Variance of options returns

Question

Let's say we write a standard call option on $S_t$ which pays $Max[0, S_t-K] \,\forall \, t \in T $. Given that $\frac{dS}{S} = \mu \,dt + \sigma \,dW_t$, and, $V_T = (S_T -K)_+$, we can solve this under the Black-Scholes framework as:

$$V_t[S_t,K,\sigma_S,r,t] = S_t \varPhi[d_1] - K e^{-r (T-t)} \varPhi[d_1 - \sigma \sqrt{T-t}]$$

where: $\varPhi[x]$ is a cumulative distrbution function; and,

$$d_1 = \frac{\ln\left(\frac{S_t}{K}\right)+{(r+\sigma^2/2)(T-t)} }{\sigma \sqrt{T-t}}$$

What is the expected variance of this option's returns, $\sigma_V$? I.e., how does the process $E \left[ V_t \right]$ evolve wrt time?

Intuitively, the logarithmic variance should be defined if we constrain that the option must take non-zero, positive values.

I ask because I am trying to assess what might be called a compound option in which the parameters are adapted for $V_t$.

Answer 1

First off the $\mathbf E[V_t]$ is either redundant or not what you are after. If you mean to write $\mathbf E[V_t|\mathcal F_0]$, it is just $V_0e^{rt}$ if the interest rate is constant. I don't think this is what you want. If you mean to write $\mathbf E[V_t|\mathcal F_t]$, it is just $V_t$ and your symbol is redundant.

I think, though it is not clearly stated, you want the latter, i.e., are asking what the instantaneous variance $V_t$ is for the price process of an option. This is clear from the hedging argument of the option. It is delta hedged with the underlying $S_t$ to cancel the stochasticity or volatility (variance). So, the instantaneous volatility of $dV_t$ has to be that contained in $\Delta dS_t$ which is $|\Delta|\sigma S$. If we need the instantaneous volatility to be for the form $\frac{dV_t}{V_t}$, it is $|\Delta|\frac{S_t}{V_t}\sigma$. All the above argument is valid not only for the Black-Scholes setting but for the most general option setting. One can just as easily argue this with Ito's lemma.

The ratio of the option volatility over the underlying volatility is the percentage change ratio of the option over that of the underlying, or $\frac S V\Big|\frac{\partial V}{\partial S}\Big|$. Suppose all the stochastic process involved is a continuous path diffusion; the stochastic processes of the drift, volatility and correlation are all independent of the value of the underlying. Then the value $V$ is a convex function of $S$ when the payoff is. The proof of this proposition for the general case (not restricted to the Black-Scholes setting) is actually not trivial. So $\frac{\partial V}{\partial S}\Big|_{S_t}>\frac{V(S_t)-V(S=0)}{S_t-0}$ for a convex payoff. Call and put payoffs are convex functions of the underlying. For a general call, $V(S=0)=0$, which implies $\frac S V\frac{\partial V}{\partial S}>1$. But for a put, $V(S=0)=K\neq0$. The derivation breaks down. In fact, it is not necessarily true that $\frac S V\Big|\frac{\partial V}{\partial S}\Big|>1$ for the put, particularly when the put is deep in the money. It is also false for other options even for those where $V(S=0)=0$, so long as the payoff is not convex, e.g., a call spread and a digital option.