Deriving a Black-Scholes Strike from Delta and Volatility

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


# 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 :

```
𝐾=𝐹𝑡(𝑇)𝑒−(𝑁−1(Δ)+1/2)𝜎𝑇−𝑡√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 `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?

## Answer by LvM_ (score 0, accepted)

https://quant.stackexchange.com/a/73953

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.