# Black-Scholes Formula under $T$-forward measure

The Black-Scholes price of a European call option is given by $$C_0^{BS}(T, K) = \mathbb{E}_Q[e^{-rT}(S_T - K)_+] = S_0 \Phi(d_1) - Ke^{-rT}\Phi(d_2) ,$$

where $$d_{1,2} = \frac{\log\big(\frac{S_0}{K}\big) + (r\pm \frac{1}{2}\sigma^2)T}{\sigma \sqrt{T}},$$

and the underlier $$S_t$$ has the following dynamics under $$Q$$:

$$dS_t = rS_tdt + \sigma S_t dW^Q_t$$

I'm familiar with the derivation of this formula. Is there a similar formula for pricing under a different measure? In particular, I am concerned with the $$T$$-forward measure, $$Q^T$$.

For example, if I want to price a derivative which has the value $$C_0(T, K) = P(0, T) \mathbb{E}_{Q^T}[(S_T - K)_+],$$ can I derive a similar Black-Scholes formula?

Here's my attempt:

Given that $$\frac{dQ^T}{dQ} = \frac{1}{P(0, T)B(T)}$$, then under Black-Scholes assumptions (constant short rate) $$\frac{dQ^T}{dQ} = 1$$. Hence, the dynamics of $$S_t$$ under $$Q^T$$ are: $$dS_t = rS_tdt + \sigma S_t dW^{Q^T}_t$$ Then, one can imitate the proof of the Black-Scholes formula: \begin{align} C_0(T, K) &= \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}(S_0\exp\{(r-\frac{1}{2}\sigma^2)T + \sigma\sqrt{T}z\} - K)_+ e^{-\frac{z^2}{2}} \end{align} then, the integrand is only non-zero when $$z > \frac{\log{\frac{K}{F}} + \frac{1}{2}\sigma^2 T}{\sigma \sqrt{T}} := -\tilde{d_2}$$ where $$F = S_0e^{rT}$$. I'll skip the rest of the proof because it's basically identical to the Black-Scholes formula derivation. This yields

$$C_0(T, K) = P(0, T) [F \Phi(\tilde{d_1}) - K\Phi(\tilde{d_2})]$$

where $$\tilde{d}_{1,2} = \frac{\log\big(\frac{F}{K}\big) \pm \frac{1}{2}\sigma^2T}{\sigma \sqrt{T}}.$$

Does this look correct?

• You are almost there: you forgot the bond price $P \left(0, T\right)$ in the expression of the call price under the $T$-forward measure. That shows that if the interest rate is constant, working under either risk-neutral or $T$-forward measure is exactly the same. – siou0107 Jul 25 at 23:05
• Could you make it an answer @siou0107? – Bob Jansen Jul 26 at 7:09
• Could you validate it then? ;) – siou0107 Jul 26 at 9:27

What that showed is that under the Black-Scholes assumption of a constant rate, working under the risk-neutral measure or under the $$T$$-forward measure is exactly the same.
When rates are stochastic, however, you do not know the value of $$B_T = e^{\int_0^T{r_t \mathrm{d} t}}$$ and to work under $$Q$$ you must compute the entire integral within the expectation, and finding a closed-form solution is difficult; using numerical methods is no easier.
However, you do know the value of $$P(0, T)$$ and the forward price $$\frac{S_t}{P(t, T)}$$ is a martingale. Note that its diffusion term is $$\sqrt{\sigma^2 + \sigma_P^2 - 2 \rho \sigma \sigma_P}$$ ; you thus require an estimate of the bond price volatility $$\sigma_P$$ and spot-bond correlation $$\rho$$, and can then use the simpler closed-form solution under the $$Q^T$$-measure.