Vega and Volatility Sensitivity

Volatility as a Price Sensitivity

Delta measures sensitivity to the underlying stock price, and gamma describes how delta changes as the stock price moves. Vega answers a different question: how does a European option’s value change when the volatility input in the Black–Scholes model changes, with the other inputs held fixed? This derivative is useful both for comparing options and for estimating the effect of a small revision to a volatility assumption.

Work on a filtered probability space $(\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})$. For pricing, let $\mathbb{Q}$ be the risk-neutral measure on the same filtered space, and let $B_t^{\mathbb{Q}}$ be a Brownian motion under $\mathbb{Q}$. Assume a non-dividend-paying stock follows $dS_t=rS_t\,dt+\sigma S_t\,dB_t^{\mathbb{Q}}$, where $r$ is constant and $\sigma>0$. In integral form, the Brownian integrand $\sigma S_t$ is progressively measurable and belongs to $\mathcal{L}^2$ on each finite horizon under $\mathbb{Q}$: its squared expectation integrated over that horizon is finite. The Black–Scholes prices below are risk-neutral prices; $\mathbb{P}$ is the physical measure and is not substituted for $\mathbb{Q}$ in the pricing formulas.

Fix a valuation time $t<T$, stock price $S=S_t>0$, strike $K>0$, and time to maturity $\tau=T-t>0$. Volatility sensitivity means differentiation with respect to $\sigma$, holding $S,K,r,\tau$ fixed. Define the standard normal density and distribution function by $\phi(x)=(2\pi)^{-1/2}e^{-x^2/2}$ and $\Phi(x)=\int_{-\infty}^{x}\phi(u)\,du$. Set

d_1=\frac{\log(S/K)+(r+\tfrac12\sigma^2)\tau}{\sigma\sqrt{\tau}}, \qquad d_2=d_1-\sigma\sqrt{\tau}.

The Black–Scholes call and put prices are $C=S\Phi(d_1)-Ke^{-r\tau}\Phi(d_2)$ and $P=Ke^{-r\tau}\Phi(-d_2)-S\Phi(-d_1)$. The discount factor and all inputs other than $\sigma$ remain fixed when taking vega.

Definition: The vega of a call or put is the partial derivative of its Black–Scholes price with respect to volatility, $\partial C/\partial\sigma$ or $\partial P/\partial\sigma$, with $S,K,r,\tau$ held fixed. Volatility is conventionally entered as a decimal: a move from $0.20$ to $0.21$ is a change of $0.01$, or one volatility percentage point.

Deriving Call and Put Vega

Theorem: If $S>0$, $K>0$, $\sigma>0$, and $\tau>0$, then the Black–Scholes call and put have the same vega: $$ \frac{\partial C}{\partial\sigma} = \frac{\partial P}{\partial\sigma} = S\phi(d_1)\sqrt{\tau} = Ke^{-r\tau}\phi(d_2)\sqrt{\tau} >0. $$

Proof. Let $a=\log(S/K)+r\tau$. Then $d_1=a/(\sigma\sqrt{\tau})+\sigma\sqrt{\tau}/2$ and $d_2=a/(\sigma\sqrt{\tau})-\sigma\sqrt{\tau}/2$. Differentiating each with respect to $\sigma$, while holding the other inputs fixed, gives $\partial d_1/\partial\sigma=-d_2/\sigma$ and $\partial d_2/\partial\sigma=-d_1/\sigma$. Also, the identity $\phi'(x)=-x\phi(x)$ implies $\Phi'(x)=\phi(x)$.

We first establish the density identity needed to simplify the derivatives. Since $d_1-d_2=\sigma\sqrt{\tau}$ and $d_1+d_2=2a/(\sigma\sqrt{\tau})$, we have $d_1^2-d_2^2=2a$. Consequently

\frac{\phi(d_1)}{\phi(d_2)} =e^{-(d_1^2-d_2^2)/2} =e^{-a} =\frac{K}{S}e^{-r\tau}, \qquad S\phi(d_1)=Ke^{-r\tau}\phi(d_2).

Applying the chain rule to the call formula yields

\frac{\partial C}{\partial\sigma} =S\phi(d_1)\frac{-d_2}{\sigma} -Ke^{-r\tau}\phi(d_2)\frac{-d_1}{\sigma} =\frac{S\phi(d_1)(d_1-d_2)}{\sigma} =S\phi(d_1)\sqrt{\tau}.

