Reading the Skew: Implied Vols From Four Quotes
A stock trades at $S = 100$. Interest rates and dividends are zero, and all options are European with $T = 0.25$ years to expiry. You see these mid quotes:
- 100-strike call: $3.99$
- 90-strike put: $1.45$
- 90-strike call: $11.45$
- 110-strike call: $0.48$
Use the Black-Scholes model, $C = S\,N(d_1) - K\,N(d_2)$ with $d_{1,2} = \dfrac{\ln(S/K) \pm \sigma^2 T/2}{\sigma\sqrt{T}}$ when $r = 0$.
(a) What is the implied volatility of the 100-strike call? (With $r = 0$ and $K = S$ the inversion can be done by hand.)
(b) What would the 90-strike put be worth under Black-Scholes at a flat volatility of $20\%$?
(c) What is the implied volatility of the 90-strike put?
(d) What implied volatility does the 90-strike call carry, and why? Together with the 110-strike call (implied volatility about $16\%$), what shape does the volatility curve have?
Hints
- At the money with $r = 0$, $d_1 = \sigma\sqrt{T}/2$ and $d_2 = -d_1$, so $C = S\,(2N(\sigma\sqrt{T}/2) - 1)$. Solve for $N(\cdot)$ first, then invert the normal CDF.
- For the put use $P = K\,N(-d_2) - S\,N(-d_1)$. If the market put costs much more than the flat-vol value, its implied volatility must be higher, since the price rises with $\sigma$.
- Bracket the put's implied vol: price it at $25\%$ and $26\%$ and interpolate. For the 90-strike call, compare $C - P$ with $S - K$.
Worked Solution
How to Think About It: Each quote maps to exactly one volatility, because the Black-Scholes price is strictly increasing in $\sigma$ (vega is positive). With $r = 0$ and $K = S$ the map can be inverted with a normal table. Away from the money you bracket and interpolate (or take Newton steps). Comparing the implied vols across strikes is what "reading the skew" means.
Quick Estimate: At the money, $C \approx 0.4\,\sigma S\sqrt{T}$, so $\sigma \approx 3.99/(0.4 \times 100 \times 0.5) \approx 0.20$. The 90 put is worth well under $1$ at $20\%$ but trades at $1.45$, so its implied vol is clearly higher, roughly in the mid-twenties.
Formal Solution:
*Step 1 -- At-the-money implied vol (a).* With $r = 0$ and $K = S$, $d_1 = \sigma\sqrt{T}/2$ and $d_2 = -\sigma\sqrt{T}/2$, so $$C = S\big(N(d_1) - N(-d_1)\big) = S\big(2N(\sigma\sqrt{T}/2) - 1\big).$$ Then $N(\sigma\sqrt{T}/2) = (1 + 3.99/100)/2 = 0.51995$, so $\sigma\sqrt{T}/2 = N^{-1}(0.51995) = 0.05003$ and $$\sigma = \frac{2 \times 0.05003}{\sqrt{0.25}} = 0.2001.$$ The 100-strike implied vol is about $20.0\%$.
*Step 2 -- The 90 put at a flat 20% (b).* With $\sigma = 0.2$, $\sigma\sqrt{T} = 0.1$: $$d_1 = \frac{\ln(100/90) + 0.005}{0.1} = \frac{0.10536 + 0.005}{0.1} = 1.1036, \qquad d_2 = 1.0036.$$ $N(-d_1) = 0.13488$ and $N(-d_2) = 0.15778$, so $$P = 90 \times 0.15778 - 100 \times 0.13488 = 14.2006 - 13.4882 = 0.712.$$ The market's $1.45$ is about twice that: the downside put is priced at a much higher volatility than the at-the-money option.
*Step 3 -- The 90 put's implied vol (c).* The put price increases with $\sigma$. At $25\%$ it is $1.3193$; at $26\%$ it is $1.4535$. The quote $1.45$ lies between them, and linear interpolation gives $$\sigma \approx 0.25 + 0.01 \times \frac{1.45 - 1.3193}{1.4535 - 1.3193} = 0.2597.$$ Bisection to full precision gives $0.25974$. A single Newton step from $20\%$ (vega $10.85$) overshoots to $26.8\%$; a second step corrects it. The 90-strike implied vol is about $26.0\%$.
*Step 4 -- The 90 call (d).* With $r = q = 0$, put-call parity reads $C - P = S - K$ for every value of $\sigma$. The quotes satisfy $11.45 - 1.45 = 10 = 100 - 90$, so the 90 call priced at the put's implied vol comes out at exactly $11.45$: it carries the same $26.0\%$. If the two vols differed, parity would be violated and a riskless trade would exist. Skew is a function of strike and expiry, not of whether the option is a call or a put; in-the-money calls carry the same vol as the out-of-the-money puts at the same strike.
*Step 5 -- The shape.* The implied vols at strikes $90$, $100$, $110$ are about $26\%$, $20\%$ and $16\%$ (the 110 call at $0.48$ inverts to $16.0\%$): a downward-sloping skew. Under Black-Scholes all three would be equal. The skew says the market's risk-neutral distribution has a fatter left tail than a lognormal with one volatility.
Answer: (a) $\sigma \approx 0.200$ ($20.0\%$). (b) $0.712$. (c) $\sigma \approx 0.260$ ($25.97\%$). (d) The same $26.0\%$ as the 90 put, because put-call parity ($C - P = S - K$ at any $\sigma$) ties the two prices together; with the 110 call at about $16\%$ the curve is a downward-sloping skew ($26\%$, $20\%$, $16\%$ for strikes $90$, $100$, $110$).
Intuition
Implied volatility is just a price quoted in different units: the one $\sigma$ that makes Black-Scholes match the quote. At the money with zero rates the formula collapses to $C = S(2N(\sigma\sqrt T/2) - 1) \approx 0.4\,\sigma S\sqrt T$, so a $3.99$ call on a $100$ stock over three months is a $20\%$ vol. The 90-strike put costs about twice its $20\%$ value, which takes about $26\%$ to justify, while the 110-strike call needs only $16\%$: a downward-sloping skew. Put-call parity holds for every $\sigma$, so a call and a put at the same strike and expiry must carry the same implied vol; the skew depends on strike, not on whether the option is a call or a put.