Black-Scholes: solve for $\sigma$ given $d_1$ and $d_2$

Question

Black Scholes valuation for european call option is:

$$C_0=S_0N(d_1)-Xe^{-rT}N(d_2)$$

where $d_1=\dfrac{\ln(\frac{S_0}{x})+(r+{\sigma^2\over2})T}{\sigma\sqrt{T}}$

and $d_2=\dfrac{\ln(\frac{S_0}{x})+(r-{\sigma^2\over2})T}{\sigma\sqrt{T}}$

I need to find $\sigma$ in $d_1$ and $d_2$

So far for $d_1$ I have gone to this point: $r(d_1\sqrt{T}-Tr)-\frac{T\sigma^2}{4}=\ln(\frac{S_0}{x})+Tr^2$ but I'm not sure how to continue solving for $\sigma$.

Answer 1

It is well known that $$d_2 = d_1 - \sigma\sqrt{T}\text{.}$$ To see why this is true, notice that (with $\delta = 0$, the dividend rate): $$\begin{align} d_1 &= \dfrac{\ln\left(\dfrac{S_0}{x}\right) +\left(r +\dfrac{\sigma^2}{2}\right)T}{\sigma\sqrt{T}} \\ &= \dfrac{\ln\left(\dfrac{S_0}{x}\right) +\left(r +\dfrac{\sigma^2}{2}+\dfrac{\sigma^2}{2}-\dfrac{\sigma^2}{2}\right)T}{\sigma\sqrt{T}} \\ &= \dfrac{\ln\left(\dfrac{S_0}{x}\right) +\left(r -\dfrac{\sigma^2}{2}\right)T+\left(\dfrac{\sigma^2}{2}+\dfrac{\sigma^2}{2}\right)T}{\sigma\sqrt{T}} \\ &= \dfrac{\ln\left(\dfrac{S_0}{x}\right) +\left(r -\dfrac{\sigma^2}{2}\right)T}{\sigma\sqrt{T}}+\dfrac{2\left(\dfrac{\sigma^2}{2}\right)T}{\sigma\sqrt{T}} \\ &= d_2+ \sigma\sqrt{T}\text{,} \end{align}$$ thus proving $d_2 = d_1 - \sigma\sqrt{T}$. So this gives $$\sigma\sqrt{T} = d_1 - d_2 \implies \sigma = \dfrac{d_1-d_2}{\sqrt{T}}\text{.} $$