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I'm trying to derive Vanna from the Black-Scholes Model (BSM) equation, but had a hook up on one of the manipulations in the formula.

Vanna, also referred to as DvegaDspot and DdeltaDvol, is a second order derivative of the option value, once to the underlying spot price and once to volatility.

Essentially it tells you how much Vega (sensitivity to volatility) will change if price changes; it's also equal to how much delta will change if volatility changes. It can be useful for hedging options against changing volatility for various strategies.

You can get it by taking the BSM equation, deriving it with respect to price (which leaves you with delta - the rate of change with respect to the underlying price), then you take the derivative of delta with respect to volatility, which leaves you with this next formula:

In the below PDF, on step 5 (page 4):

$$e^{-qt}\sqrt{T-t} N'(d_1)(d_2/\sigma)$$

Step 6, they adjust it into (with the assumption of no dividend, q=0)

$$\sqrt{T-t} N'(d_1)(1-d_1)$$

by making the substitution (from step 5):

$$d_2/\sigma = 1 - d_1$$

I can't seem to figure out how they did it. I'd be grateful if anyone had a little bit of knowledge they'd want to share. (I included a link to the actual PDF below as well.) I think I can figure out everything else, but that one adjustment is confusing me. Makes me wonder if I'm forgetting a basic rule or something.

Various Formulas:

Black-Scholes(-Merton) Model: $$C = S_tN(d_1)-Ke^{-rt}N(d_2)$$

Where:

C = Call Option Price

S = Current Stock Price

K - Strike Price

r - risk free interest rate

t - time to maturity

N = Normal CDF

$$d_1 = ((\ln(S_0/X) +t(r-q+\sigma^2/2))/\sigma\sqrt{\pi})$$

$$d_2 = D_1 -\sigma \sqrt{t}$$

N = Normal cumulative distribution function

$N^\prime$ = (Normal?) Probability Density Function

$N^{\prime \prime}$ = $$-(x*e^{-x^2/2})/(\sqrt{2}*\sqrt{pi})$$

($N^{\prime \prime}$ is my best guess, I checked with a calc, but could be wrong.

https://acfr.aut.ac.nz/__data/assets/pdf_file/0017/185300/172496-K-Huang-Vanna-Volga_Auckland.pdf

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    $\begingroup$ You may wish to include the definition of Vanna for people who aren't familiar with Greeks/sensitivities :) $\endgroup$
    – Jose Avilez
    Jan 7, 2022 at 15:44
  • $\begingroup$ Good idea! I fixed it up. $\endgroup$
    – NaturallyNick
    Jan 7, 2022 at 16:08
  • $\begingroup$ After you ask a question here, if you get an acceptable answer, you should "accept" the answer by clicking the check mark $\checkmark$ next to it. This scores points for you and for the person who answered your question. You can find out more about accepting answers here: How do I accept an answer?, Why should we accept answers?, What should I do if someone answers my question?. $\endgroup$
    – Jose Avilez
    Jan 11, 2022 at 1:03

1 Answer 1

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The claim is wrong, unless I'm missing something: $$\frac{\partial C^{\textrm{BS}}}{ \partial S_t \partial \sigma}=\frac{\partial \Delta_C^{\textrm{BS}}}{\partial \sigma}=\frac{\partial}{\partial \sigma }(e^{-q(T-t)}\Phi(d_1))=\frac{\partial d_1}{\partial \sigma}e^{-q(T-t)}\phi(d_1)$$ $$d_1=\frac{\ln(S_t/K)+(r-q+1/2\sigma^2)(T-t)}{\sigma\sqrt{T-t}}$$ $$\frac{\partial d_1}{\partial \sigma}=-\frac{\ln(S_t/K)+(r-q)(T-t)}{\sigma^2\sqrt{T-t}}+(1/2)\sqrt{T-t}=\\=\frac{-\ln(S_t/K)-(r-q)(T-t)+(1/2)\sigma^2(T-t)}{\sigma^2\sqrt{T-t}}=-\frac{d_2}{\sigma}$$ $$\boxed{\frac{\partial C^{\textrm{BS}}}{ \partial S_t \partial \sigma}=-\frac{d_2}{\sigma}e^{-q(T-t)}\phi(d_1)}$$

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    $\begingroup$ Agreed. The linked paper seems to be incorrect; this looks better! $\endgroup$
    – Jose Avilez
    Jan 7, 2022 at 19:59
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    $\begingroup$ Hello @JoseAvilez!! $\endgroup$
    – Snoop
    Jan 7, 2022 at 19:59
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    $\begingroup$ Hey @Snoop ! Thanks for taking care of the finance tag :) $\endgroup$
    – Jose Avilez
    Jan 7, 2022 at 20:02

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