# Why A Derivative With Intrinsic Arbitrage Cannot Be Valued & Hedged With Assets In Risk Neutral?

I'm attempting to concisely show why a derivative that, by nature, introduces arbitrage cannot be valued using risk neutral pricing tools.

Derivative:

Buyer is sold a 'call option', with time 0 value consistent with:

$$V(t_{0}) = C(t_{0}, T, \sigma, r, q, K)$$

Where C is the standard black scholes (merton) formula for the price of a european call option with payoff $$(S_{T} -K)^{+}$$ at maturity T.

Here is the wrinkle:

Up until $$t_{m}$$ where $$t_{0} \leq t_{i} < t_{m} < T$$, the holder can sell the call back to the writer for current B-S market value of the option $$C(t=t_{i}, T, \sigma, r, q, K), t_{i} < t_{m}$$.

At $$t_{m}$$ the writer takes, as a 'fee', $$x \%$$ of the option. Therefore the holder/buyer of the option has value function:

$$V(t_{i} < t_{m}) = C(t_{i}, T, \sigma, r, q, K)$$

$$V(t_{i} \geq t_{m}) = (1-x\%) * C(t_{i}, T, \sigma, r, q, K)$$

This obviously has intrinsic arbitrage, call prices are martingales with respect to the risk neutral measure so without compensating the holder, the fee structure introduces arbitrage.

Question:

Outside of arguments around replication, what mathematical violation prevents us from modelling the value of the derivative to the writer in risk neutral (and therefore 'hedged' taking credit for the arbitrage)?

I have a hunch there is an argument to be made with respect to the martingale measure, but it's not well formed:

$$E[V(t_{m}) | \mathcal{F}_{t_{i}}] \neq V(t_{i}), t_{i} < t_{m}$$

Any help / ideas is much appreciated!