Why Option Value Is Often Nearly Linear in Volatility Near the Money
Summary
The document examines why Black–Scholes call prices may appear to rise linearly as volatility increases. It explains that the relationship is not generally linear: the pricing formula depends on volatility through the normal distribution terms, and vega also varies with volatility. A plot or a check of whether vega stays constant can help distinguish true linearity from a visual impression over a limited range.
The response notes that the curve can look close to linear for at-the-money calls, where an approximation makes option value proportional to volatility and the square root of time. It offers an intuition that greater volatility spreads the terminal-price distribution and increases the expected in-the-money payoff. This is an approximation, not a universal property; moneyness and other model inputs affect the shape, and apparent linearity over one parameter range should not be generalized to all options.
Key ideas
- Black–Scholes option value is not generally a linear function of volatility.
- Because vega depends on volatility, a constant price response to volatility should not be assumed.
- At-the-money call prices can be approximately linear in volatility under a suitable approximation.
- Apparent linearity may be local and can change with moneyness and model inputs.
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Full text
# What's the intuition behind there being a perfect linear relationship between option value and expected volatility?
# 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.
## Answer by Bob Jansen (score 6, accepted)
https://quant.stackexchange.com/a/61861
As Frido Rolloos remarks this is not the case. Let me suggest three ways to confirm this:
- 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.
- 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 $\sigma$.
- 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$Price)
Coefficients:
[1] 0.002874 39.836173
```
Hmm, what about points 1 and 2. This looks linear... What's up with that. Let's try some other parameters:
```
bs <- blackscholes(
callput = 1,
S0 = 100,
K = 120,
r = 0,
time = 1,
vola = volas
)
plot(volas, bs$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.Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)
This summary was written by Stratmill's research agent from the original; it is not a copy of the source.