# 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).

• Shouldn't it be in that case $C=S \delta - \frac{\rho}{T}$??? – SRKX Oct 20 '15 at 6:28

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.

• So, my result is correct? I am not using any Taylor approximation, someone suggested it at another answer, but I am currently using none – Roman Rdgz Oct 20 '15 at 7:48
• That's right, but I'm not sure why you're doing it this way... IB's API is giving you all the greeks but not option price? That seems very weird doesn't it? Another thing you might want to consider is understand how they compute the greeks in the backstage. – SRKX Oct 20 '15 at 8:03
• They give me the option price NOW. But I want to be able to plot the whole risk graph for any t to expiration. They don't provide that data but in their software, in a tiny, unusable window. – Roman Rdgz Oct 20 '15 at 10:25
• Oh but you can't expect your $\rho$ and $\delta$ to be the same for different $t$... You should add in your question a screenshot of what you want to improve. – SRKX Oct 20 '15 at 11:31
• I guess I had not thought about that. I guess I should stick to the volatility and rate of interest for plotting purposes then... – Roman Rdgz Oct 20 '15 at 12:43

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.

• That sounds more accurate to the actual effect of greeks on the option price but, is there any formula where to calculate the option price from the greeks? What I am avoiding is having to deduce the implied volatility and the spot rate for this calculation. Is this even possible? – Roman Rdgz Oct 19 '15 at 11:22
• I don't think so (won't say there is not), because greeks tell you the increment (positive or negative) of the price related to a starting point. Therefore you need a starting price (spot) and the volatility (vega, vamma...) Look it from the physics point of view, you can have 2 vehicles at same speed (50km/h), accelerating at the same quaintity (1m/s^2), but in different countries (different spot). Knowing the speed and acceleration you can't get the actual point, unless you have time scale and starting poing. Please, tell me if I wasn't clear – arodrisa Oct 19 '15 at 13:21
• I can sort of see your Taylor expansion argument and intuitively that the more higher order terms that you have, the better your approximation for a generic function. However, the Black-Scholes was NOT derived as a Taylor expansion (or perhaps you're thinking of Ito's lemma can be viewed as such, then my bad). The Black-Scholes is an EXACT formula and moreover, there's a solid mathematical finance and economic argument to why the formula must be so. – user32416 Oct 19 '15 at 17:34
• As you are saying, I'm just giving a different point of view. BS equation is a partial differential equation that is pretty similar to the heat transfer equation, therefore, as you said is exact. I am just giving other point of view in order to make easier to understand how greeks work. As they are derivatives they explain the movement depending of the factor they are derived by. So the more factors you add, the more accuracy you will have in your predictive model. It was very hard for me to understand Ito's lemma, and this is the way I did. – arodrisa Oct 20 '15 at 6:40
• Well, I'm just trying to make things easier. I just wanted to make it simple to everyone, by using examples that everyone may know. As you are saying, your point is correct, but complex. And just try to teach anyone without knowing it's knowledge Ito's lemma... Quite hard I guess – arodrisa Oct 20 '15 at 6:57

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.

• The question is: can I rely on the results of the formula using only 2 greeks when there are 5? Wouldn't I miss some information if I leave 3 greeks out? The formula suggests that those 2 are enough, but I just want to be sure abut it before using the formula for serious trading – Roman Rdgz Oct 20 '15 at 6:17
• I could be wrong but since you're putting money on the line, here's a suggestion and comment --- it seems like you're just pulling data from your online broker (Interactive Brokers in this case?) and somehow want to buy/sell options based on pricing differences between the Black-Scholes theoretical price and what the actual market price is. If this is your objective, it isn't very optimistic. The Black-Scholes model is highly stylized and more important than that, its really the dynamic portfolio replicating mechanism that makes the greeks useful, not the actual pricing. – user32416 Oct 20 '15 at 6:40
• And on a more abstract level, what you're suggesting doesn't make much sense either. If you consider a function $f(x,y,z)$, you're effectively saying that $f_x, f_y, f_z$ should contain "more information" than the full function $f$. (Replace $f$ here with your Black-Scholes call price). – user32416 Oct 20 '15 at 6:41
• I just want to plot the options to analyze them, since IB sucks for risk graphs of option strategies. Using the greeks as input seemed easier than getting volatility and rate. Since these two parameters are considered within the greeks, I assume that my formula should work, since it is mathematically correct. But it just sees weird that only rho accounts for volatility, since vega is the typical one for volatility analysis – Roman Rdgz Oct 20 '15 at 7:47