For the put, differentiating both arguments $-d_2$ and $-d_1$ gives

\frac{\partial P}{\partial\sigma} =Ke^{-r\tau}\phi(d_2)\frac{d_1}{\sigma} -S\phi(d_1)\frac{d_2}{\sigma} =\frac{S\phi(d_1)(d_1-d_2)}{\sigma} =S\phi(d_1)\sqrt{\tau}.

In the second equality for each price we used $S\phi(d_1)=Ke^{-r\tau}\phi(d_2)$; in the last, $d_1-d_2=\sigma\sqrt{\tau}$. Since $S>0$, $\sqrt{\tau}>0$, and the normal density is strictly positive, this common derivative is strictly positive. The density identity also gives the alternative expression $Ke^{-r\tau}\phi(d_2)\sqrt{\tau}$. $\square$

Corollary: With $S,K,r,\tau$ fixed and $\tau>0$, both Black–Scholes call and put prices are strictly increasing functions of $\sigma$ on $(0,\infty)$.

Proof. Take any $0<\sigma_1<\sigma_2$. Each price is continuous on $[\sigma_1,\sigma_2]$ and differentiable on $(\sigma_1,\sigma_2)$. By the mean value theorem, its price difference between the endpoints equals $(\sigma_2-\sigma_1)$ times its volatility derivative at some point in that interval. The derivative is strictly positive by the theorem, and $\sigma_2-\sigma_1>0$, so the price difference is strictly positive for both options. $\square$

The positivity statement concerns a live option, meaning $\tau>0$. At expiry, the payoff depends on $S_T$ and $K$, not on an input volatility, so its volatility sensitivity is zero when considered as a payoff function. The formula with $\sqrt{\tau}$ is for positive time to maturity.

Using Vega for a Small Volatility Shift

If volatility changes from $\sigma$ to $\sigma+\delta\sigma$, with the other inputs unchanged, the first-order estimate is $\text{price change}\approx\text{vega}\times\delta\sigma$. This follows from differentiability: the price change equals $\text{vega}\times\delta\sigma+o(\delta\sigma)$ as $\delta\sigma$ tends to zero. It is a local estimate, not an exact formula for a finite volatility move. For a larger move, re-evaluate the Black–Scholes formula at the new volatility.

Worked Example: An At-the-Money Call and a One-Point Shift

Take $S=K=100$, $r=0$, $\tau=1$, and $\sigma=0.20$. Find the call vega and estimate the price change when volatility rises by $0.01$. Then reprice at the shifted volatility as a check.

First, $d_1=(0+0.20^2/2)/0.20=0.10$ and $d_2=0.10-0.20=-0.10$. Since $\phi(0.10)=(2\pi)^{-1/2}e^{-0.005}\approx0.3970$, the vega is $100(0.3970)\sqrt{1}\approx39.70$. The first-order estimated increase is $39.70(0.01)\approx0.3970$.

For the repricing check, at $\sigma=0.21$, $d_1=0.21/2=0.105$ and $d_2=0.105-0.21=-0.105$. The call price at volatility $v$ in this example is $100[2\Phi(v/2)-1]$. Thus the exact price difference expressed through the normal distribution is $200[\Phi(0.105)-\Phi(0.10)]$, which is approximately $0.3969$. The estimate $0.3970$ is close, as expected for a small shift. Both computations show an increase.

Check: The original price is $100[2\Phi(0.10)-1]\approx7.966$; the repriced value is $100[2\Phi(0.105)-1]\approx8.362$. Their difference, computed before rounding the displayed prices, is approximately $0.3969$, agreeing with the direct difference above. The small discrepancy from the linear estimate is the effect of using the initial derivative over a nonzero volatility change.

Worked Example: Put Vega and the Density Identity

Take $S=K=100$, $r=0.05$, $\tau=1$, and $\sigma=0.20$. Compute the put vega using $d_1$, check it using $d_2$, and estimate the put price change for a $0.02$ volatility increase.

