Simulating the Distribution of a Black–Scholes Call Price

Summary

The document asks for the probability distribution of a European call option's price at an intermediate time under the risk-neutral measure. The response starts from the Black–Scholes pricing function, which maps the underlying asset price and time to maturity into the call value. Since the underlying follows geometric Brownian motion under the stated constant-parameter assumptions, the option value can be viewed as a nonlinear transformation of a lognormally distributed asset price.

The response says this transformed distribution has no convenient closed-form expression and proposes estimating it by simulation: draw paths or terminal samples for the Brownian motion, calculate the corresponding underlying prices, and apply the Black–Scholes formula to each. This yields an empirical distribution rather than an analytic law. The note does not provide simulation results or discuss sample size, discretization, or distribution summaries. Its brief suggestion to sample Brownian motion over the full horizon should be interpreted consistently with the target time and pricing formula.

Key ideas

• Under constant Black–Scholes assumptions, the underlying asset price is lognormally distributed under the risk-neutral measure.
• The call price is a nonlinear function of the underlying price and time to maturity.
• The response states that the transformed call-price distribution lacks a closed-form expression.
• Monte Carlo sampling can approximate the distribution by repricing the option for simulated underlying prices.

Tags

# Distribution of Black Scholes call option price at time 0<t <T


# Distribution of Black Scholes call option price at time 0<t <T












Does anyone know how to find the probability law (distribution) under P* of a Black Scholes Call Option price $C_t$ for $0 < t < T $?

(Under P*, $ dC_t = \frac{\partial c}{\partial s}\sigma S_t dW_t^{*} + rcdt $, where $C_t = c(s,t)$, $t \in [0,T]$ )

I'm expecting it will not be geometric Brownian motion but I'm not sure how to prove it.

Thanks!

## Answer by emcor (score 3)

https://quant.stackexchange.com/a/18332

This is the Black Scholes Call Price:

\begin{align} C(S, t) &= N(d_1)S - N(d_2) Ke^{-r(T - t)} \\ d_1 &= \frac{1}{\sigma\sqrt{T - t}}\left[\ln\left(\frac{S}{K}\right) + \left(r + \frac{\sigma^2}{2}\right)(T - t)\right] \\ d_2 &= \frac{1}{\sigma\sqrt{T - t}}\left[\ln\left(\frac{S}{K}\right) + \left(r - \frac{\sigma^2}{2}\right)(T - t)\right] \\ &= d_1 - \sigma\sqrt{T - t} \end{align}

All parameters except the underlying price $S$ are assumed constant. $S$ has a lognormal distribution and follows a GBM under $Q$:

$$S_t=S_0e^{(r-\frac{\sigma^2}{2})t+\sigma W_{t}^{Q}}$$

You can directly observe from the $C(S,t)$ formula that the distribution of $C$ cannot be in closed form since $N(*)$ is not in closed form.

You can simulate the distribution of $C$ by drawing many samples from $W\sim N(0,T)$.

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