\sigma}\right)^2T\right)\right/
\Bigg(\mathbf{1}_{\{G=0\}}\frac{1+\Phi\left(\frac{a-W_T}{\sqrt{T+\delta-t}}\right)-\Phi\left(\frac{b-W_T}{
\sqrt{T+\delta-t}}\right)}{1+\Phi\left(\frac{a}{\sqrt{T+\delta}}\right)-\Phi\left(\frac{b}{\sqrt{T+\delta}}\right)}\\
&&\hspace{7cm} +\mathbf{1}_{\{G=1\}}\frac{\Phi\left(\frac{b-W_T}{\sqrt{T+\delta-t}}\right)-\Phi\left(\frac{a-W_T}{\sqrt{T+\delta-t}}\right)}{
\Phi\left(\frac{b}{\sqrt{T+\delta}}\right)-\Phi\left(\frac{a}{\sqrt{T+\delta}}\right)}\Bigg)
\end{eqnarray*}
and $\mathbb{Q}^*$ and $\alpha$ are defined in (\ref{eq1ex1})-(\ref{eq2ex1}).
The table below provides the values of the optimal $\alpha$ for $\mu=0.08$, $\sigma=0.25$, $S_0=100$, $K=110$, $T=0.25$, $\delta=0.02$, $G=1$ and different values of $\epsilon$ and endpoints of interval $[a,b]$ for $S_{T+\delta}$. In the programme we use simple fact that $\mathbb{E}[f(W_T)|G=1]=\mathbb{E}[f(W_T), W_{T+\delta}\in[a,b]]/\mathbb{P}(G=1)$ for a measurable function $f$
and to simulate numerator we choose only those trajectories for which $W_{T+\delta}\in[a,b]$.
\newline
\begin{center}
\begin{tabular}{|c|c|c|c|c|c|c|}
\hline
~ & \multicolumn{6}{|c|}{$[a,b]$}\\
\hline
~ & ~ & [109,111] & [108,112] & [107,113] & [112,114] & [106,108] \\
\hline
\multirow{6}{*}{$\epsilon$} & 0.01 & 0.272 & 0.284 & 0.296 & 0.413 & 0.135 \\
& 0.05 & 0.142 & 0.150 & 0.157 & 0.277 & 0.039 \\
& 0.10 & 0.087 & 0.088 & 0.095 & 0.209 & 0.010 \\
& 0.15 & 0.053 & 0.055 & 0.059 & 0.164 & 0.001 \\
& 0.20 & 0.032 & 0.033 & 0.034 & 0.129 & $<0.001$ \\
& 0.25 & 0.017 & 0.019 & 0.020 & 0.102 & $<0.001$ \\
\hline
\end{tabular}
\end{center}

