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

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

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

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

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

It can be below 0 or above 1.

On the subject of delta calculation take a look at this article:

https://www.linkedin.com/pulse/spot-vol ... y-klassen/

On the subject of delta calculation take a look at this article:

https://www.linkedin.com/pulse/spot-vol ... y-klassen/

@billy524

Totally agree with @frolloos. As long as the assumptions on which your model is based are met, the model**is correct** by definition.

You may be interested in**what happens when some assumptions are violated**. You are going to need a measure of the "performance" achieved by your model and find the impact in terms of this performance. Not an easy task I would say, but some typical examples are "hedging in the black-scholes world with the wrong volatility" or "disregarding second order movements in yield curves" or "reality is mean reverting, model is not".

You should always test the sensitivity of your model to the misspecification of its characteristics.

Totally agree with @frolloos. As long as the assumptions on which your model is based are met, the model

You may be interested in

You should always test the sensitivity of your model to the misspecification of its characteristics.

- complyorexplain
**Posts:**93**Joined:**

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

If you mean 'the sensitivity of observed price to change in observed underlying price', well that could be anything really. It's likely to be close to the theoretical delta for no-arbitrage reasons, but that's all.

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.

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.

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.

Ah - but this isn’t physics! 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. Assume that the current stock price S satisfies H<S<K. It is an exercise left to the reader to find parameter values such that the current call delta will be negative.

"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.

This is unphysical.

"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.

Correct! That’s exactly my point. Physics has nothing to do with finance, aside from a tradition of sharing some math. Well, that and finance being overrun by droves of physicists unable to find gainful employment in their primary field.

Who would you rather go into quantitative finance? A bunch of mathematicians who couldn't trade their way out of a paper bag?

Meh - having spent a relatively long career mostly managing quants (of various stripes, from exotic derivatives on the sell side to traditional buy side, with a bit of hedge fund, analytical software and insurance work thrown in for good measure), I find that the good ones transcend their original field of study. They may have studied math or physics or, probably my favorite, electrical engineering; but they also have taken the trouble to learn enough of the relevant finance to be able to put things in context. For some reason, this is much harder than it looks. And you have a few notable Finance guys who managed to pick up some math and programming along the way and do well for themselves in the business. But I’ll certainly concede that the supply of reasonably brilliant and hard working physicists has had a major impact on the field. And, as far as trading is concerned, few PhDs of any kind has the right temperament for the role. Having to make lots of poorly informed judgment calls under tight time pressure is not usually a good fit for those of us who like to sit back and ponder things a bit.

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