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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: $$ p_{0}=KN(-d_{2}^p)-S_{0}N(-d_{1}^p)=16N(d_1^c)-20N(d_2^c)=7.2 $$

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  • $\begingroup$ I understand this is not possible unless it is done computationally but given that this question was asked, it seems like there is a trick somewhere to solve this. $\endgroup$
    – Kai
    Dec 7, 2023 at 6:24
  • $\begingroup$ Is this an exam question? The intent might be to solve it using different methods than back-calculating vol. E.g. the strikes are nearly symmetrical to the spot price (+/- 20%) $\endgroup$
    – D Stanley
    Dec 7, 2023 at 14:56
  • $\begingroup$ Yes, it is an exam question with no calculators allowed. I have noticed the symmetry as well but I am unable to get the symmetry to amount to anything. Though I might be missing something. $\endgroup$
    – Kai
    Dec 7, 2023 at 15:22
  • $\begingroup$ So the question is not "how do we find the implied volatility" but "how do we price Option B knowing the price of a similar Option A?" (I do not know the answer but if you frame it that way it might get better responses) (Even if you did know the IV, I doubt you could price the option without a calculator) $\endgroup$
    – D Stanley
    Dec 7, 2023 at 15:25
  • $\begingroup$ See if quant.stackexchange.com/questions/53662 helps $\endgroup$
    – D Stanley
    Dec 7, 2023 at 15:31

2 Answers 2

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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.

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  • $\begingroup$ Thanks, Kai :) But I have to solve this without approximation. It does seem like there is a trick to this. $\endgroup$
    – Kai
    Dec 7, 2023 at 15:24
  • $\begingroup$ @Kai why don't you use put call parity? Actually, I still do not quite understand your question, it is worded quite confusingly. $\endgroup$
    – KaiSqDist
    Dec 7, 2023 at 17:28
  • $\begingroup$ I've posted a solution. Hopefully that makes sense. Took me a while to solve it. Thanks for looking at it! $\endgroup$
    – Kai
    Dec 8, 2023 at 18:15
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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: $$ p_{0}=KN(-d_{2}^p)-S_{0}N(-d_{1}^p)=16N(d_1^c)-20N(d_2^c)=7.2 $$

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