Hi,
Does anyone know a no-arbitrage proof that call option delta has to be between 0 and 1? We know that Black Scholes model N(d1) tells us that it is between 0 and 1, but is there a no-arbitrage argument?
Thanks
Define "correct". Aren't all models approximations? Some are good/realistic approximations, others are mathematically interesting but may not be appropriate to explain a particular phenomenon.Thanks a lot. The deeper I go into models, the more I ask myself, "how do we know our models derivations/results" are correct. Sometimes getting confused between real financial markets and mathematical models
Depends what you mean by 'delta'. If you mean N(d1), then (as you say) it is by definition between 0 and 1. Or if you mean 'the sensitivity of option price to change in the underlying price' then you have to ask what is meant by 'option price'. If it means the price given by the standard BS formula, then again it comes down to N(d1). The price given by the standard BS formula is of course established using the no-arb assumption, so there is such a proof as you mention.Hi,
Does anyone know a no-arbitrage proof that call option delta has to be between 0 and 1? We know that Black Scholes model N(d1) tells us that it is between 0 and 1, but is there a no-arbitrage argument?
Thanks
Just look at the physics! Not the first derivatives or the N(d1).
there are two extrema -
(1) S =1, K = 99, T = 1mo... this option is worthless and will expire worthless. It has no sensitivity to S, i.e. the delta = 0.
(2) S =999, K = 99, T = 1mo. This option will be exercised, you are going to own the share with probability 1. The option is just a short-term vehicle for delivering the equity, and must move 1-for-1
with S. In other words, the delta = 1.
All other cases lie somewhere in-between.
"Consider a stock price process of the BSM form, except that the first time the stock price hits the level H its volatility permanently doubles."
This is unphysical.