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I would like to use the Black Scholes model to get the strike based on delta, without having the option price.

So I read this: From Delta to moneyness or strike

Which show that we can get a strike price with :

𝐾=𝐹𝑑(𝑇)π‘’βˆ’(π‘βˆ’1(Ξ”)+1/2)πœŽπ‘‡βˆ’π‘‘βˆšT-t

But with this my software have to guess option prices to get the Sigma. I would like to guess strikes instead since I have a better first guess.

How can I solve 𝜎 out of this equation , so I can have 𝜎 = f(K), and I can then guess K to find the best 𝜎 ?

Or, to put it differently, having delta and no option price, what is my equation to guess the strikes?

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  • $\begingroup$ What exactly is the use case for this? $\endgroup$
    – AKdemy
    Commented Dec 4, 2022 at 18:38

1 Answer 1

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Let's suppose we work for a Vanilla Call Option. The formula for Delta is : $\Delta = \frac{\partial V}{\partial S} = N(d1)$

where $d1 = \frac{ln(\frac{S}{K})+(r + \frac{\sigma^2}{2})t}{\sigma \sqrt(t)}$ and $N(x)$ is the Standard normal cumulative distribution function.

Now, given the value of Delta, you can extract the value of $d1$. For example you can use P. J. Acklam algorithm for the inverse Normal CDF (accurate to 1.15E-9). (Good website: https://stackedboxes.org/2017/05/01/acklams-normal-quantile-function/)

Now you have: $N^{-1}(\Delta) = d1 = \frac{ln(\frac{S}{K})+(r + \frac{\sigma^2}{2})t}{\sigma \sqrt(t)}$

Rewriting it, you get: $\sigma = \frac{ln(\frac{S}{K})+(r + \frac{\sigma^2}{2})t}{d1 \sqrt(t)} = f(K)$. (f - function of K assuming all other parameter are fixed).


edit: Thank you to siou0107 who point out that I still had a sigma term on the RHS.


$\sigma = \frac{ln(\frac{S}{K})}{d1 \sqrt{t}}+ \frac{(r + \frac{\sigma^2}{2})t}{d1 \sqrt{t}}$

$\sigma d1 \sqrt{t} - (r + \frac{\sigma^2}{2})t= ln(\frac{S}{K})$

$ (r + \frac{\sigma^2}{2}) - \sigma \frac{d1} {\sqrt{t}}= - \frac{1}{t} ln(\frac{S}{K})$

$ \frac{\sigma^2}{2} - \sigma \frac{d1} {\sqrt{t}}= - \frac{1}{t} ln(\frac{S}{K}) - r$

$ \sigma^2 - 2 \frac{d1} {\sqrt{t}}\sigma = - \frac{2}{t} ln(\frac{S}{K}) - 2r$

Complete the square: $(a^2 - 2ab+b^2) = (a-b)^2$

$(\sigma - \frac{d1} {\sqrt{t}})^2 - \frac{d1^2} {t} = - \frac{2}{t} ln(\frac{S}{K}) - 2r$

$(\sigma - \frac{d1} {\sqrt{t}})^2 = - \frac{2}{t} ln(\frac{S}{K}) - 2r + \frac{d1^2} {t}$

$\sigma = \pm \sqrt{- \frac{2}{t} ln(\frac{S}{K}) - 2r + \frac{d1^2} {t}} + \frac{d1} {\sqrt{t}}$


Edit: What happen if $- \frac{2}{t} ln(\frac{S}{K}) - 2r + \frac{d1^2} {t} < 0$ (i.e. square root of a negative number is a complex number which is not what we want!)


For our solution to be real, we need to have $-\frac{2}{t}\ln(\frac{S}{K}) - 2r + \frac{d1^2}{t}>0$

Let's play around and check if we can prove this is ALWAYS the case (or not?)

$\Leftrightarrow \frac{d1^2}{t}>\frac{2}{t}\ln(\frac{S}{K}) + 2r $

$\Leftrightarrow d1^2>2\ln(\frac{S}{K}) + 2rt $

$\Leftrightarrow \frac{(\ln(\frac{S}{K}) + (r+\frac{\sigma^2}{2})t)^2}{\sigma^2t}>2\ln(\frac{S}{K}) + 2rt $

$\Leftrightarrow (\ln(\frac{S}{K}) + (r+\frac{\sigma^2}{2})t)^2>2\ln(\frac{S}{K}) \sigma^2t+ 2rt^2 \sigma^2$

$\Leftrightarrow (\ln(\frac{S}{K})^2 + (2\ln(\frac{S}{K})rt+ (r^2+r\sigma^2+\frac{\sigma^4}{4})t^2)>\ln(\frac{S}{K}) \sigma^2t+ 2rt^2 \sigma^2$

$\Leftrightarrow \ln(\frac{S}{K})^2 + 2\ln(\frac{S}{K})rt+ r^2t^2+\frac{\sigma^4t^2}{4}>\ln(\frac{S}{K}) \sigma^2t+ rt^2 \sigma^2$

$\Leftrightarrow (\ln(\frac{S}{K}) + rt)^2 +\frac{\sigma^4t^2}{4}>(\ln(\frac{S}{K}) + r t)\sigma^2t$

We know that S, K, r, t, $\sigma \geq 0$. The only element that can be negative is $\ln(\frac{S}{K})$ That happens when $\frac{S}{K} \in (0,1)$

The LHS only contains square, so we know that the LHS is always positive. The RHS contains $\ln(\frac{S}{K})$ which could be negative.

Therefore, $- \frac{2}{t} ln(\frac{S}{K}) - 2r + \frac{d1^2} {t} > 0$.

This is not a rigorous proof but hopefully you can see why the value inside the square root CANNOT be negative.


Regarding the $\pm$ part, please refer to my comment.

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    $\begingroup$ In your $\sigma$ formula, you still have $\sigma$ on the other side through the convexity term $+ \frac{\sigma^2}{2}t$ $\endgroup$
    – siou0107
    Commented Dec 4, 2022 at 11:19
  • $\begingroup$ Thank you @siou0107 , I edit my answer accordingly. I am going to check my answer on a piece of paper as it is the first time I came across this kind of question (which was quite fun to think about). I don't know how to save temporary draft - Can someone tell me what is the best practice when editing an existing answer? $\endgroup$
    – LvM_
    Commented Dec 4, 2022 at 11:49
  • $\begingroup$ yep that's what i just got, but wait, what do we do with the negative sign ? and ± :) $\endgroup$ Commented Dec 4, 2022 at 12:12
  • $\begingroup$ Try the code for several parameter and see how the function behave. For example, fix some parameter, S = 100, K = 100, sigma = 0.2, r = 0.05, tau = 1.0, and see if you can retrieve the expected value (i.e. 0.2) try again for multiple parameter. Regarding $\pm$ check if both answer are relevant and if not check if one of the answer is consistently correct (i.e. just keep the + or just the -). $\endgroup$
    – LvM_
    Commented Dec 4, 2022 at 13:28
  • $\begingroup$ Answer updated with the negative sign. Short answer, it cannot be negative (according to the way I played around with the formula). Feel free to check it yourself and let me know if you reach the same conclusion! HF $\endgroup$
    – LvM_
    Commented Dec 4, 2022 at 18:35

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