\section{Proofs}\label{proofs}

Before we give the proofs of the main Theorems \ref{maxprobab} and \ref{minCostQH} we present few
introductory lemmas and theorems. We start with the result of \cite{A00} and \cite{AIS98} concerning the properties of the equivalent martingale measure $\mathbb{Q}_\mathbb{G}$ for the insider. We recall that we assume that
the condition (\ref{equiv}) is satisfied.
\begin{theorem}\label{ther1}
\begin{itemize}
\item[(i)] Process $Z_t^{\mathbb{G}}:=\frac{Z^{\mathbb{F}}_t}{p_t^G}$ is a $(\mathbb{G},\mathbb{P})$-martingale.
\item[(ii)] The measure $\mathbb{Q}_\mathbb{G}$ defined in (\ref{QG}) has the following properties:
\subitem(a) $\mathcal{F}_T$ and $\sigma(G)$ are independent under $\mathbb{Q}_\mathbb{G}$;
\subitem(b) $\mathbb{Q}_\mathbb{G}=\mathbb{Q}_\mathbb{F}$ on $(\Omega,\mathcal{F}_T)$ and $\mathbb{Q}_\mathbb{G}=\mathbb{P}$ on $(\Omega,\sigma(G))$.
\end{itemize}
\end{theorem}
We are now in a position to state the theorem
which relates the martingale measures of the insider and the regular trader.
\begin{theorem}\label{ther2}
Let $X=(X_{t})_{t\geq 0}$ be an $\mathbb{F}$-adapted process. The following statements are equivalent:
\begin{itemize}
\item[(i)]$X$ is an $(\mathbb{F},\mathbb{Q}_\mathbb{F})$-martingale;
\item[(ii)]$X$ is an $(\mathbb{F},\mathbb{Q}_\mathbb{G})$-martingale;
\item[(iii)]$X$ is a $(\mathbb{G},\mathbb{Q}_\mathbb{G})$-martingale.
\end{itemize}
\end{theorem}
\begin{proof}
Equivalence (i) and (ii) follows from the fact that
$\mathbb{Q}_\mathbb{F}=\mathbb{Q}_\mathbb{G}$ on $\mathcal{F}_T$. The implication $(iii)\Rightarrow (ii)$ is a consequence of the
tower property of the conditional expectation.
Finally, taking $A=A_{s}\cap\{\omega\in \Omega: G(\omega)\in B\}$ ($A_{s}\in\mathcal{F}_s$, $B$ - Borel set), the
implication $(ii)\Rightarrow (iii)$ follows from the standard monotone class arguments and following equalities:
\begin{eqnarray*}
\lefteqn{\mathbb{E}_{\mathbb{Q}_\mathbb{G}}(\mathbf{1}_{A}X_{t})=\mathbb{E}_{\mathbb{Q}_\mathbb{G}}(\mathbf{1}_{A_s}\mathbf{1}_{\{G\in B\}}X_{t})=
\mathbb{Q}_\mathbb{G}(G\in B)
\mathbb{E}_{\mathbb{Q}_\mathbb{G}}(\mathbf{1}_{A_s}X_{t})}
\\&&=\mathbb{Q}_\mathbb{G}(G\in B)
\mathbb{E}_{\mathbb{Q}_\mathbb{G}}(\mathbf{1}_{A_s}X_{s})=
\mathbb{E}_{\mathbb{Q}_\mathbb{G}}(\mathbf{1}_{A}X_{s}),\qquad s\leq t,
\end{eqnarray*}
where in the second equality we use Theorem \ref{ther1}(ii).
\end{proof}
\begin{remark}
We need the Theorem \ref{ther2} to guarantee the martingale representation for the insider's replicating strategy. Moreover, from this representation and the Theorem \ref{ther1} we can deduce that the cost of perfect
hedging for the insider is the same as for the regular trader, that is $\mathbb{E}_{\mathbb{Q}_\mathbb{G}}[H|\mathcal{G}_0]=\mathbb{E}_{\mathbb{Q}_\mathbb{F}}H$.
\end{remark}
\begin{remark}
In general the Theorem \ref{ther2} is not true for the local martingales, since a localizing sequence $(\tau_{n})$ of $\mathbb{G}$-stopping times is not a sequence of $\mathbb{F}$-stopping times.
\end{remark}

