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Loss Probability for an Unhedged Short Call in Black–Scholes

Article Quant Q&A · Author: Landscape
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Summary

The document frames the probability that an unhedged short European call loses money at expiration after being sold above its risk-neutral price. The premium, including the extra amount received, is invested at the risk-free rate, while the seller remains exposed to the call payoff. Under the stated zero-rate Black–Scholes setup, the stock follows geometric Brownian motion under the real-world probability measure.

The question highlights a key step: the loss event requires the call payoff to exceed the premium received. Since the call payoff is zero below the strike and linear above it, the probability must be handled piecewise rather than by removing the positive-part operator unconditionally. The document supplies the setup but does not give the completed probability formula or a numerical result. Any probability depends on the real-world drift, volatility, maturity, strike, and premium; the risk-neutral option price alone does not determine it.

Key ideas

  • The short call’s expiration loss event is defined by its payoff exceeding the premium received.
  • The call payoff’s positive-part function makes the event piecewise in the terminal stock price.
  • The stock price is modeled as geometric Brownian motion under the real-world measure.
  • The document poses the probability calculation but does not complete its derivation.

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# What is the P-probability of an unhedged call-arbitrage to lose money at expiration


# What is the P-probability of an unhedged call-arbitrage to lose money at expiration












Assume that the Risk Neutral Price (under the $\mathbb{Q}$-measure) of an European Call Option with expiration date $T$ has a price of $F(S_0,0)$ at time $t=0$ in the single asset Black-Scholes model (without dividens and zero interest rate, $r=0$).

Suppose that we are somehow "skillfull" egnouh to sell this option for $F(S_0,0) + \varepsilon$ at time $t=0$, invest the money into the risk-free asset, and do not hedge our risk. Then I want to find the probability (under the $\mathbb{P}$-measure) of our "portfolio" denoted by $V$ to expire with a loss, i.e. $\mathbb{P}(V(T) < 0)$. We clearly have that $$V(T)=F(S_0,0)+\varepsilon -(S_T-K)^+ \quad, \\ \quad \text{where } (\cdot)^+ \equiv \max( \;\cdot \; , \;0) \text{ is the payoff function for the call-option}.$$

I get the idea of how to perform these calculation but I do now know how to handle to payoff function in the probability. As a starting point I know that the stock price is a GBM with SDE $$dS_t = \mu S_t dt + \sigma S_t dW_t \quad , \quad S_0 = s.$$ As this is a GBM we know that the solution is given by $$S_t=S_0e^{\left(\mu-\frac{\sigma^2}{2}\right)t+\sigma W_t}.$$

Plugging this into our problem yields $$\mathbb{P}(V(T)<0) = \mathbb{P}\left(F(S_0,0)+\varepsilon-(S_T-K)^+ <0\right)$$ $$=\mathbb{P}\left(F(S_0,0)+\varepsilon-\left(S_0e^{\left(\mu-\frac{\sigma^2}{2}\right)T+\sigma W_T}-K\right)^+ <0\right)$$

I do not kow how to continue from here. However, I would like to argue (i do not know how) that i can look apart from the $+$, then isolate for $W_T$ and use that $W_T = z\cdot\sqrt{T}, \;z\sim N(0,1)$. This would give me an expression similar to something like $N(d_2)$ in the Black-Scholes formula.

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This summary was written by Stratmill's research agent from the original; it is not a copy of the source.