# The reason behind the selection of a 1 standard deviation movement for self financing delta hedge

I'm learning this material and I can follow the derivation of the BSM PDE fairly well. The only problem I have is there is an assumption in the derivation (that I am reading) that a stock price movement of exactly 1 standard deviation will produce a self financing delta hedge and thus should earn the risk free rate and everything falls into place from there. Where did the assumption of 1 standard deviation come from? It seems like it could have been any movement (y) and the PDE would adjust so then the option price formula would adjust and the market maker would then have a self financing portfolio when the movements are always (y). Am I missing some part of Ito's Lemma that makes the one standard deviation the magic number?

• Where did you find this "assumption of 1 standard deviation"? There should be no such assumption. Please provide a referenece. – Alexey Kalmykov Jul 12 '13 at 8:26

The self-financing model that leads to the Black Scholes formula generally only makes distributional assumptions not assumptions about the absolute variability of the underlying assets.

Was such assumption part of the discretization approach? Because then infinitesimal asset value changes without changes in positions in the assets would form the definition of a self-financing portfolio and then such infinitesimal change may include bounds or specific values of asset variability.

But the way you said it was stated I would say is incorrect because a self-financing portfolio can include an asset that has any kind of variability as long as the model distributional assumptions are respected. The assumed distribution is of course wrong but market accepted standard in pricing derivatives that utilize the derived B-S framework.

Your questions lacks a bit of detail. However: Since you are referring to a PDE it appears as if the Black Scholes formula is proved by considering a discrete model (1 standard-deviation move per time-step), then taking a limit "time-step size to zero".

For example, in a tree, you are essentially approximating the normal distributed increment with a Bernoulli distributed increment (you can reach only two states per time-step and node) and the relation of these two states is that "one standard-deviation". In the limit this will converge to the normal distribution with the respective standard deviation (Theorem of Donsker).

I you would use some other variability it would converge to some other normal distribution (and not to your model).

Put differently: It is assumed that the approximation has only two states and given the time step, it is necessary that these two states differ by one standard deviation to obtain the desired normal distribution in the limit.

(This is just a guess, since your questions lacks detail on how the proof was actually done).

For any option, as you make your time increment shorter, a quadratic approximation of the sensitivity of the option to S will be a better and better fit to the actual potential swing in the option price based on S's behavior over the time increment. Basically, if you adjust your hedge often enough, more complex curvature in the option price won't matter because you're going to catch changes before they move into a region with a different slope.

A hedge that is delta neutral and nothing else is based on a linear approximation. You have to pay attention to a quadratic approximation because S accumulates variance at a rate of sigma squared, which can be viewed as a deterioration in your hedge at a rate of sigma. To control the situation you have to maintain your hedge at a rate that matches the rate of deterioration. This is an issue that sticks around no matter how short you make the time interval. You can wash out other details, but you have to make a provision for this.

The other details come from market equilibrium arguments. A radius of one standard deviation results in the actual outcome washing out over many time increments and any other radius is going to generate a surplus or shortfall over time. You argue that if one standard deviation is statistically risk free over time, it has to return the risk free rate. This puts you in the position that hedging in a way that produces a radius greater than one standard deviation gives you a position that is still statistically risk free but returns less than the risk-free rate, and that hedging to produce a radius shorter than one standard deviation gives you a position that has a risk exposure.

I figured this out after studying. I'll try and do a better job of communicating my question in my answer.

My basic question was when a market maker is in a delta-hedged position after writing a vanilla call option why is it that a stock movement of one standard deviation produces a profit of \$0 (approximately), ie the movement in value of calls = the movement in value of stock position.

I understood that it was due to interpreting the gamma coefficient in the BS PDE as a stock price movement, which would produce a movement in the call values exactly equal to the movement in the stock values.

What I didn't understand was why the gamma coefficient had to be sigma^2 x S^2, which if interpreted as a stock price movement produced that magic number effect. Most of the derivations I saw just went from one line to the next without explaining where that coefficient came from.

When I was working through the derivation of Ito's Lemma I figured it out, that coefficient is necessary if the underlying asset's distribution is modeled as a geometric browinan motion, from the (ds)^2 term.

Hopefully that makes sense.

• This should probably be an amendment to your question rather than an answer. – chrisaycock Jul 15 '13 at 19:29