\begin{lemma}\label{NPL}
Let $k$ be a positive $\mathcal{G}_t$-measurable random variable. For every $A\in\mathcal{F}$ such that
$$\mathbb{Q}^*\left(\left.A\right|\mathcal{G}_t\right)\leq\mathbb{Q}^*
\left(\left.\left.\frac{{\rm d}\mathbb{Q}^*}{{\rm d}\mathbb{P}}\right|_{\mathcal{G}_T}\leq k\right|\mathcal{G}_t\right),\qquad \mathbb{P}-{\rm a.s.}$$
we have:
$$\mathbb{P}\left(\left.A\right|\mathcal{G}_t\right)\leq\mathbb{P}
\left(\left.\left.\frac{{\rm d}\mathbb{Q}^*}{{\rm d}\mathbb{P}}\right|_{\mathcal{G}_T}\leq k\right|\mathcal{G}_t\right),\qquad \mathbb{P}-{\rm a.s.}$$
Similarly, if
$$\mathbb{P}\left(\left.A\right|\mathcal{G}_t\right)\geq\mathbb{P}
\left(\left.\left.\frac{{\rm d}\mathbb{Q}^*}{{\rm d}\mathbb{P}}\right|_{\mathcal{G}_T}\leq k\right|\mathcal{G}_t\right),\qquad \mathbb{P}-{\rm a.s.}$$
then
$$\mathbb{Q}^*\left(\left.A\right|\mathcal{G}_t\right)\geq\mathbb{Q}^*
\left(\left.\left.\frac{{\rm d}\mathbb{Q}^*}{{\rm d}\mathbb{P}}\right|_{\mathcal{G}_T}\leq k\right|\mathcal{G}_t\right),\qquad \mathbb{P}-{\rm a.s.}$$
\end{lemma}
\begin{proof}
Denote $\tilde{A}:=\left\{\left.\frac{{\rm d}\mathbb{Q}^*}{{\rm d}\mathbb{P}}\right|_{\mathcal{G}_T}\leq k
\right\}$. Note that $$(\mathbf{1}_{\tilde{A}}-\mathbf{1}_A)
\left(k-\left.\frac{{\rm d}\mathbb{Q}^*}{{\rm d}\mathbb{P}}\right|_{\mathcal{G}_T}\right)
\geq 0.$$ Thus
\begin{eqnarray*}
\left.\frac{\;{\rm d}\mathbb{Q}^*}{\;{\rm d}\mathbb{P}}\right|_{\mathcal{G}_t}\left(\mathbb{Q}^*(\tilde{A}|\mathcal{G}_t)-\mathbb{Q}^*(A|\mathcal{G}_t)\right)
&=&\left.\frac{\;{\rm d}\mathbb{Q}^*}{\;{\rm d}\mathbb{P}}\right|_{\mathcal{G}_t}\mathbb{E}_{\mathbb{Q}^*}\left((\mathbf{1}_{\tilde{A}}-\mathbf{1}_{A})\left.\right|\mathcal{G}_t\right)\\
&=&\mathbb{E}_\mathbb{P}\left(\left.\left.\frac{{\rm d}\mathbb{Q}^*}{{\rm d}\mathbb{P}}\right|_{\mathcal{G}_T}(\mathbf{1}_{\tilde{A}}-\mathbf{1}_{A})\right|\mathcal{G}_t\right)\\
&\leq&k\left(\mathbb{P}(\tilde{A}|\mathcal{G}_t)-\mathbb{P}(A|\mathcal{G}_t)\right),
\end{eqnarray*}
which completes the proof.
\end{proof}

\begin{lemma}\label{equalone}
The following holds true:
$$\left.\frac{{\rm d}\mathbb{Q}^*}{{\rm d}\mathbb{Q}_\mathbb{G}}\right|_{\mathcal{G}_0}=1.$$
\end{lemma}
\begin{proof}
Note that for $A=\{\omega\in\Omega: G\in B\}\in \mathcal{G}_0$ ($B$ is a Borel set) we have
\begin{eqnarray*}
\lefteqn{\mathbb{Q}^*(A)=\mathbb{E}_{\mathbb{Q}_{\mathbb{G}}}\left[\frac{{\rm d}\mathbb{Q}^*}{{\rm d}\mathbb{Q}_\mathbb{G}}\mathbf{1}_{A}\right]=\frac{\mathbb{E}_{\mathbb{Q}_{\mathbb{G}}}\left[H\mathbf{1}_{A}\right]}{\mathbb{E}_{\mathbb{Q}_{\mathbb{G}}}H}}\\
&&=
\frac{\mathbb{E}_{\mathbb{Q}_{\mathbb{G}}}\left[H\right]\mathbb{Q}_{\mathbb{G}}(A)}{\mathbb{E}_{\mathbb{Q}_{\mathbb{G}}}H}=\mathbb{Q}_{\mathbb{G}}(A),
\end{eqnarray*}
where in the last but one equality we use Theorem \ref{ther1}.
\end{proof}

