[1] The Binomial Model (The One Period Model)

The one-period model of the binomial model (-> simplist "toy" model)

(Notatioin)

• $Z$ is $u$ with probability $p_u$ and $d$ with probability $p_d$ be a random variable such that $p_u+p_d=1$.

• $B_t =$ the price of a bond at time t
  Then, $B_0=1$ and $B_1 = 1+R$

• $S_t =$ the price of a stock at time t
  Then, $S_0 = s$ and $S_1 = \begin{cases} su,\;with\;p_u\\ sd,\;with\;p_d \end{cases}$

(1) Portfolio

A vector $h=(x,y)$ is portfolio such that $x$ is the value of bonds and $y$ is the number of stocks.

(Assumption)

• $h = (x, y) \in \mathbb{R}^2$ is allowed : dept, fraction, $\cdots$
• selling price = buying price
• no transaction cost
  
  
  

(2) Value process $V_t^h$

It means the total asset price of $h$ at time $t$
Let $V_0^h=x+ys$, then
$V_1^h=\begin{cases} x(1+R)+ysu,\; with \;p_u \\ x(1+R)+ysd,\;with\;p_d\end {cases}$

(3) Arbitrage portfolio

A portfolio $h$ is said to be an arbitrage portfolio if

• today : $V_0^h=0$
• tomorrow : $P(V_1^h \geq 0) = 1$ & $P(V_1^h>0)>0$

It means $h$ cannot lose but gain.
($P(V_1^h=0)=1$ is impossible.)

In the real market, It is almost impossible for arbitrage portfolio to exist. We call arbitrage free when there is no arbitrage portfolio.

(Proposition)

An one-period model is arbitrage free $\Leftrightarrow$ $d<1+R<u$

proof)
($\rightarrow$) Suppose $d\geq 1+R$ or $u \leq 1+R$.
Then we can take a arbitrage portfolio. But It is a contradiction.
($\leftarrow$) Suppose that there is a arbitrage porfolio $h=(x,y)$.
But It does not satisfy the condition of arbitrage porfolio.

(4) Martingale measure

The martingale measure is the probability such that these conditions.

• $q_u + q_d =1$
• $E^Q[S_1] = (su)q_u + (sd)q_d = s(1+R)$

So, maringale measure $Q=(q_u, q_d) = (\frac{(1+R)-d}{u-d}, \frac{u-(1+R)}{u-d})$

(5) Contigent claim (Financial derivative)

$\Phi : \mathbb{R} \to \mathbb{R}$ is a contract function from $x$ to $\Phi(x)$.
It means the value of the contigent claim depending on the stock price $x$.

For example, the value of the European call option is $\Phi(x) = (x-K)_+$.

For hedging risk, we always assume that $sd<K<su$.

(Notation)

$\Pi(t;X)$ is the price of $X$ at time $t$ such that $X = \Phi(x)$.

In the case of European call option, $\Pi(1;X)=\Phi(S_1)=(S_1-K)_+$

(6) Hedging portfolio

A contigent claim $X$ is called reachable if there exists a portfolio $h=(x,y)$ such that $V_1^h=X$. and such $h$ is called hedging portfolio(replicating portfolio). If all claims are reachable, the market is said to be complete.

We want to find hedging portfolio $h=(x,y)$ such that
$\Phi(S_1)=V_1^h.$

$\begin {cases} \Phi(su)=x(1+R)+ysu \\ \Phi(sd)=x(1+R)+ysd\end {cases}$

$\therefore h=(x,y) = (\frac{1}{1+R}\frac{u\Phi(sd)-d\Phi(su)}{u-d}, \frac{1}{s}\frac{\Phi(su)-\Phi(sd)}{u-d})$

(7) Risk neutral evaluation

Because $\Phi(S_1)$ is replicated by hedging portfolio $h=(x,y)$, It is reasonable to say that today's fair price of $\Phi(S_1)$ is eqaul to $V_0^h$.

$\Pi(0;X) = V_0^h = x+sy = \frac{1}{1+R}\frac{u\Phi(sd)-d\Phi(su)}{u-d} + s \frac{1}{s}\frac{\Phi(su)-\Phi(sd)}{u-d}$

$\to \Pi(0;X)=\frac{1}{1+R}\{\frac{(1+R)-d}{u-d}\Phi(su)+ \frac{u-(1+R)}{u-d}\Phi(sd)\}$

$\to \Pi(0;X) = \frac{1}{1+R} \{q_u\Phi(su)+q_d\Phi(sd)\}$

$\to \Pi(0;X) = \frac{1}{1+R}E^Q[X]$

It is surprising conclution. 😲

$\Pi(t;X) = \begin {cases} \Phi(S_1), \; t=1 \\ \frac{1}{1+R}E^Q[\Phi(S_1)], \; t=0\end {cases}$

The fair relates to martingale measure $(q_u, q_d)$, not to real probablility $(p_u, p_d)$.