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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.

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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.

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  • $\begingroup$ 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. $\endgroup$ – Josh Pilipovsky Feb 6 at 14:58
  • $\begingroup$ NA means no Arbitrage. ATM means at the money, and by that I mean that the strike equals the forward price. $\endgroup$ – siou0107 Feb 6 at 16:03
  • $\begingroup$ Got it, thanks again! $\endgroup$ – Josh Pilipovsky Feb 6 at 17:30

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