# What's the intuition behind there being a perfect linear relationship between option value and expected volatility?

I modelled option prices using the BS model at different levels of volatility. Surprisingly, I came out with a perfectly linear relationship. As volatility rises, so does the option value, which is expected. But what I don't get is the intuition behind it being a linear relationship.

• There is no perfect linear relationship between the Black-Scholes-Merton option price and volatility. I think you made an error in your calculation. Mar 20, 2021 at 8:45

As Frido Rolloos remarks this is not the case. Let me suggest three ways to confirm this:

1. Look at the Black-Scholes formula (from Wikipedia): $$C(S, t) = N(d_1)S_t - N(d_2)PV(K)$$ where $$d_1 = \frac{1}{\sigma \sqrt{T-t}}\Big[\ln(S_t/K) + (r + \sigma^2/2)(T-t)\Big]$$, $$d_2 = d_1 - \sigma \sqrt{T-t}$$ and $$PV(K) = Ke^{-r(T-t)}$$ and $$N(\cdot)$$ the cumulative distribution of the standard normal. This certainly doesn't look linear in $$\sigma$$. You could substitute in $$N(\cdot)$$ and do some algebra and come the conclusion that indeed it isn't.
2. You can look at Vega, the derivative of the option value with respect to volatility and check whether it does not depend on $$\sigma$$. If that is the case the value of the option would be linear in volatility. However, looking up Vega: $$\mathcal{V} = S e^{-q (T-t)} N(d_1) \sqrt{T-t}.$$ we notice that it depends on $$N(d_1)$$ which depends on $$\sigma$$.
3. We can compute a bunch of option values and plot the result:
library(ragtop)
volas <- (1:100)/500
bs <- blackscholes(
callput = 1, # 1 for calls, -1 for puts
S0 = 100,
K = 100,
r = 0,
time = 1,
vola = volas
)
plot(volas, bs$Price)  line(volas, bs$$Price) # More about this below. Call: line(volas, bs$$Price) Coefficients:  0.002874 39.836173 Hmm, what about points 1 and 2. This looks linear... What's up with that. Let's try some other parameters: bs <- blackscholes( callput = 1, S0 = 100, K = 120, r = 0, time = 1, vola = volas ) plot(volas, bs$Price) The relation is certainly not linear but can appear linear locally. Also see this answer for different plots or play around with the R code above.

Kevin pointed to a useful approximation for ATM call options: $$C(S, t) = \frac{2}{5}Se^{-rT}\sigma\sqrt{T}$$ which is indeed linear in $$\sigma$$. Note that the coefficient found using the R line() function matches with the $$0.4S$$ in the formula.

In conclusion, the relation is not perfectly linear but in some cases is close to linear. My intuition for the linear behaviour for ATM options is that the distribution of $$S_T$$ scales approximately linearly with volatility. The expected value of the in the money part is therefore also approximately linear.

• To put some equations to @Bob's answer (+1), it's well-known that you can approximate ATM Black Scholes call option prices by $$\frac{2}{5}Se^{-rT}\sigma\sqrt{T},$$ which is indeed linear in volatility, see this answer. Mar 20, 2021 at 9:42
• Thank you so much! Really appreciate the extensive application. Makes a lot of sense now. Mar 20, 2021 at 19:25