# Zero Volatility Options Pricing

Suppose an asset evolves in time according to the SDE

$$dS = \mu S dt + \sigma S dW,$$ where $$\mu>0,\sigma>0$$ are fixed constants and $$dW$$ is a Wiener process. To price options for this underlying, you have the whole procedure of using delta hedging or risk-neutral pricing to get the BS equation and eventually the value of an option $$V(S,t)$$. Now, I am thinking about what would happen if there was zero volatility, i.e. $$\sigma = 0$$. My question is: How would you price options in that case?

The dynamics are completely deterministic now, since $$dS = \mu S dt$$, and we can find the stock as a function of time as

$$S(t) = S_0e^{\mu t}$$

So given some starting price $$S_0$$, the stock at expiration $$T$$ will always be $$S(T) = S_0e^{\mu T}$$. To price the options, you have to think about arbitrage opportunities. Let's start with calls first: From regular arbitrage arguments, you already know that $$S(t)-Ee^{-r(T-t)}\leq C(t) \leq S(t)$$. This comes from creating a portfolio long a stock and short a call, and looking at the bounds at $$t = T$$. Applying this to our weird deterministic case gives

$$S_0e^{\mu t}-Ee^{-r(T-t)} \leq C(t) \leq S_0e^{\mu t}$$

I know how the call is bounded, but I want an exact formula. If you think about it, since we know $$S(T) = S_0e^{\mu T}$$, then any call option with a strike price $$E \geq S_0e^{\mu T}$$ will be worthless at expiration, because $$C(T) = \max(S(T)-E,0)$$. Thus, this gives

$$C(t) \equiv 0,\ \forall E \geq S_0e^{\mu T}$$

For $$E < S_0e^{\mu t}$$, the option will most certainly be nonzero, but I am having trouble finding what it would be. I know you have to take into account the risk-free rate somehow, but since any portfolio you construct is risk free, wouldn't there be unlimited arbitrage? What I mean by this: Say you buy an option with strike $$E < S_0e^{\mu T}$$. Then, your profit at expiration will always be

$$\text{Profit}_T = S_0e^{\mu T} - E$$

Therefore, the value should be at least this much, i.e. $$C(t) \geq S_0e^{\mu T} - E$$. Discounting for time gives $$C(t) = e^{-r(T-t)}(S_0e^{\mu T}-E)$$. However, this doesn't seem right. Any thoughts would be appreciated.

The only missing point is that, by NA, if an asset has zero volatility, it is riskless and must therefore grow at the risk-free interest rate: $$\mu \equiv r$$ (Else, you buy the highest yielding asset and sell the lowest yielding).
In such situation, the valuation of an option is straightforward: it is the discounted payoff $$e^{-r\left(T - t\right)} \left[S_t e^{r\left(T - t\right)} - K \right]^+ = \left[S_t - Ke^{-r\left(T - t\right)} \right]^+$$.
That makes every OTM/ATM option worthless, and every ITM option worth exactly the PV of its known payoff, when the “money” is defined as $$K = S_t e^{r\left(T - t\right)}$$, the forward price.
• I'm not sure I understand exactly. What do you mean by "NA"? Also, I understand that the option would be the discounted payoff, that makes sense. So, what you're saying is that the options value is just $S_0e^{\mu t} - Ke^{-r(T-t)}$ only when it is ITM. If it goes OTM, it becomes worthless, so it's like a piecewise continuous function. The last thing, I'm not sure what you mean by "money" here. Feb 6, 2020 at 14:58
• You are missing that if the stock has zero volatility, it will grow at the risk-free rate. The spot price that will apply to your payoff is not 100, but $100 e^{5\% \times 1}$. Feb 8, 2021 at 8:00