{\it Proof of Theorem \ref{maxprobab}}

Consider the value process $V_{t}=\alpha\mathbb{E}_{\mathbb{Q}_\mathbb{G}}H+\int_{0}^{t}\xi_{u}\;{\rm d}S_{u}$ for
any strategy $(\alpha\mathbb{E}_{\mathbb{Q}_\mathbb{G}}H,\xi)\in\mathcal{A}^\mathbb{G}$. Note that for its success set $A$ defined in (\ref{succset})
we have:
$$V_T\geq H\mathbf{1}_A.$$
Moreover, by the Theorem \ref{ther2} the process $V_{t}$ is a nonnegative $(\mathbb{G},\mathbb{Q}_\mathbb{G})$-local martingale, hence it is a $(\mathbb{G},\mathbb{Q}_\mathbb{G})$-supermartingale and
$$\alpha\mathbb{E}_{\mathbb{Q}_\mathbb{G}}H=\alpha V_0\geq \mathbb{E}_{\mathbb{Q}_\mathbb{G}}(V_T|\mathcal{G}_0)\geq \mathbb{E}_{\mathbb{Q}_\mathbb{G}}(H\mathbf{1}_{A}|\mathcal{G}_0).$$
Thus, from Lemmas \ref{equalone} and \ref{NPL},
$$\mathbb{Q}^*(A|\mathcal{G}_0)\leq\frac{\alpha}{\left.\frac{{\rm d}\mathbb{Q}^*}{{\rm d}\mathbb{Q}_\mathbb{G}}\right|_{\mathcal{G}_0}}=\alpha,\qquad \mathbb{P}-{\rm a.s.}$$
and therefore
\begin{eqnarray}
\mathbb{P}(A|\mathcal{G}_0)\leq\mathbb{P}\left(\left.\left.\frac{{\rm d}\mathbb{Q}^*}{{\rm d}\mathbb{P}}\right|_{\mathcal{G}_T}\leq k\right|\mathcal{G}_0\right),\qquad \mathbb{P}-{\rm a.s.} \label{P_ineq}
\end{eqnarray}
It remains to show that $\left(\mathbb{E}_{\mathbb{Q}_\mathbb{G}}\left[H \left.\mathbf{1}_{\left\{\left.\frac{{\rm d}\mathbb{Q}^*}{{\rm d}\mathbb{P}}\right|_{\mathcal{G}_T}\leq k\right\}}\right|\mathcal{G}_0\right],\tilde{\xi}\right)\in\mathcal{A}^\mathbb{G}$, and that this strategy attains the upper bound
(\ref{P_ineq}). The first statement
follows directly from the definition of $\tilde{\xi}$:
$$
\mathbb{E}_{\mathbb{Q}_\mathbb{G}}\left(H \left.\mathbf{1}_{\left\{\left.\frac{{\rm d}\mathbb{Q}^*}{{\rm d}\mathbb{P}}\right|_{\mathcal{G}_T}\leq k\right\}}\right|\mathcal{G}_0\right)
+\int_0^t\tilde{\xi}_u\;{\rm d}S_u=\mathbb{E}_{\mathbb{Q}_\mathbb{G}}\left(H \left.\mathbf{1}_{\left\{\left.\frac{{\rm d}\mathbb{Q}^*}{{\rm d}\mathbb{P}}\right|_{\mathcal{G}_T}\leq k\right\}}\right|\mathcal{G}_t\right)\geq 0.$$
Moreover,
\begin{eqnarray}
\lefteqn{\mathbb{P}\left(\left.\mathbb{E}_{\mathbb{Q}_\mathbb{G}}\left[H \left.\mathbf{1}_{\left\{\left.\frac{{\rm d}\mathbb{Q}^*}{{\rm d}\mathbb{P}}\right|_{\mathcal{G}_T}\leq k\right\}}\right|\mathcal{G}_0\right]
+\int_0^T\tilde{\xi}_u\;{\rm d}S_u\geq H\right|\mathcal{G}_0\right)}\nonumber\\&&=
\mathbb{P}\left(\left.H \mathbf{1}_{\left\{\left.\frac{{\rm d}\mathbb{Q}^*}{{\rm d}\mathbb{P}}\right|_{\mathcal{G}_T}\leq k\right\}}\geq H\right|\mathcal{G}_0\right)\geq
\mathbb{P}\left(\left. \left.\frac{{\rm d}\mathbb{Q}^*}{{\rm d}\mathbb{P}}\right|_{\mathcal{G}_T}\leq k\right|\mathcal{G}_0\right),\nonumber
\end{eqnarray}
which completes the proof in view of (\ref{P_ineq}).
\vspace{3mm} \hfill $\Box$

