# Arbitrage free implies complete market?

In Tomas Björk's Arbitrage Theory in Continuous Time (or here), $\exists$ this proposition It seems that to show that the model is complete, we must show that the claims are reachable. That is, we must find replicating portfolios.

Which part of finding the replicating portfolio makes use of the assumption?

You can view the arbitrage-free statement as being about infinitely liquid trading.

If one has to trade $x$ by paying a half-spread $\nu$, and you have a trivial payoff $\Phi(\cdot) \equiv 1+R$ then there is no solution. You would be trying to solve $$(1+R)x - \nu + suy = (1+R)x = (1+R)x - \nu + sdy$$

• What is half spread? Anyway I think I got this. Just plug in x and y into value process then show h is not an arb portfolio – BCLC Jul 17 '14 at 18:54
• One often thinks of the cost of entering a position as the quantity $q$ times the difference $\nu$ between the offer (bid) and the midpoint. If the spread is $h$ then $\nu=h/2$, so I call it a half-spread. – Brian B Jul 17 '14 at 20:03
• Hmmm...according to the book bid=ask so spread=0...so your equation implies u=d which contradicts d<1+R< u following from the no arb assumption...wait I think I get it... – BCLC Jul 17 '14 at 20:06
• Is it that the only reason we are able to solve the system is because of d<1+R<u? – BCLC Jul 17 '14 at 20:08

The only reason we are able to solve the system is because of $d<u$ which follows from $d<1+R<u$, which follows from absence of arbitrage by Prop 2.3

• Hmm, I don't see how this answers your question in toto: "Which part of finding the replicating portfolio makes use of the assumption?". Can you elaborate? – Bob Jansen Jul 17 '14 at 20:20
• @Bob if d<u then we can solve the system. I didn't see it right away until I considered Brian's equations without the v. It says in the proof " Since by assumption..." though that is not obvious. From absence of arbitrage we infer the inequality and thus we can solve the system and obtain the replicating portfolio – BCLC Jul 17 '14 at 20:21
• Still not quite sure I follow, can you explain Prop 2.3? – Bob Jansen Jul 17 '14 at 20:23
• @Bob market is free of arbitrage iff the inequality holds. The explanation of the inequality is that the return on the bond is not allowed to be greater than the return on the stock or vice versa. If it were ($d<u \leq 1+R$ or $1+R \leq d<u$), then we can come up with arbitrage (and if we can come up with arbitrage then it must be that the one of the other inequalities is true). I forgot to mention that my prof uses definitions diff from author. Prop 2.3 in book is not a strict inequality, which I never really got. I think it would have an impact on martingale prob computations – BCLC Jul 17 '14 at 20:32

A market is arbitrage-free, if riskneutral measure Q exists (under which the discounted stockprice becomes martingale).

A market is complete, when the riskneutral measure Q is unique.

Therefore, any market with a riskneutral measure Q is arbitrage-free, and if Q is unique it is also complete.

The riskneutral probabilities $q$ are unique for the binomial model, so it is arbitrage-free and complete.

For any multinomial model with $m>2$ stockmoves, Q has infinite solutions hence market is incomplete, but it is still arbitrage-free because Q's exist.

One can show how the existence and uniqueness of the riskneutral measure Q for a Call in the Binomial Model implies existence of a replicating portfolio as follows:

At time $t=0$, the asset price is $S_0$ and the call option price is $C_0$, to be determined. At time $t=1$, there are two possible asset prices $S_{1u} = S_0(1+u)$ with probability $p$ and $S_{1d} = S_0(1+d)$ with probability $1-p$. The payoff of the option at expiration is

$$C_1 = \max(S_1-K,0),$$

where the random variable $S_1$ is $S_{1u}$ or $S_{1d}$ depending on the future state of the market:

$$C_{1u} = \max(S_{1u}-K,0)\\\ C_{1d} = \max(S_{1d}-K,0)$$

The value of the option at time $t=0$ is the discounted expected value of the payoff. However this expected value is not calculated with real probability $p$ -- but rather the risk-neutral probability $\hat{p}$ that preclude arbitrage opportunities.

We can determine the no-arbitrage risk-neutral probability $\hat{p}$ by showing it is possible to construct a hedged portfolio of the option and the asset that is risk free -- it has the same value in both future states. Hence the value of the portfolio grows in time at the risk-free rate of interest $i$.

Suppose the portfolio is long $1$ call option and short $\Delta$ shares of the asset. The value at time $t=1$ is

$$V_t= C_t-\Delta S_t.$$ We can solve for the hedge ratio $\Delta$ so that the value of the portfolio at time $t=1$ is independent of the state of the market:

$$C_{1u}-\Delta S_{1u}=C_{1d}-\Delta S_{1d},$$

or

$$C_1-\Delta S_1=C_{1u}-\Delta S_0(1+u)=C_{1d}-\Delta S_0(1+d).$$

This value of the hedge ratio is independent of the probabilities for the value of the asset at time $t=1$:

$$\Delta = \frac{C_{1u}-C_{1d}}{S_0(u-d)}$$

Consequently, in the absence of arbitrage, the portfolio grows at the risk-free rate:

$$C_1-\Delta S_1 = (C_0 - \Delta S_0)(1+i)$$

and we can solve for the value of the call option at time $t=0:$

$$C_0 -\Delta S_0= \frac{C_{1u}-\Delta S_0 (1+u)}{1+i},$$

$$C_0 = \frac{\Delta S_0 (1+i) +C_{1u}-\Delta S_0 (1+u)}{1+i},$$

$$C_0 = \frac{-\Delta S_0 (u-i) + C_{1u}}{1+i}.$$

Substituting for $\Delta$, we obtain

$$C_0 = \frac{\hat{p}C_{1u}+(1-\hat{p})C_{1d}}{1+i}=\frac{1}{1+i}\{\hat{p}\max[S_0(1+u)-K,0)]+(1-\hat{p})\max[S_0(1+d)-K,0)]\},$$

where

$$\hat{p} = \frac{i-d}{u-d}.$$

This has the form of an expected value with a different probability -- the risk-neutral probability. The fair value of the call option is the discounted expected value under the risk-neutral probability measure.