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


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.


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!


1 Answer 1


Valuing something to the writer vs valuing it to the buyer makes no difference. We just value the instrument.

In this case the buyer surely would prevent the writer from collecting the fee, by exercising the right to sell back the option to the buyer. (If there was a desire to continue to hold the option, he can just buy another one in the market).

So this question of arbitrage does not arise. One just analyzes the behavior of participants using the no arbitrage framework.

  • $\begingroup$ I think this is the obvious reaction, but I'm looking for a measure theoretic answer as to why call prices valued on the market and the specific call sold to this buyer cannot be valued under the same measure without necessitating a behavior argument. My sense is that because calls are martingales, any measure that makes the 'value' of the option equal to a martingale in the P measure does not exist, because calls are q-martingales. $\endgroup$ Commented Jan 29 at 0:53

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