3
$\begingroup$

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 $V_t$.

$\endgroup$
  • 1
    $\begingroup$ You need to define your terms and symbols, which are unconventional to say the least. What is $\|$? Is your $>=$ supposed to be $\ge$ or the right angle bracket followed by a $>$? What is $f[V_t;x]$? What is $E|Vt|$? Is that the same as $\mathbf E[V_t]$? On what filtration is the expectation taken on, at time $0$ or time $t$ or some other time? What is that specific probability measure $\varPhi$? Is it supposed to be the cumulative standard Gaussian distribution? What is $X$ the logarithmic moneyness? $\endgroup$ – Hans Feb 12 '18 at 7:23
  • $\begingroup$ This is a really sloppily and vaguely composed question. Do you really understand each symbol and what meaning it conveys, and what meaning you attempt to convey using it, when you write it down? $\endgroup$ – Hans Feb 12 '18 at 7:32
  • $\begingroup$ @Hans. I cleaned it up a little to use standard notation. I didn't know the LaTeX notation for "conditioned" or "distributed", so I probably used the wrong stuff. $\endgroup$ – David Addison Feb 12 '18 at 16:21
  • $\begingroup$ Well, obviously you did not read what you have written afterward. You would have said the same, had you done that. $\endgroup$ – Hans Feb 12 '18 at 17:52
  • $\begingroup$ @Hans Not clear what you mean by that. $\endgroup$ – David Addison Feb 12 '18 at 17:58
5
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Thank you for the elegant response. It intuitively makes sense the instantaneous volatility of the option has to be higher than for $S_t$, especially because $V_t \le S_t \forall t$. Also, I apologize for omitting the filtration, but you are correct in your assumption that $V_t$ contains information up to t. $\endgroup$ – David Addison Feb 12 '18 at 19:12
  • $\begingroup$ @DavidAddison: You are welcome. Your intuition is correct but only for a call and your rationale is insufficient. I have added a proof for this proposition for a call and the related counterexamples. $\endgroup$ – Hans Feb 13 '18 at 3:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.