# Find strike of an option based on a delta without option price

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 :

๐พ=๐น๐ก(๐)๐โ(๐โ1(ฮ)+1/2)๐๐โ๐กโT-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 ๐ out of this equation , so I can have ๐ = f(K), and I can then guess K to find the best ๐ ?

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

• What exactly is the use case for this? Dec 4, 2022 at 18:38

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.

• In your $\sigma$ formula, you still have $\sigma$ on the other side through the convexity term $+ \frac{\sigma^2}{2}t$ Dec 4, 2022 at 11:19
• 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?
– LvM_
Dec 4, 2022 at 11:49
• yep that's what i just got, but wait, what do we do with the negative sign ? and ± :) Dec 4, 2022 at 12:12
• 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 -).
– LvM_
Dec 4, 2022 at 13:28
• 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
– LvM_
Dec 4, 2022 at 18:35