Risk-Neutral Pricing and a Fee in a Callable-Back Option
Summary
The document considers a European call that the holder may sell back to its writer at its Black–Scholes value before a specified time. At that time, the writer imposes a fee by reducing the holder's option value by a percentage. The question asks whether the fee creates an arbitrage that invalidates risk-neutral valuation or the martingale property.
The answer argues that the supposed arbitrage does not arise as described: before the fee takes effect, the holder can exercise the right to sell the option back at market value, avoiding the reduction. If the holder still wants exposure, they can purchase another option in the market. The discussion emphasizes analyzing participant behavior within a no-arbitrage framework. It is a concise conceptual answer and does not develop a formal pricing model or explore contract details that could alter the exercise incentives.
Key ideas
- The proposed contract permits the holder to sell the call back at its current market value before the fee date.
- The answer says the holder can use that right before the fee reduces the option value.
- A holder who wants continued exposure could replace the option by buying another in the market.
- The response treats participant behavior and contract rights as central to the no-arbitrage analysis.
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Full text
# Why A Derivative With Intrinsic Arbitrage Cannot Be Valued & Hedged With Assets In Risk Neutral?
# 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!
## Answer by dm63 (score 2)
https://quant.stackexchange.com/a/78141
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.Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)
This summary was written by Stratmill's research agent from the original; it is not a copy of the source.