Black-Scholes call option formula, which probability measure

Question

The stock and bond under the Black-Scholes framework, no dividends: $$S_t=S_0e^{\sigma W_t+\mu t}=S_0e^{\sigma \tilde{W}_t +(r-\frac{1}{2}\sigma^2)t}$$ $$B_t=e^{rt}$$ where $\tilde{W}_t$ is $\mathbb{Q}$-Brownian motion. Thus, the risk-neutral stock price dynamics: $$S_T = LN_\mathbb{Q}(\ln{S_0}+(r-\frac{1}{2}\sigma^2)T,\sigma^2T)$$ Black-Scholes call option formula: \begin{align} V_0&=e^{-rT} \mathbb{E}_\mathbb{Q}[(S_T-k)^+]\\ &=e^{-rT}\mathbb{E}_\mathbb{Q}(S_T1_{S_T>k})-ke^{-rT}\mathbb{Q}(S_T>k)\\ &=S_0\Phi (d_1)-ke^{-rT}\Phi(d_2) \end{align} where $$d_1=\frac{\ln{\frac{S_0}{k}}+(r+\frac{1}{2}\sigma^2 )T}{\sigma\sqrt{T}}$$ $$d_2=\frac{\ln{\frac{S_0}{k}}+(r-\frac{1}{2}\sigma^2 )T}{\sigma\sqrt{T}}$$ My question is, are $\Phi (d_1)$ and $\Phi (d_2)$ computed under the risk-neutral measure $\mathbb{Q}$ or the real world measure $\mathbb{P}$? And is it $$e^{-rT}\mathbb{E}_\mathbb{Q}(S_T1_{S_T>k})-ke^{-rT}\mathbb{Q}(S_T>k)$$ or $$e^{-rT}\mathbb{E}_\mathbb{Q}(S_T1_{S_T>k})-ke^{-rT}\mathbb{P}(S_T>k)$$

It seems obvious to me that they should be computed under $\mathbb{Q}$ due to the replication pricing strategy which utilises the risk-neutral stock price dynamics. However, in textbook exercises, finding the explicit value of the call option involves using the table containing probabilities for the standard normal distribution, which are evidently computed under the real world measure $\mathbb{P}$.

Edit: Related Understanding the solution of this integral

Answer 1

Indeed,these probabilities are obtained under different probability measures but we should change the measure $\mathbb{Q}$ to another measure $\mathbb{Q}^S$. Evaluating $\mathbb{E}_{t}^{\mathbb{Q}}\left[S_T1_{S_T>K}\right]$ requires changing the measure $\mathbb{Q}$:

Consider the Radon-Nikodym derivative $$\frac{d\mathbb{Q}}{d\mathbb{Q}^S}=\frac{B_T/B_t}{S_T/S_t}$$ where $$B_t=\exp\left(\int_{0}^{t}r\,du\right)=e^{rt}$$ as a result $${{\mathbb{Q}}^{S}}({{S}_{T}}>K)=\int\limits_{K}^{+\infty }{d{{\mathbb{Q}}^{S}}}=\frac{{{e}^{-r(T-t)}}}{{{S}_{t}}}\int\limits_{K}^{+\infty }{{{S}_{T}}\,d\mathbb{Q}}=\frac{{{e}^{-r(T-t)}}}{{{S}_{t}}}\int\limits_{K}^{+\infty }{{{S}_{T}}{{f}_{{{S}_{T}}}}(x)dx} $$ we have $$\mathbb{Q}^S(S_T>K)=\frac{e^{-r(T-t)}}{S_t}E^\mathbb{Q}[S_T|S_T>K]=N\left(\frac{\ln \left(\frac{X_t}{K}\right)+\left( r+\frac{1}{2}\sigma ^{2} \right)(T-t)}{\sigma^2\sqrt{T-t}}\right)$$ Indeed $$\mathbb{Q}^S(S_T>K)=N(d_1)$$ on the other hand $$V(t,S_t)=e^{-r(T-t)}\mathbb{E}_{t}^{\mathbb{Q}}\left[\,\max\{S_T-K\},0\,\right]$$ it is obvious $$\max\{S_T-K,0\}=(S_T-K)\mathbb{1}_{\{S_T>K\}}$$ then $$V(t,S_t)=e^{-r(T-t)}\mathbb{E}_{t}^{\mathbb{Q}}\left[S_T\mathbb{1}_{\{S_T>K\}}\right]-e^{-r(T-t)}\mathbb{E}_{t}^{\mathbb{Q}}\left[K\mathbb{1}_{\{S_T>K\}}\right]$$ as a result $$V(t,S_t)=X_t\,\mathbb{E}_{t}^{\mathbb{Q}}\left[\frac{S_T/S_t}{B_T/B_t}\mathbb{1}_{\{X_T>K\}}\right]-Ke^{-r(T-t)}\mathbb{E}_{t}^{\mathbb{Q}}\left[\mathbb{1}_{\{S_T>K\}}\right]$$ in other words $$V(t,S_t)=S_t\mathbb{E}_{t}^{\mathbb{Q}^S}\left[\mathbb{1}_{\{S_T>K\}}\right]-Ke^{-r(T-t)}\mathbb{E}_{t}^{\mathbb{Q}}\left[\mathbb{1}_{\{S_T>K\}}\right]$$ so $$\color{red}{V(t,S_t)=S_t\mathbb{Q}^S(S_T>K)-Ke^{-r(T-t)}\mathbb{Q}(S_T>K)}$$ Finally $$V(t,S_t)=S_tN(d_1)-Ke^{-r(T-t)}N(d_2)$$

Answer 2

I think you might confuse two things here.

In the Black-Scholes formula, the term

\begin{equation} \Phi \left( d_2 \right) = \mathbb{Q} \left( \left. S_T > K \right| \mathfrak{F}_t \right) \end{equation}

is the conditionally probability of ending up in-the-money under the risk-neutral probability measure $\mathbb{Q}$. Similarly,

\begin{equation} \Phi \left( d_1 \right) = \mathbb{S} \left( \left. S_T > K \right| \mathfrak{F}_t \right) \end{equation}

is the conditionally probability of ending up in-the-money under an auxiliary measure where the underlying asset is used as the numeraire. Neither of them is the real-world probability of ending up in-the-money.

The way I understand your question is that you now seem to assume that the evaluation of the corresponding normal distribution function depends on some measure. This is not the case. They are just functions and your lookup tables for them are independent of the respective probabilistic interpretation.