# Is the binomial model wrong?

In the standard MBA one-period binomial model, the value of an option is

$v = \frac{1}{R}\bigl(\frac{u - R}{u - d}V(sd) + \frac{R - d}{u - d}V(su)\bigr)$

where $R$ is the realized return over the period and the stock goes from $s$ down to $sd$ or up to $su$, where $d\lt R\lt u$, and $V$ is the option payoff. Note

$\frac{dv}{ds} = \frac{1}{R}\bigl(\frac{u - R}{u - d}V'(sd)d + \frac{R - d}{u - d}V'(su)u\bigr)$

is the "delta" hedge. Suppose $V$ is a call spread consisting of long a call struck at slightly higher than $sd$ and short a call struck slightly lower than $su$, then $V'(sd) = V'(su) = 0$, hence $dv/ds = 0$.

Wat?! How can that be???

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If price of underlined changes , does it change Option Price ? Yes rt. So without looking at Binomial or Black Scholes or "my own model" dv/ds != 0 . So if thats the case then going backwards what you defined as detla hedge for Binomial model is incorrect. –  ash May 10 '12 at 12:59

OK, I'm going to be slightly facetious but I hope marginally helpful, by quoting legendary Prof. George E. P. Box: "all models are wrong, but some are useful."

So, yes, certainly, the binomial model is wrong. It is also incredibly useful!

Where you are going wrong is using it as a continuous model (and using differential calculus) when it is intended to be a simplified discrete model of the world. The only changes that matter in this discrete model are discrete changes. If you look at everything as differences in value between t to t+1 then it all makes sense. Anything between t and t+1 (including at time t+$\delta$t) is undefined.

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I agree that models are unperfected by their very nature. –  edouard Feb 12 '13 at 0:26
But in a binomial model where $s$ can go to $S^-$ or $S^+$ the option value is $v = ((S^+ - Rs)V(S^-) + (Rs - S^-)V(S^+)/R(S^+ - S^-)$ and $dv/ds = (V(S^+) - V(S^-))/(S^+ - S^-)$. I have to confess I am trolling a little here. See kalx.net/ftapd.pdf for the explanation. Also note that $d(Rv)/dR = s(V^+ - V^-)/(S^+ - S^-)$ is the dollar delta in both models. –  Keith A. Lewis May 8 '12 at 17:00
Relax, I'm a friendly troll. :-) Here is another one that I don't have a good answer for. Let $F = fe^{-\sigma^2t/2 + \sigma B_t}$. The (forward) value of a log contract is $v = E[\log F]$ and so $dv/df = 1/f$. No surprise. Now consider the payof $v = E[\log F/f]$. Now $dv/df = 0$! It is easy to see what is going on in this example, but how do you know with more complicated parameterizations when the correct delta is not the derivative of value with respect to underlying? Hope you find this puzzle more interesting. –  Keith A. Lewis May 9 '12 at 20:26