Take the 2-minute tour ×
Quantitative Finance Stack Exchange is a question and answer site for finance professionals and academics. It's 100% free, no registration required.

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?

share|improve this question

1 Answer 1

up vote 1 down vote accepted

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 $t=1$.

share|improve this answer
    
Thanks! I now know how better to phrase future/related questions to my prof. ^_^ Putting Math and Finance together is really sudden. –  BCLC Jun 23 at 14:02

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.