Dimensional Consistency in the Black–Scholes Formula

Summary

The note explains why the Black–Scholes call pricing formula is dimensionally consistent, despite an apparent problem involving the square root of time in the normal cumulative distribution inputs. The option price, underlying price, and strike are all measured in currency, while the normal distribution terms are dimensionless.

In the expression for the standardized variables, the log price ratio is dimensionless, and the annualized rate and variance terms become dimensionless when multiplied by time. Volatility has units of inverse square root of time, so multiplying it by the square root of time also produces a dimensionless denominator. The explanation assumes consistent annualization and time units; it does not address other modeling assumptions or pricing limitations.

Key ideas

• The Black–Scholes option price has currency units because its price terms are multiplied by dimensionless probabilities.
• The ratio of underlying price to strike is dimensionless, so its logarithm has no units.
• An annualized variance rate multiplied by time is dimensionless.
• Volatility has units of inverse square root of time, making volatility times square root of time dimensionless.

Tags

# Why does Black Scholes formula give inconsistent dimensional analysis result?


# Why does Black Scholes formula give inconsistent dimensional analysis result?












For example, distance = speed * time, m = m/s * s.

But this technique gives wrong answer on the Black Scholes formula. The square root in the denominator gives wrong unit inside of the culumulative probability function.

Is this because some assumptions used in the equation fundamentally changed the dimension? What is the fundamental reason for the dimension to be inconsistant?

## Answer by bhutes (score 9)

https://quant.stackexchange.com/a/45534

$C= S_0 N(d_1) - K e^{-rT} N(d_2)$

$C$, $S_0$ and $K$ have units of currency (e.g. USD).

$N(d1)$ and $N(d_2)$ are unit-less (dimensionless), the formula is dimensionally correct.

Considering,

$d1 = \frac {ln{\frac {S_0} K} + r T + \frac {\sigma^2} {2} T} {\sigma \sqrt T }$

$r$ and $\sigma^2$ have units of "per year", as they are stated on an annualized basis.

So, $\sigma$ has unit of "square root of "per year"".

Hence, $d1$ is also dimension-less.

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This summary was written by Stratmill's research agent from the original; it is not a copy of the source.