{\it Proof of Theorem \ref{minCostQH}}

Observe that for any $(\alpha\mathbb{E}_{\mathbb{Q}_\mathbb{G}}H,\xi)\in\mathcal{A}^\mathbb{G}$ we
have: \begin{eqnarray}
\lefteqn{\mathbb{Q}^*\left(\left.\alpha\mathbb{E}_{\mathbb{Q}_\mathbb{G}}H+\int_0^T\xi_u\;{\rm
d}S_u\geq H\right|\mathcal{G}_0\right)}\nonumber\\&&
=\frac{1}{\left.\frac{{\rm d}\mathbb{Q}^*}{{\rm
d}\mathbb{Q}_\mathbb{G}}\right|_{\mathcal{G}_0}}
\mathbb{E}_{\mathbb{Q}_\mathbb{G}}\left(\left.\left.\frac{{\rm d}\mathbb{Q}^*}{{\rm
d}\mathbb{Q}_\mathbb{G}}\right|_{\mathcal{G}_T}\mathbf{1}_{\left\{\alpha\mathbb{E}_{\mathbb{Q}_\mathbb{G}}H+\int_0^T\xi_u\;{\rm
d}S_u\geq H\right\}}\right|\mathcal{G}_0\right)\nonumber\\&&
\leq\frac{\mathbb{E}_{\mathbb{Q}_\mathbb{G}}\left(\left.\alpha\mathbb{E}_{\mathbb{Q}_\mathbb{G}}H+\int_0^T\xi_u\;{\rm
d}S_u\right|\mathcal{G}_0\right)}{\mathbb{E}_{\mathbb{Q}_\mathbb{G}}H}=\alpha.\nonumber \end{eqnarray}
Applying second part of Lemma \ref{NPL} for the success set
$A=\left\{\alpha\mathbb{E}_{\mathbb{Q}_\mathbb{G}}H+\int_0^T\xi_u\;{\rm d}S_u\geq
H\right\}$ and using required inequality (\ref{P2inequality}) and
definition of $k$ given in (\ref{eqk2}) we derive: \begin{eqnarray}
\alpha&\geq&\mathbb{Q}^*\left(\left.\alpha\mathbb{E}_{\mathbb{Q}_\mathbb{G}}H+\int_0^T\xi_u\;{\rm d}S_u\geq H\right|\mathcal{G}_0\right)\nonumber\\
&\geq&\mathbb{Q}^*\left(\left.\left.\frac{{\rm d}\mathbb{Q}^*}{{\rm d}\mathbb{P}}\right|_{\mathcal{G}_T}\leq k\right|\mathcal{G}_0\right).\nonumber
\label{rhs}
\end{eqnarray}
We prove now that for this particular minimal choice of $\alpha$ being the rhs of (\ref{rhs})
the strategy $\left(\mathbb{E}_{\mathbb{Q}_\mathbb{G}}\left[H \left.\mathbf{1}_{\left\{\left.\frac{{\rm d}\mathbb{Q}^*}{{\rm d}\mathbb{P}}\right|_{\mathcal{G}_T}\leq k\right\}}\right|\mathcal{G}_0\right],\tilde{\xi}\right)$ satisfies the inequality (\ref{P2inequality}) of the Problem \ref{P2}:
\begin{eqnarray}
\lefteqn{\mathbb{P}\left(\left.\alpha\mathbb{E}_{\mathbb{Q}_\mathbb{G}}H+\int_0^T\tilde{\xi}_u\;{\rm d}S_u\geq H\right|\mathcal{G}_0\right)}\nonumber\\
&&=\mathbb{P}\left(\left.\mathbb{E}_{\mathbb{Q}_\mathbb{G}}\left.\left[H \mathbf{1}_{\left\{\left.\frac{{\rm d}\mathbb{Q}^*}{{\rm d}\mathbb{P}}\right|_{\mathcal{G}_T}\leq k\right\}}\right|\mathcal{G}_0\right]
+\int_0^T\tilde{\xi}_u\;{\rm d}S_u\geq H\right|\mathcal{G}_0
\right)\nonumber\\
&&=
\mathbb{P}\left(\left. H \mathbf{1}_{\left\{\left.\frac{{\rm d}\mathbb{Q}^*}{{\rm d}\mathbb{P}}\right|_{\mathcal{G}_T}\leq k\right\}}
\geq H\right|\mathcal{G}_0
\right)
\nonumber\\
&&\geq\mathbb{P}\left(\left.\left.\frac{{\rm d}\mathbb{Q}^*}{{\rm d}\mathbb{P}}\right|_{\mathcal{G}_T}\leq k\right|\mathcal{G}_0\right)=1-\epsilon.\nonumber
\end{eqnarray}
This completes the proof.
\vspace{3mm} \hfill $\Box$

