# Arbitrage Strategy Proof in Bjork

In Tomas Bjork's Arbitrage Theory in Continuous Time (or here), $\exists$ this proposition

Proposition 2.9 Suppose that a claim X is reachable with replicating portfolio h. Then any price at t=0 of the claim X, other than $V_0^{h}$ will lead to an arbitrage possibility.

My prof uses $V_0({\phi})$ instead of $V_0^{h}$, but $\phi$ still refers to the portfolio.

Let $\Pi(t;x)$ be the price of the contingent claim at time t. Then, $\Pi(0;x)$ must = $V_0({\phi})$.

This is the proof written on the board:

Suppose $\Pi(0;x)$ > $V_0({\phi})$.

The arbitrage strategy is:

Sell (or short sell) the claim for $\Pi(0;x)$, and obtain the portfolio $\phi$ worth $V_0({\phi})$.

Left over amount is $\Pi(0;x)$ - $V_0({\phi})$.

At t = 1, the payoff for the claim X w/c you will be liable for will be covered the value of the portfolio $V_1({\phi})$ at t=1.

Suppose $\Pi(0;x)$ < $V_0({\phi})$.

The arbitrage strategy is:

Sell (or short sell) the portfolio worth $V_0({\phi})$. Use that amount to buy claim worth $\Pi(0;x)$.

Left over amount is $V_0({\phi}) - \Pi(0;x)$.

At t = 1, you will get payoff X, w/c is equal to $V_1({\phi})$.

Soooo I tried constructing the arbitrage strategy for the first part to see the exact profit, but I seem to be missing a step.

Suppose $\Pi(0;x)$ > $V_0({\phi})$.

At t = 0,

transaction $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \$ cash flow

1 short/sell claim for $\Pi(0;x)$ $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \$ +$\Pi(0;x)$

2 Buy $\phi$ $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \$ -$V_0({\phi})$

3 Invest $\Pi(0;x)$ - $V_0({\phi})$ at R until t=1 $\ \ \ \ \ \ \ \ \ \ \ \ \ \$ -($\Pi(0;x)$ - $V_0({\phi})$)

$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \$ 0

At t = 1,

transaction $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \$ cash flow

1 collect investment $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \$ +($\Pi(0;x)$ - $V_0({\phi})$)(1+R)

2 Portfolio grows in value $\phi$ $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \$ +$V_1({\phi})$

3 [...]? $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \$ -$V_1({\phi})$

4 Close short position if needed $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \$ 0

Profit $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \$ +($\Pi(0;x)$ - $V_0({\phi})$)(1+R)

Where does the $V_1({\phi})$ go? I was thinking that we were supposed to borrow $\frac{V_1({\phi})}{1+R}$ at t=0 so it would look something like:

Suppose $\Pi(0;x)$ > $V_0({\phi})$.

At t = 0,

transaction $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \$ cash flow

1 short/sell claim for $\Pi(0;x)$ $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \$ +$\Pi(0;x)$

2 Buy $\phi$ $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \$ -$V_0({\phi})$

3 Borrow $\frac{V_1({\phi})}{1+R}$ at R until t=1 $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \$ +$\frac{V_1({\phi})}{1+R}$

4 Invest $\Pi(0;x)$ - $V_0({\phi})$ + $\frac{V_1({\phi})}{1+R}$ at R until t=1 $\ \ \ \ \ \ \ \ \ \ \ \ \ \ $$-(\Pi(0;x) - V_0({\phi}) + \frac{V_1({\phi})}{1+R}) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0 At t = 1, transaction \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ cash flow 1 collect investment \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \$$(\Pi(0;x)$ - $V_0({\phi})$ + $\frac{V_1({\phi})}{1+R})(1+R)$

2 Pay debt $\phi$ $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \$ -$V_1({\phi})$

3 Close short position if needed $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \$ 0

Profit $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \$ +($\Pi(0;x)$ - $V_0({\phi})$)(1+R)

Or is there some action for the missing step that makes both of them equivalent?

If $V_0(\phi) < \Pi(0,x)$

at $t=0$

1. You sell short the claim and collect $\Pi(0,x)$
2. You buy the portfolio $\phi$ for $V_0(\phi)$
3. You put the money $\Pi(0,x) - V_0(\phi)$ in your risk-free instruments

at $t=1$

1. At $t=1$ you'll be liable the payoff of the claim you have shorted. The money you owe the counterpart long the claim is $\Pi(1,x)$.
2. $\phi$ is the replicating portfolio hence $V_1(\phi) = \Pi(1,x)$. You can sell the portfolio $\phi$ and get $V_1(\phi)$
3. Therefore, you are left with $(\Pi(0,x) - V_0(\phi)) (1+R)$ in your bank account.

I don't think you'll need to borrow $V_1(\phi) / (1+R)$ at $t=0$: you can already finance the long position in the portfolio with part of the proceeds from the short sale of the claim.

In any case, you don't know the value $V_1(\phi)$ at $t=0$. All you know is that $V_1(\phi) = \Pi(1,x)$ at $t=1$.

• Thanks! I now know how better to phrase future/related questions to my prof. ^_^ Putting Math and Finance together is really sudden.
– BCLC
Commented Jun 23, 2014 at 14:02