The following integral represents an expected value of a geometric brownian motion for $S_T>K$ (i.e. part of the Black-Scholes call option price): $$\int_{z^*} (S_te^{\mu\tau-\frac{1}{2}\sigma^2\tau+\sigma\sqrt{\tau}z})\frac{1}{\sqrt{2\pi}}e^{-\frac{z^2}{2}}dz =e^{\mu\tau}S_tN(d_1^*)$$ with $z^*=\frac{\ln\frac{K}{S_t}-(\mu-\sigma^2/2)\tau}{\sigma\sqrt{\tau}}$, $d_1^*=\frac{\ln\frac{S_t}{K}+(\mu+\sigma^2/2)\tau}{\sigma\sqrt{\tau}}$, and $N(\cdot)$ cumulative Standardnormal distribution.

Can you please explain how this equality is obtained?


2 Answers 2


Let $\tau = T-t$. Then \begin{align*} S_T = S_t e^{(\mu - \frac{1}{2}\sigma^2) \tau + \sigma \sqrt{\tau}\, Z}, \end{align*} where $Z$ is a standard normal random variable, independent of $\mathcal{F}_t$. Moreover, \begin{align*} E\left(S_T 1_{\{S_T >K\}}\mid \mathcal{F}_t \right) &= E\left(S_t e^{(\mu - \frac{1}{2}\sigma^2) \tau + \sigma \sqrt{\tau}\, Z}\, 1_{\left\{S_t e^{(\mu - \frac{1}{2}\sigma^2) \tau + \sigma \sqrt{\tau}\, Z} >K\right\}}\mid \mathcal{F}_t \right)\\ &=\int_{-\infty}^{\infty}S_t e^{(\mu - \frac{1}{2}\sigma^2) \tau + \sigma \sqrt{\tau}\, z}\, 1_{\left\{S_t e^{(\mu - \frac{1}{2}\sigma^2) \tau + \sigma \sqrt{\tau}\, z} >K\right\}} \frac{1}{\sqrt{2\pi}}e^{-\frac{z^2}{2}} dz\\ &=\int_{z^*}^{\infty}S_t e^{(\mu - \frac{1}{2}\sigma^2) \tau + \sigma \sqrt{\tau}\, z}\frac{1}{\sqrt{2\pi}}e^{-\frac{z^2}{2}} dz\\ &=\int_{z^*}^{\infty}\frac{1}{\sqrt{2\pi}} S_t e^{(\mu - \frac{1}{2}\sigma^2) \tau + \sigma \sqrt{\tau}\, z - \frac{z^2}{2}}dz\\ &=\int_{z^*}^{\infty}\frac{1}{\sqrt{2\pi}} S_t e^{(\mu - \frac{1}{2}\sigma^2) \tau -\frac{1}{2}(z-\sigma \sqrt{\tau})^2 +\frac{1}{2}\sigma^2 \tau }dz\\ &=S_t e^{u\tau} \int_{z^* - \sigma \sqrt{\tau}}^{\infty} \frac{1}{\sqrt{2\pi}}e^{-\frac{x^2}{2}}dx\\ &=S_t e^{u\tau}N(\sigma \sqrt{\tau}-z^*)\\ &=S_t e^{u\tau}N(d_1^*), \end{align*} where \begin{align*} d_1^* &= \sigma \sqrt{\tau}-z^*\\ &=\frac{\ln\frac{S_t}{K}+(\mu+\sigma^2/2)\tau}{\sigma\sqrt{\tau}}. \end{align*}

  • 1
    $\begingroup$ +1. You can also perform a change of measure to express the above expectation under an EMM using the stock price as numeraire, in which case the calculations are easier. $\endgroup$
    – Quantuple
    Commented Apr 29, 2016 at 15:14
  • $\begingroup$ @Quantuple: Thanks for your comments. That technique was discussed at quant.stackexchange.com/questions/19038/…. $\endgroup$
    – Gordon
    Commented Apr 29, 2016 at 15:40
  • $\begingroup$ damn I wish I had read this before writing my answer :( $\endgroup$
    – Quantuple
    Commented Apr 29, 2016 at 16:08
  • $\begingroup$ Can you please explain how you come to the step from $\int_{z^*}^{\infty}\frac{1}{\sqrt{2\pi}} S_t e^{(\mu - \frac{1}{2}\sigma^2) \tau -\frac{1}{2}(z-\sigma \sqrt{\tau})^2 +\frac{1}{2}\sigma^2 \tau }dz =S_t e^{u\tau} \int_{z^* - \sigma \sqrt{\tau}}^{\infty} \frac{1}{\sqrt{2\pi}}e^{-\frac{x^2}{2}}dx$? $\endgroup$
    – emcor
    Commented Apr 30, 2016 at 1:24
  • 1
    $\begingroup$ You combine the constant terms, you obtain the term $\mu\tau $. Then you make the substitution $x=z-\sigma\sqrt {\tau} $. $\endgroup$
    – Gordon
    Commented Apr 30, 2016 at 2:38