\section*{Acknowledgements}
This work is partially supported by the Ministry of Science and Higher Education of Poland under the grant N N2014079 33 (2007-2009).

\bibliographystyle{abbrv}


\section{Juggling with one or two balls}
Mathematically, juggling is studied via siteswap sequences (see \cite{jugggle}) which are a sequence of numbers $t_1t_2\ldots $ which indicates a throw of a ball to ``height'' $t_i$ at time step $i$ (i.e., so the ball will land at time $i+t_i$).  A periodic juggling pattern is one which repeats in which case we only need to include enough of the sequence to get us started.  A siteswap with period $n$ then will be a sequence $t_1t_2\ldots t_n$ where at time $i+kn$ we throw a ball to height $t_i$.  One simple assumption to make is that no two balls {\em land}\/ at the same time (in multiplex juggling this condition is relaxed).  This translates into the condition
\[
\{1,2,\ldots,n\}=\bigcup_{i=1}^n\{i+t_i\pmod n\}.
\]
A simple consequence of this is that $(t_1+t_2+\cdots+t_n)/n$ is integer valued and corresponds to the number of balls needed to juggle the sequence.  (When $t_i=0$ then this indicates an empty hand or a no throw.)

Another type of juggling has recently been introduced by Akihiro Matsuura \cite{spherical} which involves juggling inside of a sphere where balls are ``thrown'' from the north pole on great circles to return back after some number of time steps.  Such juggling sequences are again denoted by a siteswap $t_1t_2\ldots t_n$ and in addition to the condition that $i+t_i \pmod n$ are distinct (avoiding collisions at the north pole) we also need that $i+(t_i/2) \pmod n$ are distinct for the $t_i\neq 0$ (to avoid collisions at the south pole).

We now  count the number of one ball siteswap sequences of length $n$ for both forms of juggling, these have the added advantage of being impossible to have collisions.  In this case since there is only one ball then we have that $t_1+t_2+\cdots+t_n=n$, i.e., no balls can be thrown further than length $n$.  In particular, the only thing that we need to know is at what times we have the ball in our hands and at what time our hand is empty.  This can be done by associating such juggling sequences with necklaces with two colors of beads (with position $1$ at the top and increasing clockwise).  The black beads will indicate when the ball is in our hand and the white beads will indicate when the ball is in the air.  For example, in Figure~\ref{fig:jug} we have a necklace with black and white beads, to translate this into a juggling pattern we associate each white bead with a $0$ and each black bead with the number of steps until the next black bead (note in one ball juggling we must throw to the next occurrence of a black bead).  So this particular necklace is associated with the juggling sequence $30020300$.

