Why Black-Scholes Price Can Be Expressed Using Delta and Rho

# Can I get Black-Scholes option price from greeks?


# Can I get Black-Scholes option price from greeks?












I am unpleased with current Interactive Brokers risk graph for option strategies, so I'm planning on writing an application myself to plot it.

My initial idea is to get the option greek values from the broker's data feed, so I would have the following data:

- Strike price

- Current underlying price

- Time to expiration

- Delta, Gamma, Theta, Rho, Vega

Since the Black-Scholes formula is as follows: $$C=SN(d_1)-e^{-rT}KN(d_2)$$

And assuming the greeks formulas as described in this paper, I can conclude that: $$N(d_1)=\delta$$ $$e^{-rT}N(d_2)=\frac{\rho}{KT}$$ And therefore I can calculate the Black-Scholes formula knowing only delta, rho, current underlying price, strike price and time to expiration: $$C=S \delta-K \rho$$

The problem is that of course this must be wrong. It cannot be possible that I am able to calculate option price using only 2 greeks, or at least it looks hard to believe from what I know.

So, which assumption of those I'm taking is wrong? Is there any resource somewhere of how to calculate the option price from greeks (I searched but couldn't find one, that's why I started playing with these equations).

## Answer by SRKX (score 5, accepted)

https://quant.stackexchange.com/a/21325

You're actually pricing your call option with all known inputs here, so the fact that you need only $\delta$ and $\rho$ is just an analytical result.

You use the greeks to take a Taylor approximation approach, where the goal is to estimate the value of the call if one of the input changes (the bigger the change the more greeks you'll need to estimate the change in call price accurately), but if the inputs stay the same, you then all the greeks are ignored anyway.

## Answer by arodrisa (score 1)

https://quant.stackexchange.com/a/21300

I understand Greeks in option pricing as the Taylor Theorem, therefore, the more Greeks you have, the more explanatory your function will be. This is the same idea, you need to approximate the price to a curve (volatility), and depending on the degree of the equation (greeks) you will obtain more accuracy.

## Answer by user32416 (score 0)

https://quant.stackexchange.com/a/21311

I don't understand the source of confusion. If you go back to the classical Black-Scholes (1974) paper, or in effect, any other current textbook derivation of the model, your equation of call option price $C$ is EXACTLY the Black-Scholes. Once you have the analytical solution $C = C(s,\sigma,T,K,\cdots)$ as a function of the other parameters, then you take a first order derivative to get the comparative statics (i.e. "Greeks" if you may; but this process of getting the value function and then perturbing the parameters, this is a very, very standard procedure in finance and economic theory. Essentially, an economist is interested in knowing how the solution to a model changes as the underlying parameter changes --- which is precisely the point of the Black-Scholes greeks anyway).

Once you take those derivatives and get the greeks, you have what you have. And all I can say is that you've found an alternative expression for the Black-Scholes formula, but that's potentially not that surprising nor interesting --- given that the $\delta$ gives you the hedge ratio and $e^{-rT} N(d_2)$ tells you how many bonds you need to hold. This is the precise replicating portfolio argument to pricing a derivative.