# 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!

• is this homework or an assignment?
– Matt
Jun 11, 2015 at 4:01
• Neither, revision question. Jun 11, 2015 at 4:38

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 $W\sim N(0,T)$.

• Why is your $Q$ equal to the measure $P^*$ in the question?
– g g
Jun 13, 2015 at 4:46
• @gg Because I assume the author uses $P^*$ for $Q$ (and $P$ for $P$). Jun 13, 2015 at 7:57