Constant 30% Volatility or Random Volatility With Mean 30%?
You price a European call on a non-dividend stock two ways. In the first, the volatility is a constant $30\%$. In the second, the volatility is random: at time $0$ a value $\sigma$ is drawn from a distribution with mean $30\%$ (independent of the stock's Brownian motion), and the stock then follows geometric Brownian motion with that volatility until expiry, so the option price is the expectation of the Black-Scholes price over the draw of $\sigma$.
Which of the two option prices is higher? Does the answer depend on the strike? Justify your answer using the shape of the Black-Scholes price as a function of $\sigma$.
Hints
- With the volatility drawn independently at time $0$, the random-volatility price is $E_\sigma[c_{\text{BS}}(\sigma)]$. Jensen's inequality compares $E[c(\sigma)]$ with $c(E[\sigma])$ and the direction depends on whether $c$ is convex or concave in $\sigma$.
- Differentiate vega: $\partial^2 c/\partial\sigma^2 = \text{vega}\cdot d_1 d_2/\sigma$ (volga). Its sign is the sign of $d_1 d_2$.
- $d_1 > 0 > d_2$ (concave region) exactly when $-\sigma^2\tau/2 < \ln(S/K) + r\tau < \sigma^2\tau/2$, a narrow band around the money; outside it $d_1 d_2 > 0$ and the price is convex, so the random-volatility price is higher.
Worked Solution
How to Think About It: Under the random-volatility model the price is the average of Black-Scholes prices over the draw of $\sigma$. Comparing an average of a function with the function of the average is Jensen's inequality, so the question reduces to the curvature of $c_{\text{BS}}$ in $\sigma$. Compute volga and read off its sign.
Quick Estimate: Let $S = 100$, $r = 0$, $\tau = 1$, and let the random $\sigma$ be $20\%$ or $40\%$ with equal probability (mean $30\%$). At $K = 130$: constant-vol price $3.57$, random-vol price $\tfrac12(1.01 + 6.94) = 3.97$; random wins. At $K = 100$: constant $11.92$, random $\tfrac12(7.97 + 15.85) = 11.91$; constant wins by a hair. So the answer flips near the money.
Formal Solution:
*Step 1 -- Set-up.* Conditional on the drawn $\sigma$, the stock is a geometric Brownian motion with that volatility, so the conditional price is $c_{\text{BS}}(\sigma)$ and the unconditional price is $E[c_{\text{BS}}(\sigma)]$. The constant-volatility price is $c_{\text{BS}}(E[\sigma]) = c_{\text{BS}}(0.3)$.
*Step 2 -- Jensen.* If $c_{\text{BS}}$ is convex in $\sigma$ over the support of the draw, then $E[c_{\text{BS}}(\sigma)] \ge c_{\text{BS}}(E[\sigma])$ and the random-volatility price is higher; if concave, the constant-volatility price is higher.
*Step 3 -- Curvature: volga.* Vega is $\mathcal{V} = S\sqrt{\tau}N'(d_1)$. Differentiating again with respect to $\sigma$, using $\partial d_1/\partial\sigma = -d_2/\sigma$ and $N''(x) = -xN'(x)$, $$\frac{\partial^2 c}{\partial\sigma^2} = S\sqrt{\tau}N'(d_1)\cdot(-d_1)\cdot\left(-\frac{d_2}{\sigma}\right) = \mathcal{V}\,\frac{d_1 d_2}{\sigma}.$$ Since $\mathcal{V} > 0$, the price is convex in $\sigma$ where $d_1 d_2 > 0$ and concave where $d_1 d_2 < 0$.
*Step 4 -- Where is $d_1 d_2 < 0$?* Because $d_1 > d_2$, the product is negative only when $d_1 > 0 > d_2$, i.e. $$-\tfrac12\sigma^2\tau < \ln(S/K) + r\tau < \tfrac12\sigma^2\tau,$$ a band of half-width $\sigma^2\tau/2$ in log-moneyness around the forward. For $\sigma = 0.3$, $\tau = 1$ this is $|\ln(Se^{r\tau}/K)| < 0.045$, roughly strikes within $4.5\%$ of the forward. Outside this band, both $d_1$ and $d_2$ have the same sign, the price is convex in $\sigma$, and the random-volatility price is higher.
*Step 5 -- Conclusion.* For most strikes (anything more than a few percent away from the forward) the option priced with random volatility is worth more. Near the money the Black-Scholes price is almost linear in $\sigma$ (recall $c \approx 0.4\,\sigma S\sqrt{\tau}$) and slightly concave, so the constant-volatility price is marginally higher, as the numerical example shows ($11.92$ versus $11.91$). This is exactly the mechanism by which stochastic volatility produces a smile: wings up, at-the-money almost unchanged.
Answer: Usually the random-volatility price is higher, because the Black-Scholes price is convex in $\sigma$ (volga $= \mathcal{V}d_1 d_2/\sigma > 0$) whenever $d_1 d_2 > 0$, and Jensen's inequality gives $E[c(\sigma)] > c(E[\sigma])$. The exception is a narrow band around the money where $d_1 > 0 > d_2$, the price is concave in $\sigma$, and the constant-volatility price is (slightly) higher. So the answer depends on the strike.
Intuition
A call's price is increasing in volatility but not linearly: far from the money it is convex (a little extra volatility does a lot for a lottery-ticket option), while right around the money it is nearly linear and in fact slightly concave. Jensen's inequality then says randomizing volatility around the same mean raises the price wherever the curve is convex, which is almost everywhere except a thin band near the forward. This is why stochastic-volatility models generate a smile: they raise the wings relative to a flat-volatility model and leave the at-the-money price almost unchanged. Volga, the second derivative in $\sigma$, is the Greek that measures this exposure and is what "vol of vol" traders monitor.