Here $d_1=(0+0.05+0.20^2/2)/0.20=0.35$, and $d_2=0.35-0.20=0.15$. The first expression gives $100\phi(0.35)\sqrt{1}\approx100(0.3752)=37.52$. For the second expression, $e^{-0.05}\approx0.9512$ and $\phi(0.15)\approx0.3945$, so $100e^{-0.05}\phi(0.15)\sqrt{1}\approx100(0.9512)(0.3945)\approx37.52$. This numerical agreement checks the density identity for these inputs.

The first-order estimate for the put price change is $37.52(0.02)\approx0.7504$. The theorem says that the call with the same inputs has the same vega, so its first-order estimate for this same shift is also approximately $0.7504$. This does not say the call and put prices are equal; it says their derivatives with respect to volatility are equal.

Check: The put estimate is positive because both the calculated vega and the volatility increase are positive. If the shift were instead $-0.02$, the first-order estimate would be $37.52(-0.02)\approx-0.7504$, consistent with strict increase in volatility.

Worked Example: Vega for a Lower-Spot Call

Choose $K=100$, $r=0$, $\tau=1$, $\sigma=0.20$, and $S=100e^{-0.42}$. Compute $d_1,d_2$, vega, and the estimated call price change for a $0.01$ volatility increase.

By construction, $\log(S/K)=-0.42$. Therefore $d_1=(-0.42+0.20^2/2)/0.20=(-0.42+0.02)/0.20=-2$, and $d_2=-2-0.20=-2.2$. The spot is $100e^{-0.42}\approx65.70$, and $\phi(-2)=\phi(2)\approx0.0540$. Hence the vega is $100e^{-0.42}\phi(2)\sqrt{1}\approx65.70(0.0540)\approx3.548$. The estimated increase for a $0.01$ shift is $3.548(0.01)\approx0.03548$.

Check: The alternative formula gives $100\phi(-2.2)\sqrt{1}\approx100(0.03547)\approx3.547$, matching the first calculation up to rounding. The vega is positive, but smaller than the at-the-money vega in the first example. The exact amount depends on the inputs through $S\phi(d_1)\sqrt{\tau}$; moneyness alone does not replace that formula.

Interpreting and Applying the Number

Vega is a derivative with respect to decimal volatility. If a displayed vega is $40$, a move of one volatility percentage point means $\delta\sigma=0.01$, giving a first-order price change of about $0.40$ in the same currency units as the option price. Multiplying by $1$ instead would estimate the change for a full $100$-percentage-point volatility move, not for a one-point move.

Vega is positive for both vanilla calls and puts under the stated assumptions, so an increase in volatility raises their Black–Scholes prices and a decrease lowers them. It is not generally constant as volatility changes: $d_1$, and hence the density $\phi(d_1)$, depends on volatility. Therefore the linear estimate is most useful for small changes; for a substantial shift, calculate the new price directly.

Common mistake: differentiating only the visible density terms. A tempting but incorrect step is to differentiate the call formula as though $d_1$ and $d_2$ were fixed. For the first example, this would give $S\phi(d_1)-Ke^{-r\tau}\phi(d_2)=0$, since $S=K=100$, $r=0$, and $d_1=0.10$, $d_2=-0.10$ imply equal densities. That answer is wrong: the normal distribution functions also change as $\sigma$ changes. The correction is to include $\partial d_1/\partial\sigma=-d_2/\sigma$ and $\partial d_2/\partial\sigma=-d_1/\sigma$ in the chain rule. Including those factors gives $S\phi(d_1)(d_1-d_2)/\sigma=S\phi(d_1)\sqrt{\tau}$, which is approximately $39.70$ in this example, not zero.

Check Your Understanding

Use the Black–Scholes definitions and hold all inputs fixed except volatility unless a question says otherwise.

For $S=K=80$, $r=0$, $\tau=1$, and $\sigma=0.30$, calculate $d_1$, $d_2$, and the call vega.
Explain why the call and put vegas are equal even though their prices need not be equal.
If vega is $25$, what is the first-order price change for a volatility increase of $0.005$?
State the assumptions under which the theorem gives strictly positive vega, and explain what happens to the payoff’s volatility sensitivity at expiry.
Identify the missing terms in the incorrect calculation that treats $d_1$ and $d_2$ as constants when differentiating price with respect to volatility.

Vega and Volatility Sensitivity

What You'll Learn