Arbitrage free implies complete market in general binomial model?

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, i.e. we must find replicating portfolios for each claim.

Which part of finding the replicating portfolio makes use of the assumption, where I understand the assumption is equivalent to $$d < 1+R < u$$ (or $$d \le 1+R \le u$$, but $$d)? All I see so far is the $$d, but that's nothing really special: of course the stock price $$u$$ if it would go up from its current price should be higher than the stock price $$d$$ if it would go down from the its current price.

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.

• i don't get it. 1 - arbitrage-free is equivalent to $d < 1+R < u$ right? 2 - where exactly is $d < 1+R < u$ used in the proof (either your proof or the one in the book) (besides of course the weaker $d < u$) ? is it in that $0 < \hat{p} < 1$? I think I get it if $i=1+R$. is it? it seems like $0 < \hat{p} < 1$ relies on $d < 1+R < u$...maybe is even equivalent...?
– BCLC
Sep 1 '21 at 13:06
• @BCLC In this formulation the jump is "1+u", hence p only exists (0<p<1) for u>d and r>d. From 1-p>0, one can derive that (u-r)/(u-d)>0 for r<u. The question is in a way misleading, as this is not an assumption but a consequence of the condition for p to exist. Sep 2 '21 at 14:18
• A - by r>d do you mean 1+R>d so r=1+R?
– BCLC
Sep 14 '21 at 3:58
• B - by '1-p>0' do you mean $1-\hat p>0$ ?
– BCLC
Sep 14 '21 at 3:58
• C - emcor, is this question relevant? $E[F_T] = F_0 \ \rightarrow \ \text{or} \ \leftarrow \ p = \frac{1-d}{u-d}$. thanks for your answer and replies thus far
– BCLC
Sep 14 '21 at 3:59

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 $$(1+R)x - \nu + suy = (1+R)x = (1+R)x - \nu + sdy$$

• 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. Jul 17 '14 at 20:03
• i don't get it. 1 - arbitrage-free is equivalent to $d < 1+R < u$ right? 2 - where exactly is $d < 1+R < u$ used in the proof (besides of course the weaker $d < u$) ?
– BCLC
Sep 1 '21 at 13:04