Pricing a Put from a Related Call Using Black–Scholes Symmetry
Summary
The document considers whether a call option's quoted price can determine the price of a put with a different strike, when both share the same underlying, expiry, and volatility assumptions. In the stated zero-rate example, the call has spot 20, strike 25, expiry one year, and price 0.90; the target put has strike 16. The proposed approach avoids first solving numerically for implied volatility.
Under Black–Scholes, the two strike ratios are reciprocal, which makes the call and put pricing terms correspond through symmetry in their normal-distribution arguments. The solution therefore rewrites the put value using the call's pricing terms and gives 7.2. This result depends on the specified model assumptions, including shared volatility and zero interest rates. A second response cautions that implied volatility generally requires numerical approximation, underscoring that this shortcut relies on the particular relationship between the strikes.
Key ideas
- The example assumes Black–Scholes pricing, zero rates, and a common volatility for both options.
- Reciprocal spot-to-strike ratios create a symmetry between the call and put pricing terms.
- The stated put value is obtained by transforming the call formula without explicitly solving for volatility.
- The shortcut depends on the example's special strike relationship and model assumptions.
Tags
Full text
# Price Option B Knowing The Price of a Similar Option A
# Price Option B Knowing The Price of a Similar Option A
How do we find the implied volatility from the price in a call option and apply it to another option without a calculator? Or is there actually a better way? For example, given a 25-strike 1.0-expiry call option on $S$ has time-0 price 0.90 with $r = 0$, with $S_0 = 20$. Find the price of a 16-strike 1.0-expiry put option on $S$. How do we solve this question mathematically and not numerically?
Solution:
I've found the answer for this.
Assuming $r = 0$ and using Black Scholes: For our 25-strike call option, we know that $$ 20N(d_1^c)-25N(d_2^c) = 0.90, \text{ where } d_{1,2}^c=\frac{\log{20/25}}{\sigma}\pm\frac{\sigma}{2}=\frac{\log{4/5}}{\sigma}\pm\frac{\sigma}{2} $$ Similarly, for our 16 strike put, we know that: $$ d_{1,2}^p=\frac{\log{20/16}}{\sigma}\pm\frac{\sigma}{2}=-\frac{\log{4/5}}{\sigma}\pm\frac{\sigma}{2} $$ Using this, we can see that $d_{1,2}^p=-d_{2,1}^c$. Hence: $$ p_{0}=KN(-d_{2}^p)-S_{0}N(-d_{1}^p)=16N(d_1^c)-20N(d_2^c)=7.2 $$
## Answer by Kai (score 1, accepted)
https://quant.stackexchange.com/a/77641
Solution:
I've found the answer to this.
Assuming $r = 0$ and using Black Scholes: For our 25-strike call option, we know that $$ 20N(d_1^c)-25N(d_2^c) = 0.90, \text{ where } d_{1,2}^c=\frac{\log{20/25}}{\sigma}\pm\frac{\sigma}{2}=\frac{\log{4/5}}{\sigma}\pm\frac{\sigma}{2} $$ Similarly, for our 16-strike put, we know that: $$ d_{1,2}^p=\frac{\log{20/16}}{\sigma}\pm\frac{\sigma}{2}=-\frac{\log{4/5}}{\sigma}\pm\frac{\sigma}{2} $$ Using this, we can see that $d_{1,2}^p=-d_{2,1}^c$. Hence: $$ p_{0}=KN(-d_{2}^p)-S_{0}N(-d_{1}^p)=16N(d_1^c)-20N(d_2^c)=7.2 $$
## Answer by KaiSqDist (score 1)
https://quant.stackexchange.com/a/77628
Welcome Kai, I am Kai. Hopefully this answers your question?
A Review on IV Calculations https://www.sciencedirect.com/science/article/pii/S0377042717300602
I don't think there are exact mathematical solutions, but much rather approximates. If you want exact solutions, they should be numerical.Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)
This summary was written by Stratmill's research agent from the original; it is not a copy of the source.