Black-Scholes formula is a monotonic increasing function of the volatility. Proof?

Question

I'm trying to show this statement: that Black-Scholes formula is a monotonic increasing function of the volatility ($\sigma$). I need to proof it from the Black-Scholes formula which is:

\begin{equation} X=N(d_1)S_0-N(d_2)Ke^{-rT} \end{equation} Where: \begin{equation} d_1=\frac{1}{\sigma \sqrt{T}}[\ln(\frac{S_0}{K})+(r+\frac{\sigma ^2}{2})T] \end{equation} \begin{equation} d_2=d_1-\sigma\sqrt{T} \end{equation} And $N$ is the cumulant function for the normal distribution.

Clearly, as $\sigma$ increases, the difference between $d_1$ and $d_2$ increases. However, the value depends on how centered is (near zero) $d_1$ , since the cumulant funcion have his "maximum grows" around zero.

Summarizing, I can't proof the statement. I hope someone could give some idea or some help. Thanks in advance.

Answer 1

In the Black-Scholes model, the underlying price $S_0$ is positive.

Then for $T>0$, "vega", the partial derivative of the option price with respect to volatility is positive $$\frac{\partial X}{\partial \sigma} = S_0N'(d_1)\sqrt{T}=\frac1{\sqrt{2\pi}}\exp(-d_1^2/2)S_0\sqrt{T}>0$$