# Binomial model arbitrage

I've recently started studying math finance from Shreve's Stochastic calculus text. In the binomial model, there is no arbitrage $$\iff d<1+r. To show that no arbitrage implies $$1+r, suppose $$1+r\geq u$$. Short sell $$x$$ stock and invest in the money market, so after time $$1$$, $$x(1+r)\geq u>d$$.

Again, I'm new to these financial concepts, but my understanding is at time zero, you assume you have zero wealth. So you'd have to borrow $$x$$ stock, but then wouldn't there be some type of interest that you would incur?

In theory, we do not suppose there are transaction costs (or costs for short selling or even buying a security). In practice, effectively, you will have to pay the people that lend you the security you want to short (this activity is called security lending).

What we notice, all the cases, is that if there exists a risk free rate such that $$1 + r > u$$ and that the short selling costs are negligeable compare to the difference $$1 + r - u$$, you still should proceed with the arbitrage. Otherwise not. So, we notice that costs on short selling reduce the possibility of arbitrage.

The reason why we often do not consider transaction costs in mathematical finance, is that, they depend on each agent ( your size, your credit rating, the past relationship with your broker etc..).

A different way of thinking is risk-neutral pricing. Recall that a market is free of arbitrage if and only if there exists (at least) one risk-neutral probability measure. This is the (first) Fundamental Theorem of Asset Pricing.

The risk-neutral probability is frequently set to be $$q=\frac{e^{r\Delta t}-d}{u-d}.$$ This expression defines a valid probability measure if $$q$$ takes only values between zero and one. Firstly, since $$u>d$$, you need $$e^{r\Delta t}> d$$ for positivity. Secondly, you need $$e^{r\Delta t}< u$$ for the fraction to be less than one. Thus, $$d< e^{r\Delta t}< u.$$ Of course, the binomial tree lives in discrete time such that discrete compounding is more appropriate. Thus, translating everything into discrete time, you obtain $$d<1+R

• Would the argument around q being a valid prob measure need tweaking for what Shreve calls the uninteresting case : u=d? Sep 12 '19 at 9:18
• Well, of course, for $u=d$, the probability $q$ is not well-defined but in that case, you also do not have a Binomial model or indeed any model at all. Because the stock price would be completely deterministic and there would be no randomness at all. As you and Shreve said: "quite uninteresting". Hence, we typically assume $d<u$. Sep 12 '19 at 9:28
• d=1+r=u arbitrage argument works right? Sep 12 '19 at 9:33
• Yes, if there exists no well-defined equivalent martingale measure, then there exist arbitrage strategies. If there are two assets (stock and risk-free asset), with same drift, no uncertainty ($u=d$) but different prices, we can construct an arbitrage strategy by purchasing the cheap one, short selling the expensive asset and pocketing the difference. Sep 12 '19 at 9:46
• I would disagree though, a good pricing theory should cover all cases- including the trivial ones! But that is a side topic. As we have spaced out, and I think all the key considerations have been covered in our comments, I am going to stop here. It was pleasure exchanging ideas with you, and thanks for the dynamic exchange! Sep 12 '19 at 11:29

Say the stock price is x, you short sell one unit of stock, and the proceeds earn r.

In the u state, the stock investment will be worth $$-ux$$, and your bank account will be worth $$x\left(1+r\right)$$. So total profit and loss will be:

$$PL=-x u + x\left(1+r\right)= x\left(1+r-u\right)$$

And in the d state, your PL will be:

$$PL=-x d+ x\left(1+r\right)= x\left(1+r-d\right)$$

If $$1+r$$ is greater than $$u$$ and $$d$$, you make a profit.

Similarly you can argue that $$1+r$$ cannot be lower than both u and d etc.

Hope this helps!