Jumping into Black-Scholes III: Calculating the Expected Value

We take off from where Jumping into Black-Scholes II: Know your contract and Asset left off,
in that we have established a probability distribution for our future asset price \(S(T)\). With this, let's take a look at the expected value of our options payoff. Let \(p(Z)\) be the pdf (probability density function) of \(Z\). Thus, using our prob-know-how, we have
\begin{eqnarray}
\mathbf{E}\left[  \max\left\{K-S_T,0\right\}\right] &= &
\mathbf{E}\left[  \max\left\{K-S_0 e^{(r-\sigma^2/2)T+\sigma\sqrt{T} Z},0\right\} \right] \nonumber \\
 &= &  \int_{-\infty}^{\infty} \max\left\{K-S_0 e^{(r-\sigma^2/2)T+\sigma\sqrt{T} Z},0\right\} p(Z) dZ \nonumber \\
&= & \int^{-d_2}_{- \infty} \left(K-S_0 e^{(r-\sigma^2/2)T+\sigma\sqrt{T} Z}\right) p(Z) dZ. \label{eq:BS1}
\end{eqnarray}

In this last step suddenly a \(d_2\) appeared. This constant \(-d_2\) has to be chosen so that for \(Z \leq -d_2\)  implies
\begin{equation} \label{findd2} \left(K-S_0 e^{(r-\sigma^2/2)T+\sqrt{T} Z}\right) \geq 0, \end{equation}
 so we could remove that infernal \(\max(\cdot)\) function. Well that's easy enough to calculate, by equating the above~(\ref{findd2}) to zero, our variable \(Z\) would be forced to equal \[d_2 = \frac{\ln(S_0/K) +(r-\sigma^2/2)T}{\sigma \sqrt{T}}.\]
I chose to call it \(d_2\) because that's what everyone else calls it. Why, out of the universe of greek letters and symbols ``they'' chose \(d_2\)? I haven't the foggiest idea, but to not be a notation anarchist, let's stick with the same symbol.


Turning back to~(\ref{eq:BS1}). Remembering that integration is a linear function, in that you can break up sums, throw constants on the "outside" of the integral so
\begin{eqnarray}
\int^{-d_2}_{- \infty} \left(K-S_0 e^{(r-\sigma^2/2)T+\sqrt{T} Z}\right) p(Z) dZ &= &
K\int^{-d_2}_{- \infty} p(Z)dZ -S_0e^{(r-\sigma^2/2)T} \int^{-d_2}_{- \infty} e^{\sigma\sqrt{T} Z} p(Z) dZ \nonumber \\
&= & \underbrace{K \Phi(-d_2)}_I -S_0e^{(r-\sigma^2/2)T} \underbrace{\int^{-d_2}_{- \infty} e^{\sigma\sqrt{T} Z} p(Z) dZ}_{II} \nonumber  \label{eq:BS2} .
\end{eqnarray}
In the I part, \(\Phi\) is the cumulative distribution function of the normal distribution. The last thing we need to deal with is the integral denoted by II. Scratching your noodle and/or looking at wiki, we see that the pdf of a standard normal distribution \(p(Z) = e^{-Z^2/2}\). With this we can solve the integral II, check out the following steps on how to do this
\begin{eqnarray*}
\int^{-d_2}_{- \infty} e^{\sigma\sqrt{T} Z} p(Z) dZ &= & \int^{d_2}_{- \infty} e^{\sigma\sqrt{T} Z} e^{-Z^2/2} dZ\\
&= & \int^{-d_2}_{- \infty} e^{-(Z -\sigma\sqrt{T})^2/2 +\sigma^2 T/2} dZ \quad [\mbox{Completing the square}] \\
&= & e^{\sigma^2 T/2} \int^{-d_2+\sigma\sqrt{T}}_{-\infty} e^{-y^2/2} dy  \quad [\mbox{Changing variables } y = Z-\sigma\sqrt{T}] \\
&= & e^{\sigma^2 T/2} \Phi\left(-d_2+\sigma\sqrt{T}\right) \quad [\mbox{Boom! We're done.}]
\end{eqnarray*}

Returning to~(\ref{eq:BS1}), we have reached the grand finale. The price of our put option is the expected value of its payoff, discounted in time by multiplying \(e^{-rT}\),
\[e^{-rT}\mathbf{E}\left[ \mbox{payoff}\right] =  K e^{-rT} \Phi(-d_2)+S_0 \Phi(-d_2+\sigma\sqrt{T}).\]
All these calculations may have hurt. But now you can price all sorts of options using a completely analogous argument. Naturally, if you work or are looking to work as a Quant, no one will ask you deduce the price of a Plain Vanilla option. But what happens when some Trader/Salesman/(Front desk Dude) decides, in a frenzy, to sell an option to a client that has a payoff that depends on two strikes \(K_1\) and \(K_2\) with
\[
\mbox{payoff} =
\begin{cases}
K_1 & \mbox{if }S_T \leq K_1,\\
S_T & \mbox{ if } K_1 \leq S_T \leq K_2, \\
0 & \mbox{ if } K_2 \leq S_T.
\end{cases}
\]
Well how much does that cost? Go on, aren't you the Quant, didn't we hire you for this? You can either blankly stare at your boss, jaw open as a sparkle of drool develops in the corner of your mouth, or you can sit down and confidently calculate the expectancy of this payoff. Get to it and post me your solution.

This approach works in general for options whose payoff depends on one time frame \(T\), known as European options, e.g, Basket Options, Binary Options, Quanto Options...etc. This approach can also  be adapted  when there are a handful of time frames. Things are not so simple when the payoff is time dependent, such as American Options. Then you really need to take the dive into stochastic calculus.