Another take on the question which uses stochastic calculus


Assume deterministic and constant rates without loss of generality. Also assume the absence of arbitrage opportunities and market completeness

Let $B_t$ denote the time-$t$ value of a risk-free money market account in which 1 unit of currency $C$ has been invested at $t=0$: \begin{align} & dB_t = rB_t dt,\ \ B(0)=1 \\ \iff& B_t = e^{rt} \end{align}

Under the risk-neutral measure $\mathbb{Q}_B$ associated to the numéraire $B_t$, for any tradable asset $V_t$ $$ \frac{V_t}{B_t} \text{ is a } \mathbb{Q}_B\text{-martingale} $$

Now, assume that the stock price follows a GBM under $\mathbb{Q}_B$ \begin{align} &\frac{dS_t}{S_t} = r dt + \sigma dW_t^{\mathbb{Q}_B},\ \ S(0)=S_0 > 0 \\ \iff& S_T = S_0 e^{(r-\frac{\sigma^2}{2})T + \sigma W_T^{\mathbb{Q}_B}} \end{align}

Define an EMM $\mathbb{Q}_S$ which uses the stock price $S_t$ as numéraire, then for any tradable asset $V_t$ $$ \frac{V_t}{S_t} \text{ is a } \mathbb{Q}_S\text{-martingale} $$

From the 2 EMM definitions we simultaneously have $$ \frac{V_0}{B_0} = E^{\mathbb{Q}_B}\left[\frac{V_T}{B_T} \vert \mathcal{F}_0\right] $$ $$ \frac{V_0}{S_0} = E^{\mathbb{Q}_S}\left[\frac{V_T}{S_T} \vert \mathcal{F}_0\right] $$ re-arranging we see that $$ E^{\mathbb{Q}_B}\left[\frac{V_T B_0}{B_T} \vert \mathcal{F}_0\right] = E^{\mathbb{Q}_S}\left[\frac{V_T S_0}{S_T} \vert \mathcal{F}_0\right] (\ = V_0) $$ in other words, the Radon-Nikodym derivative of the change of measure writes: $$ \left. \frac{d\mathbb{Q}_S}{d\mathbb{Q}_B} \right\vert \mathcal{F}_t = \frac{S_T B_0}{S_0 B_T} = e^{-\frac{\sigma^2}{2}t + \sigma W_t^{\mathbb{Q}_B}} $$ which is a Doléans-Dade exponential $$ \left. \frac{d\mathbb{Q}_S}{d\mathbb{Q}_B} \right\vert \mathcal{F}_t = \mathcal{E}(\sigma W_t^{\mathbb{Q}_B}) $$

Using Girsanov theorem we can write that \begin{align} W_{t}^{\mathbb{Q}_S} &= W_{t}^{\mathbb{Q}_B} - \langle W_{t}^{\mathbb{Q}_B}, \sigma W_t^{\mathbb{Q}_B} \rangle_t \\ &= W_{t}^{\mathbb{Q}_B} - \sigma t \end{align}

Thus the dynamics of $S_t$ under $\mathbb{Q}^S$ reads \begin{align} &\frac{dS_t}{S_t} = (r + \sigma^2) dt + \sigma dW_t^{\mathbb{Q}_S},\ \ S(0)=S_0 > 0 \end{align}


Now, the expectation:

$$ e^{-rT} E^{\mathbb{Q}_B} \left( S_T 1_{\{S_T >K\}}\mid \mathcal{F}_0 \right) = E^{\mathbb{Q}_B} \left( \frac{S_T 1_{\{S_T >K\}} B_0}{B_T} \mid \mathcal{F}_0 \right) $$

can be written (change of measure)

$$ S_0 E^{\mathbb{Q}_S}\left(1_{\{S_T >K\}}\mid \mathcal{F}_0 \right) $$

and this expectation is easy to compute since we have shown that $S_T$ remains lognormal under $\mathbb{Q}_S$.

In fact, it is exactly the same derivation as for $E^{\mathbb{Q}_B}(1_{\{S_T >K\}}) = \mathbb{Q}_B(S_T > K) = N(d_2)$, where one just has to replace $r$ by $r+\sigma^2$, hence the definitin of $d_1$.

  • 1
    $\begingroup$ With the abstract Bayes lemma, you can also compute the conditional expectation. $\endgroup$
    – Gordon
    Commented Apr 29, 2016 at 18:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.