Exercise: Why the Vol Smile Falsifies Black-Scholes

Prerequisites: Implied Volatility, Risk-Neutral Measure

Problem

Black-Scholes assumes $S_T$ is log-normally distributed under $\mathbb{Q}$ with a single volatility parameter $\sigma$. Under this assumption the implied vol $\sigma_{\text{imp}}(K)$ for a fixed maturity $T$ would be a flat function of strike — every option reads back the same $\sigma$. Real markets display non-flat $\sigma_{\text{imp}}(K)$ curves (smiles, skews). This exercise explains why.

Consider the S&P 500 skew on a day when:

$S_0 = 100$, $T = 0.25$, $r = 0$.

1. Using the Black-Scholes formula with $\sigma = 0.18$, compute the $\mathbb{Q}$-probabilities $\mathbb{Q}(S_T < 80)$ and $\mathbb{Q}(S_T > 120)$ under the ATM-implied log-normal distribution.
2. Now compute the market's $\mathbb{Q}$-probability of $\mathbb{Q}(S_T < 80)$ as implied by the $80$-strike put's price. Repeat for $\mathbb{Q}(S_T > 120)$ using the $120$-strike call's price. Compare to Part 1.
3. Argue that the vol smile is equivalent to the statement: the market's implied risk-neutral density is not log-normal. Specifically, if $\sigma_{\text{imp}}(K)$ is decreasing in $K$ (as in the equity skew), what does this say about the left vs right tails of the risk-neutral distribution compared to log-normal?
4. Use the relation $\mathbb{Q}(S_T < K) = e^{rT}\frac{\partial P_{\text{market}}(K)}{\partial K}$ (Breeden-Litzenberger, for puts) to sketch how the full risk-neutral density $q(s)$ can in principle be read off from the full vol-smile curve.

Hint

For part 1, use the standard formulas: $\mathbb{Q}(S_T < K) = \Phi(-d_2)$ and $\mathbb{Q}(S_T > K) = \Phi(d_2)$ where $d_2 = (\ln(S_0/K) + (r - \sigma^2/2)T)/(\sigma\sqrt{T})$. For part 2, solve for $\Phi(-d_2(K))$ using each strike's own $\sigma_{\text{imp}}(K)$.