What happens to the BSM call price as volatility goes to zero?

As $\sigma \to 0$, the BSM call price collapses to the discounted intrinsic value of a forward on the stock. The formula handles this edge case beautifully, and walking through it will cement your understanding.

Setup:

With $\sigma = 0$, the stock has no uncertainty. Its terminal value is known with certainty:

This is the no-arbitrage forward price (under continuous compounding with dividend yield $q$).

The two cases:

Case 1: $S_0 \cdot e^{(r-q) \cdot T} > K$ (forward is in-the-money)

The call will definitely pay off $S_T - K$ at expiry, which we can compute today:

So the call price equals $S_0 \cdot e^{-qT} - K \cdot e^{-rT}$. Plugging into BSM with $\sigma = 0$:

• $d_1$ and $d_2 \to +\infty$ (because the deterministic forward is comfortably above $K$)
• $N(d_1) \to 1$ and $N(d_2) \to 1$
• $c = S_0 \cdot e^{-qT} \cdot 1 - K \cdot e^{-rT} \cdot 1 = S_0 \cdot e^{-qT} - K \cdot e^{-rT}$ $\checkmark$

Case 2: $S_0 \cdot e^{(r-q) \cdot T} < K$ (forward is out-of-the-money)

The call will definitely expire worthless:

• $d_1$ and $d_2 \to -\infty$
• $N(d_1) \to 0$ and $N(d_2) \to 0$
• $c = 0$ $\checkmark$

Why this is a useful sanity check:

When you build a BSM spreadsheet, set $\sigma = 0.001$ (not literally zero, to avoid division by $\sigma$) and verify the call price equals discounted intrinsic value of the forward. If it does not, your formula is wrong.

The "optionality" interpretation:

Volatility is what gives options value beyond their forward-intrinsic worth. The difference

is the time value of the call, and it scales roughly with $\sigma\sqrt{T}$. As vol goes up, time value goes up. As $\sigma$ goes to zero, time value goes to zero, and the option becomes equivalent to a forward.

This is also why vega ($\partial c / \partial \sigma$) is always non-negative for vanilla options. Optionality only adds value.