# Understanding the solution of this integral

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?

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. 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. – Quantuple Apr 29 '16 at 15:14
• @Quantuple: Thanks for your comments. That technique was discussed at quant.stackexchange.com/questions/19038/…. – Gordon Apr 29 '16 at 15:40
• damn I wish I had read this before writing my answer :( – Quantuple Apr 29 '16 at 16:08
• Can you please explain how you come to the step from $\int_{z^*}^{\infty}\frac{1}{\sqrt{2\pi}} S_t e^{(\mu - \frac{1}{2}\sigma^2) \tau -\frac{1}{2}(z-\sigma \sqrt{\tau})^2 +\frac{1}{2}\sigma^2 \tau }dz =S_t e^{u\tau} \int_{z^* - \sigma \sqrt{\tau}}^{\infty} \frac{1}{\sqrt{2\pi}}e^{-\frac{x^2}{2}}dx$? – emcor Apr 30 '16 at 1:24
• You combine the constant terms, you obtain the term $\mu\tau$. Then you make the substitution $x=z-\sigma\sqrt {\tau}$. – Gordon Apr 30 '16 at 2:38

Another take on the question which uses stochastic calculus

[Digression]

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}

[Calculation]

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 $d_1$.

• With the abstract Bayes lemma, you can also compute the conditional expectation. – Gordon Apr 29 '16 at 18:54