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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

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  • $\begingroup$ It is indeed the real world measure $\mathbb{P}$ $\endgroup$ – Jan Sila Oct 18 '16 at 13:31
  • $\begingroup$ You did not say what $d_1$ and $d_2$ are and that is very important in understanding what the measure used in the BSM formula actually is (real or risk neutral). What is used in $d_1,d_2$, $r$ or $\mu$? $\endgroup$ – noob2 Oct 18 '16 at 13:37
  • $\begingroup$ Added specifications for the Black-Scholes framework. $\endgroup$ – none Oct 18 '16 at 14:43
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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 $$V(t,S_t)=S_tN(d_1)-Ke^{-r(T-t)}N(d_2)$$

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    $\begingroup$ This is it. Very informative. $\endgroup$ – none Oct 19 '16 at 10:43
  • $\begingroup$ I believe the Radon-Nikodym derivative should be $\frac{dQ^S}{dQ}=\frac{\frac{S_T}{S_t}}{\frac{B_T}{B_t}}$. $\endgroup$ – none Oct 24 '16 at 7:20
  • $\begingroup$ Yes it was typo. $\endgroup$ – user16651 Oct 24 '16 at 19:30
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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.

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  • $\begingroup$ Interesting ideas. This is what I'm thinking, by transforming the variable $S_T$ to the standard normal variable $Z$, I am changing numeraires, whilst preserving the condition that $S_T>k$. Then the standard normal distribution function simply provides a transformation to a number between 0 and 1, and this function works regardless of the probability measure under which the numeraire is defined. $\endgroup$ – none Oct 18 '16 at 14:54

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