\begin{figure}[hftb]
\centering
\includegraphics{necklaces.eps}
\caption{A necklace corresponding to the juggling siteswap sequence $30020300$.}
\label{fig:jug}
\end{figure}

Since we must have at least one black bead and there are $n$ beads then in total we have $2^n-1$ different possible necklaces (i.e., everything but the all white bead necklace) and so $2^n-1$ different siteswap sequences (note that we consider rotations of necklaces distinct just as rotations of siteswap sequences are considered distinct).  This analysis also works for spherical juggling.

Now let us consider the number of two ball juggling siteswap sequences with the added condition that a non-zero throw is made at each time step (i.e., $t_i\geq 1$ for all $i$).  There is a natural correspondence between one ball juggling siteswaps and two ball juggling siteswaps with a non-zero throw made at each time step.  Namely if $t_1t_2\ldots t_n$ is a one ball juggling siteswap sequence then we associate it with $t_1't_2'\ldots t_n'$ by letting $t_i'=t_i+1$.  Clearly this does not add any collisions where two balls land at the same time and so we can conclude that the number of two ball siteswap sequences with no non-zero throws of length $n$ is $2^n-1$.

For spherical juggling however the map $t_1t_2\ldots t_n \mapsto t_1't_2'\ldots t_n'$ {\em might}\/ create collisions at the south pole.  For example the sequence $20\mapsto 31$, this latter sequence is not a spherical juggling sequence since $0+3/2\equiv 1+1/2\pmod n$.  Let us determine which one ball juggling sequences will produce valid two ball juggling sequences where a throw is made at every time step for spherical juggling.  Our only problem will be the creation of a collision at the south pole.

Suppose that we have the necklace associated with with the one ball siteswap sequence $t_1t_2\ldots t_n$.  The halfway points between two consecutive black beads marks the times where the ball is at the south pole.  Now when we go to $t_1't_2'\ldots t_n'$ all the occurrences of the ball at the south pole will shift clockwise by exactly a half step {\em and}\/ every white bead will add a new ball at the south pole exactly a half step after the white bead.  So collisions at the south pole will exist precisely when the halfway point between two consecutive black beads lands on top of a white bead.

Therefore, in terms of necklaces, we need to have at least one black bead and between any two consecutive black beads the number of white beads must be {\em even}.

\section{Counting admissible necklaces}
So to find the number of two ball spherical juggling siteswap sequences of length $n$ with a throw made at each time we need to count the number of necklaces on $n$ beads where (i) there is at least one black bead; (ii) between any two (consecutive) black beads there is an even number of white beads; and (iii) rotations/flippings of necklaces are considered distinct.  To solve this we will find a recurrence and this is done in the following more general theorem.

\begin{theorem}\label{mainresult}
Let $a_q(n)$ be the number of arrangements of black and white beads on a necklace with a total of $n$ beads satisfying the following:
(i) there is at least one black bead;
(ii) between any two black beads the number of white beads is divisible by $q$;
(iii) rotations/flippings of a necklace are considered distinct.
Then $a_q(0)=0$, $a_q(1)=\cdots=a_q(q)=1$ and for $n\ge q+1$
\begin{equation}\label{recurq}
a_q(n)=\left\{\begin{array}{l@{\qquad}l}
a_q(n-1)+a_q(n-q)+q&\mbox{if }n\equiv 1\pmod{q};\\
a_q(n-1)+a_q(n-q)&\mbox{otherwise.}
\end{array}\right.
\end{equation}
\end{theorem}

