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\textbf{Question}

Consider the simple binomial model for asset price changes:

The asset price at time $t=0$ os $S+0$ and at time $t=1$ it can either
be $S_2$ with the probability $0 < p < 1$ or $S_1 < S_2$ with
probability $1-p$. Show that unless
$$S_1 < S_0e^r < S_2$$
where $r$ is the risk free rate, this model is arbitragabile. (You may
assume that short sales are allowed.)

A derivative security, $V$, is written on this asset. At time $t=0$
its value is $V_0$. At time $t=1$ its value can be either $V_1$, if
the underlying's price is $S_1$, or $V_2$ if the underlying's price
is $S_2$.

\begin{description}
%Question 2a
\item{(a)}
Let $V_0^p$ denote the present value (at time $t=0$) of the expected
value of $V$ at time $t=1$. Give a formula for $V_0^p$ in terms of
$p$, $r$, $V_1$ and $V_2$.

%Question 2b
\item{(b)}
Construct a risk free portfolio containing bot $V$ and $S$ and use an
arbitrage argument to show that this leads to a ``fair price'' for
$V_0$, say $V_0^{\Delta}$, in terms of $S_0$, $S_1$, $S_2$, $V_1$ and
$V_2$ but \textit{not} $p$.

%Question 2c
\item{(c)}
Construct a replicating strategy, in terms of $S$ and cash invested at
the risk-free rate $r$, which leads to an arbitrage free price for
$V_0$, say $V_0^R$, in terms of $S_0$, $S_1$, $S_2$, $V_1$ and $V_2$
but \textit{not} $p$.

%Question 2d
\item{(d)}
Deduce from wither (b) or (c) that there is a number $q$, which may be
regarded as a ``risk neutral'' probability, associated with the
underlying's price, such that the ``fair value'' of $V_0$ is the
present value (at $t=0$) of the expected value of $V$ at time $t=1$.

%Question 2e
\item{(e)}
Assuming that either $V_2 < V_1$ or $V_2 > V_1$, and that $p>q$, show
that a trader using either of the prices $V_0^{\Delta}=V_)^T$ from (b)
or (c) would necessarily be able to arbitrage a trader using the price
$V_0^p$ from (a).
\end{description}

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\textbf{Answer}

If $S_1,\ S_2>S_0e^r$ then whatever happens, risky asset grows faster
than risk free rate $\Rightarrow$ borrow $S_0$ now buy risky asset,
payback loan next time step $\Rightarrow$ arbitrage

If $S_1,\ S_2<S_0e^{rst}$, short sell asset now, invest in bank, use
$S_0e^r>S_1,\ S_2$ to close out short sale next time step
$\Rightarrow$ arbitrage.

\begin{description}
%Question 2a
\item{(a)}
 $V_0^p=e^{-r}(pV_2+(1-p)V_1)$

%Question 2b
\item{(b)}
 $\Pi=V-\Delta S$ so $\Pi_0=V_0-\Delta S_0$.

At time 1, $\Pi = \left \{ \begin{array}{ll} \Pi_1 & if S_0\rightarrow
S_1\\ \Pi_2 & if S_0\rightarrow S_2 \end{array} \right.$

where $\Pi_1=V_1-\Delta S_1$, $\Pi_2=V_2-\Delta S_2$. Eliminate risk
free by setting $\Pi_1=\Pi_2$.
\begin{eqnarray*} \Rightarrow V_1-\Delta S_1 & = & V_2 - \Delta S_2\\
\Delta & = & \displaystyle \frac{V_2-V_1}{S_2-S_1} 
\end{eqnarray*}

This gives $\Pi_0=e^{-r}\Pi_1=e^{-r}\pi_2$ since $\Pi$ is riskless and
hence
\begin{eqnarray*} V_0 & = & e^{-r}\pi_1\\ \Rightarrow V_0 & = &
e^{-r}(V_1- \Delta S_1)+\Delta S_0.
\end{eqnarray*}

Price is ``fair'' in the sense that if you think true price is
$\overline{V}_0 > V_0$, then soon can sell you option for
$\overline{V}_0$, construct risk free portfolio and close out at $t=1$,
making $\overline{V}_0 -V_0$ risk free profit. Conversely, if someone
is willing to sell for $\hat{V}_0 < V_0$, you should buy, short sell
risk free portfolio and close out at $t=1$ making $V_0-\hat{V}_0 >0$
risk free profit.

%Question 2c
\item{(c)}
Let $\Pi=\phi S_0+ \psi$ where $\psi$ is invested at risk free rate.
\begin{eqnarray*} \Pi_0 & = & \phi S_0 +\psi,\\ \Pi_1 & = & \phi S_1 +
\psi e^r,\\ \Pi_2 & = & \phi S_2+ \psi e^r.
\end{eqnarray*}

setting $\Pi_1=V_1$, $\Pi_2=V_2$, i.e. $$\phi S_1+ \psi e^r= V_1,\ \ \
\phi S_2 +\psi e^r  V_2$$

$\Rightarrow \displaystyle \phi =\frac{V_2-V_1}{S_2-S_1}$, $\psi =
e^{-r}(V_1-\phi S_1)=e^{-r}(V-2- \phi S_2)$.

Then, since $\Pi=V$ at time $1$, with certainty, $\Pi_0=V_0$, hence
$$V_0= e^{-r}(V_1- \phi S_1)+ \phi S_0$$

(note $\phi$ in (c) $=\Delta$ in (b))

%Question 2d
\item{(d)}
From either (b) or (c)
\begin{eqnarray*} V_0 & = & \displaystyle e^{-r}(V_1 -
\frac{V_2-V_1}{S_2-S_1}s_1) + \frac{V_2-V_1}{S_2-S_1}S_0\\ & = & \displaystyle
\frac{1}{S_2-S-1}[e^{-r}(V-1(S_2-S_1)-(V_2-V_1)S_1)\\
& & + (V_2-V_1)S_0]\\ &
= & \displaystyle \frac{e^{-r}}{S_2-S_1}[V_1S_2- V_1S_1-V_2S_1+
V_1S_1+ V_2S_0e^r- V_1S_0e^r]\\ & = & \displaystyle
\frac{e^{-r}}{S_2-S_1}[ (S_0e^r-S_1)V_2+ (S_2-S_0e^r)V_1]\\ & = &
e^{-r}[qV_2+(1-q)V_1]
\end{eqnarray*}
$$\displaystyle q=\frac{S_0e^r-S_1}{S_2- S_1}, \ \ \
1-q=\frac{S_2-S_0e^r}{S_2-S_1}$$

Can think of q as a probability since $S_1<S_0e^r,S_2$ so $0<q<1$.

%Question 2e
\item{(e)}
Suppose $V_2 \ne V_1$. Then $V_0^p=e^{-r}[pV_2+(1-p)V_1] \ne 0$, for
if $p>q$, $V_0^p=V_0$
\begin{eqnarray*}
\Rightarrow pV_2+(1-p)V_1 & = & V_2+(1-q)V_1\\ \Rightarrow {p-q}V_2 &
= & (p-q)V_1\\ \Rightarrow V_2 & = & V_1 \ \rm{if} \ \ p>q. \ \
(\rm{indeed \ if\ }p\ne q)
\end{eqnarray*}

Thus $V_0^p \ne V_0$, so a trader selling at $V_0^p$ can be arbitraged
by either selling at $V_0^p$ if $V_0^p>V_0$ and carrying out risk free
strategy (b) or (c) or buying at $V_0^p$ if $V_0^p<V_0$ and shorting
the risk free strategy (b) or (c).

\end{description}


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