For a pure equity process (with interest rate, dividend, etc., being zero) not necessarily the geometric Brownian motion, is the delta of a European call option always no higher than $1$? I am NOT asking for the Black-Scholes delta, but a model free general property of the European call delta. We can consider this question with and without the martingale property that the expected underlying price should be the current price.
I have also formulated this question a more formal fashion here as a calculus of variation or linear programming problem.
Answer
It is false. Here is an example. Let $$ dS_t = rS_t dt + f(S_0) S_t dW_t, $$ $$ dB_t = r dt. $$ The price is then the Black-Scholes price with volatility $f(S_0).$ The delta is the BS delta plus $$ f'(S_0) \times \operatorname{BS Vega}. $$ Picking $f$ appropriately, we can make the Delta as big as we like.
Note that the example is highly artificial in that volatility is a function of $S_0$ rather than $S_t.$
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