# Why does the price term in Vega disappear for a European call option?

In my course, I have been asked to prove a number of statements about "the Greeks" from the Black-Scholes model for pricing a European call option with no dividends and a strike price of $K$. One of the problems was to prove that $\nu$ (Vega) was positive. To prove the fact, I began by differentiating the Black-Scholes model: $$C_t = S_t\Phi(d_1(t)) - K e^{-r(T - t)}\Phi(d_2(t)),$$ where all of the variables are defined as is done typically, and $S_t$ is equal to $$S_t = S_0\exp\left(\left(b - \frac 1 2 \sigma^2\right)t + \sigma B_t\right),$$ with $B_t$ a Brownian motion. Now, when I take the derivative w.r.t. $\sigma$ of $C_t$, I find that $$\frac{\partial C_t}{\partial\sigma} = \frac{\partial S_t}{\partial \sigma}\Phi(d_1(t)) + S_t\Phi'(d_1(t))\frac{\partial d_1(t)}{\partial \sigma} - K e^{-r(T - t)}\Phi'(d_2(t))\frac{\partial d_2(t)}{\partial \sigma}.$$ Using the fact that $S_t\Phi'(d_1(t)) = K e^{-r(T - t)}\Phi'(d_2(t))$, and that $$\frac{\partial d_1}{\partial \sigma} - \frac{\partial d_2}{\partial \sigma} = \sqrt{T - t},$$ I find that $$\frac{\partial C_t}{\partial\sigma} = \frac{\partial S_t}{\partial \sigma}\Phi(d_1(t)) + \sqrt{T - t} K e^{-r(T - t)}\Phi'(d_2(t))\frac{\partial d_2(t)}{\partial \sigma}.$$

Obviously, I now need to solve $\frac{\partial S_t}{\partial \sigma}$. In so doing, I find that $$\frac{\partial S_t}{\partial \sigma} = (B_t - \sigma) S_t,$$ which makes my final answer that $$\frac{\partial C_t}{\partial\sigma} = (B_t - \sigma) S_t\Phi(d_1(t)) + \sqrt{T - t} K e^{-r(T - t)}\Phi'(d_2(t))\frac{\partial d_2(t)}{\partial \sigma}.$$ However, all the sources that I can find (e.g. here, here and here) report that Vega is just equal to $$\frac{\partial C_t}{\partial\sigma} = \sqrt{T - t} K e^{-r(T - t)}\Phi'(d_2(t))\frac{\partial d_2(t)}{\partial \sigma}.$$ Is there something that I am doing wrong? I suspect that I am doing something wrong w.r.t. the inclusion of $B_t$, as that seems strange. Any help would be appreciated.

• How does stock price change with respect to volatility? – Calvin Lin Mar 29 '14 at 5:18
• Isn't the change in stock price w.r.t. volatility given by the $\frac{\partial S_t}{\partial \sigma} = (B_t - \sigma)S_t$ term? – fbt Mar 29 '14 at 16:44
• $S_t$ is the stock valuation's price at time $t$, which is a given / constant. You do not model it as the expected effect of Brownian motion. – Calvin Lin Mar 29 '14 at 16:59
• Oh- that makes sense. We would only model its evolution over time as the expected effect of Brownian motion? – fbt Mar 29 '14 at 17:22