We are interested in the case $q=2$ which produces the sequence
\[
0, 1, 1, 4, 5, 11, 16, 29, 45, 76, 121, 199, 320, 521, 841, 1364, 2205, 3571, 5776, 9349, 15125,\ldots.
\]
These are the ``associated Mersenne numbers'' ({\tt A001350} in Sloane \cite{sloane}), which satisfy the recurrence
\[
a_2(n)=a_2(n-1)+a_2(n-2)+1-(-1)^n.
\]
Unsurprisingly these are connected to Lucas numbers, $L_n$, which satisfy a similar recurrence.  In particular the Lucas numbers start
\[
2, 1, 3, 4, 7, 11, 18, 29, 47, 76, 123, 199, 322, 521, 843, 1364, 2207, 3571, 5778, 9349, 15127,\ldots.
\]
Comparing the two sequences it is easy to see (and show) that
\[
a_2(n)=\left\{\begin{array}{l@{\qquad}l}L_n&\mbox{if $n$ is odd;}\\
L_n-2&\mbox{if $n$ is even.}\end{array}\right.
\]
In particular, our the rate of growth of the number of two ball spherical juggling siteswap sequences of length $n$ with a throw made at each time step is approximately $c\phi^n$ where $\phi$ is the golden ratio.  By comparison the number of two ball juggling siteswap sequences of length $n$ with a throw made at each time step grows at a rate of $c2^n$, significantly more.

For $n$ a prime, besides the one necklace with all black beads the remaining necklaces can be grouped into sets related by rotation.  This gives the following result.

\begin{corollary}\label{prime}
If $n$ is prime then $a_q(n)\equiv 1\pmod n$.  In particular, $L_n\equiv 1\pmod n$ for $n$ a prime.
\end{corollary}
\begin{proof}
We already have shown that $a_q(n)\equiv 1\pmod n$ for primes.   To establish the result for Lucas numbers we note that for $n=2$ we have $L_2=1\equiv1\pmod2$; otherwise, $L_n=a_2(n)\equiv1\pmod n$.
\end{proof}

This result is one of the well known properties of Lucas numbers (see \cite{bicknell}) and this proof is essentially the same as the proof given by Benjamin and Quinn \cite{count} where they relate Lucas numbers to tilings of necklaces with curved squares (which correspond to black beads) and curved dominoes (which correspond to  pairs of white beads).

\begin{proof}[Proof of Theorem~\ref{mainresult}]
We observe the initial conditions are satisfied.  First, $a_q(0)=0$ since an admissible necklace must have at least one black bead.  Similarly, $a_q(1)=\cdots=a_q(q)=1$ since the only admissible necklace in these cases is one with all black beads (i.e., there is not enough space to put $q$ white beads and at least one black bead in the necklace).

There is a $1$-$1$ correspondence between necklaces on $n-q$ beads and necklaces on $n$ beads with white beads in positions $1,2,\ldots,q$.  Namely, given such a necklace on $n$ beads we remove the white beads in position $1,2,\ldots,q$ to get an admissible necklace on $n-q$ beads, this process can also be reversed.

Similarly there is a $1$-$1$ correspondence between necklaces on $n-1$ beads and necklaces on $n$ beads with $2$ or more black beads and at least one black bead in one of the first $q$ entries. Namely, given such a necklace we remove one of the black beads in position $1,2,\ldots,q$ to get an admissible necklace on $n-1$ bead, this process can be reversed.

(Note it is not as clear that the reverse process results in a unique necklace. To see this we note that for the necklace on $n-1$ beads there are essentially two cases that can occur among the first $q-1$ beads: (1) if all of the beads are white then to satisfy the condition that between any two black beads the number of white beads is a multiple of $q$ there is a unique positions where it can be placed; (2) if some of the beads are black then it must be placed among/adjacent to the existing black beads so that again the condition that the number of white beads is multiple of $q$ is satisfied.  In either case the reverse process results in a unique necklace.)

Finally, we count what is left.  These are necklaces on $n$ beads with one black bead and it occurs in one of the positions $1,2,\ldots,q$.  This is only possible if $n\equiv 1\pmod q$ (since there are exactly $n-1$ white beads between the black bead and itself).  So there are $q$ remaining necklaces when $n\equiv 1\pmod q$ and $0$ otherwise.

Combining this altogether we can conclude that the recursion given in \eqref{recurq} holds.
\end{proof}