Why the $Vega$ of the Black Scholes Model is at its maximum for at-the-money options?

Question

In my course script, it is said that the Vega of the Black Scholes Model is at its maximum for at-the-money options.

In order to verify this, I did the following calculations:

In the Black Scholes Model, the price of Call option with dividends (q) is given by $$C_t=S_te^{-q(T-t)}\mathscr N(d_1)-Ke^{-r(T-t)}\mathscr N(d_2)$$ where $d_1=\frac{ln(\frac{S_t}{K})+(r-q+{\sigma^2 \over 2})(T-t)}{\sigma\sqrt{T-t}} \qquad d_2=d_1-\sigma\sqrt{T-t}$

I computed the Vega for this model, which is

$$ Vega_t=\frac{\partial C_t}{\partial \sigma}=S_te^{-q(T-t)}\phi(d_1)\sqrt{T-t}$$

Then I take the first order derivative of Vega w.r.t. $S_t$ and found that the Vega is maximized when $d_1=0$, which is equivalent to $$S_t=Ke^{-(r-q+{\sigma^2 \over 2})(T-t)}\qquad (\ast)$$

This result shows that the Vega is not maximized when $S_t=K$, contradicting the statement above.

I don't know whether the reason for this contradiction is because of my wrong derivation or my misunderstanding on the definition of ATM option. Anyway, can anybody help me with this problem?

Thanks!

Answer 1

I asked my lecturer this question, and he gave me the following explanation, which I think makes sense:

Basically, assets can either have forward price or spot price. Therefore, there are two definitions for ATM options. If we use the spot price of a stock, then we say an option is at-the-money when the strike price($K$) is equal to the spot price($S_t$); If we use the forward price of a stock, then an option is ATM when the strike price equals to the forward stock price, which is $S_te^{(r-q)(T-t)}$ in this case.

Now if we arrange $(\ast)$, we immediately get:

$$K=S_te^{(r-q)(T-t)}e^{{\sigma^2 \over 2}(T-t)}$$

where $S_te^{(r-q)(T-t)}$ is the forward price of stock.

Now, let's consider the real market. Since there is only a fixed number of strike prices in the market while the forward stock prices change continuously, it is rare that the strike price is the same as the forward stock price. Therefore, we can relax the ATM forward definition above to reflect this reality.

Relaxed definition: In the case of the forward price of stock, an option is at-the-money if the strike price $K\in(S_te^{(r-q)(T-t)}-\epsilon, S_te^{(r-q)(T-t)}+\epsilon)$.

Therefore, given that $e^{{\sigma^2 \over 2}(T-t)}$ is relatively small because of a small $\sigma$, it is reasonable to assume that $$K=S_te^{(r-q)(T-t)}e^{{\sigma^2 \over 2}(T-t)}\in(S_te^{(r-q)(T-t)}-\epsilon, S_te^{(r-q)(T-t)}+\epsilon)$$

which confirms the statement that the $Vega$ of the Black Scholes Model is at its maximum for ATM options.

In summary, this question is answered base on the ATM defined on the forward stock price, and the relaxed definition of ATM forward that is consistent with reality.

I hope this explanation given by my lecturer is helpful for those who have the same question as mine. If you have different ideas, please leave a comment!

Answer 2

Let $\tau := T - t$ then by differentiating the function $C_{BS}(t,T,x,K,r,\sigma)$ with respect to $\sigma$ we obtain \begin{align} \partial_{\sigma}C_{BS} = x \frac{1}{\sqrt{2\pi}}e^{-\frac{d_{1}^{2}}{2}}\partial_{\sigma}d_{1} - e^{-r\tau}K\frac{1}{\sqrt{2\pi}}e^{-\frac{d_{2}^{2}}{2}}\partial_{\sigma}d_{2}(\ast) \end{align} so \begin{align} \partial_{\sigma}d_{1} &= \frac{-1}{\sigma^{2}\sqrt{\tau}} \left\{\log{\left(\frac{x}{K}\right)}+\left(r+\frac{\sigma^{2}}{2}\right)\tau\right\} + \sqrt{\tau} \\ &= \frac{-1}{\sigma^{2}\sqrt{\tau}} \left\{\log{\left(\frac{x}{K}\right)}+\left(r-\frac{\sigma^{2}}{2}\right)\tau\right\} \\ &= \frac{-d_{2}}{\sigma} \end{align} and \begin{align} \partial_{\sigma}d_{2} &= \partial_{\sigma}d_{1}-\sqrt{\tau} &= -\frac{d_{2} + \sigma \sqrt{\tau}}{\sigma} \\ &= \frac{-d_{1}}{\sigma} \end{align} hence \begin{align} (\ast) &= x \frac{1}{\sqrt{2\pi}}e^{-\frac{d_{1}^{2}}{2}} \frac{-d_{2}}{\sigma} + e^{-r\tau}K\frac{1}{\sqrt{2\pi}}e^{-\frac{d_{2}^{2}}{2}}\frac{d_{1}}{\sigma} \\ &= \frac{1}{\sqrt{2\pi}}e^{-\frac{d_{1}^{2}}{2}} \left\{x\frac{-d_{2}}{\sigma} + e^{-r\tau}K\frac{d_{1}}{\sigma}e^{\frac{1}{2}(d_{1}^{2}-d_{2}^{2})}\right\} \\ &= \Phi'(d_{1}) \left\{x\frac{-d_{2}}{\sigma} + e^{-r\tau}K\frac{d_{1}}{\sigma}e^{\frac{1}{2}(d_{1}^{2}-d_{2}^{2})}\right\}(\ast \ast) \end{align} it follow \begin{align} d_{1}^{2}-d_{2}^{2} &= (d_{1}-d_{2})(d_{1}+d_{2}) \\ &= \sigma \sqrt{\tau}(2d_{1}-\sigma \sqrt{\tau}) \\ &= 2\log{\left(\frac{x}{K}\right)} + 2\left(r+\frac{\sigma^{2}}{2}\right)\tau - \sigma^{2}\tau \\ &= 2\log{\left(\frac{x}{K}\right)} + 2r\tau \end{align} and \begin{align} (\ast \ast ) & = \Phi'(d_{1}) \left\{x\frac{-d_{2}}{\sigma} + e^{-r\tau}K\frac{d_{1}}{\sigma}e^{\log{\left(\frac{x}{K}\right)}+r\tau}\right\} \\ &= \Phi'(d_{1}) \left\{x\frac{-d_{2}}{\sigma} + x\frac{d_{1}}{\sigma}\right\} \\ &= \Phi'(d_{1}) \left\{x \frac{d_{1}-d_{2}}{\sigma}\right\} \\ &= \Phi'(d_{1}) x \sqrt{\tau} \end{align} From the above relation \begin{align} \mathcal{V}_{t} = S_{t} \sqrt{\tau} \Phi'(d_{1}(\tau,S_{t},K,r,\sigma)). \end{align} $\Phi'(z) = e^{-z^{2}/2}/\sqrt{2\pi}> 0 $ then $\Phi$ is increasing, (assuming $ x> 0 $) and finally \begin{align} \partial_{\sigma}C_{BS} = x \sqrt{\tau} \Phi'(d_{1}) > 0. \end{align})