Why does the $N'(d_1) \cdot S \cdot e^{-qT} = N'(d_2) \cdot K \cdot e^{-rT}$ identity hold in the BSM Greeks derivation?

Yes — this identity is one of the most beautiful algebraic facts in derivative pricing, and it is the reason all the BSM Greeks have such clean closed forms. The proof is purely mechanical, and the intuition is connected to the lognormal nature of the stock distribution.

The identity:

Where $N'(d) = (1/\sqrt{2\pi}) \cdot e^{-d^2/2}$ is the standard normal PDF.

The proof in three steps:

Step 1: Write out the ratio:

Step 2: Compute the ratio of PDFs:

Step 3: Use the fact that $d_2^2 - d_1^2 = (d_2 - d_1)(d_2 + d_1)$ and $d_2 - d_1 = -\sigma \sqrt{T}$:

Substituting $d_1 = [\ln(S/K) + (r - q + \sigma^2/2) T] / (\sigma \sqrt{T})$, the algebra simplifies (it really does) to:

So:

Plugging back into the original ratio:

So the two sides are equal. QED.

Intuitive interpretation:

The identity reflects the change-of-numeraire relationship between the risk-neutral measure (where you discount at $r$ and use $N(d_2)$) and the stock-numeraire measure (where you use $S$ as numeraire and use $N(d_1)$). Both measures must be consistent with the same forward price, and that consistency forces the identity.

Why it matters:

In the delta derivation, when you apply the chain rule to both $N(d_1)$ and $N(d_2)$ terms, the $N'(d)$ factors emerge multiplied by the same $\partial d / \partial S = 1/(S \sigma \sqrt{T})$. When you subtract the two terms, the $N'(d)$ parts cancel because of this identity, leaving the elegant result $\Delta = e^{-qT} N(d_1)$.

The same identity makes:

• Gamma: $\Gamma = N'(d_1) e^{-qT} / (S \sigma \sqrt{T})$ clean.
• Vega: $\nu = S e^{-qT} \sqrt{T} N'(d_1)$ clean.
• Theta: Theta has multiple terms but they organise around the same identity.

So this identity is the algebraic engine of the entire BSM Greeks toolkit.