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Where can I find a good explanation (perhaps with a brief derivation) of N(d1) and N(d2) from Black-Scholes? Just trying to understand the general idea about these 2 probability functions and how they work...

(Thinking they probably work by trying to predict the probability of the cost or profit part of the BS equation by estimating the probability of being at a particular point of a normal distribution but I don't know how that is being achieved)

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There are several interpretations for $\Phi(d_1)$ and $\Phi(d_2)$. As you know, \begin{align*} C(t,S_t)=S_te^{-q(T-t)}\Phi(d_1) -Ke^{-r(T-t)}\Phi(d_2). \end{align*}

Exercise Probabilities

We can show that \begin{align*} \mathbb{Q}_S[\{S_T\geq K\}]&=e^{-q(T-t)}\Phi(d_1), \\ \mathbb{Q}[\{S_T\geq K\}] &=e^{-r(T-t)}\Phi(d_2). \end{align*} Thus, $\Phi(d_i)$ may be seen as probabilities of the option being in the money at maturity $T$. Here, $\mathbb{Q}$ is the equivalent martingale measure using a risk-free bank account as numeraire and $\mathbb{Q}_S$ uses the stock as numeraire. As you hedge the call option with trading into the stock and a bond, it is intuitive to have these exercise probabilities here.

Hedging Statistics

Alternatively, \begin{align*} \Delta = \frac{\partial C(t,S_t)}{\partial S_t} =e^{-q(T-t)}\Phi(d_1), \\ \kappa = \frac{\partial C(t,S_t)}{\partial K} =e^{-r(T-t)}\Phi(d_2). \end{align*} If you recall the idea of a dynamic $\Delta$ hedge, this interpretation of $\Phi(d_1)$ tells you how much you need to invest in the stock in order to hedge the call. In this sense, $\kappa$ tellls you the cost of such a hedge.

Price of Binary (Digital) Options

You can see $\Phi(d_1)$ and $\Phi(d_2)$ also as prices of binary options

  • $S_te^{-q(T-t)}\Phi(d_1)$ refers to the price of a European-style asset-or-nothing call option,
  • $e^{-r(T-t)}\Phi(d_2)$ to the price of a European-style cash-or-nothing call option.

Derivation

By risk-neutral pricing, \begin{align*} C(t,S_t) &= e^{-r(T-t)}\mathbb{E}^\mathbb{Q}[\max\{S_T-K,0\}\mid\mathcal{F}_t]\\ &= e^{-r(T-t)}\mathbb{E}^\mathbb{Q}[(S_T-K)\mathbb{1}_{\{S_T\geq K\}}\mid\mathcal{F}_t]\\ &= e^{-r(T-t)}\left(\mathbb{E}^\mathbb{Q}[S_T\mathbb{1}_{\{S_T\geq K\}}\mid\mathcal{F}_t] - K\mathbb{E}^\mathbb{Q}[\mathbb{1}_{\{S_T\geq K\}}\mid\mathcal{F}_t]\right). \end{align*} From here, you can immediately see the decomposition into exercise probabilities and binary options.

The first expectation is typically solved by a change of numeraire. In order to compute the second probability, note that \begin{align*} \mathbb{E}^\mathbb{Q}[\mathbb{1}_{\{S_T\geq K\}}\mid\mathcal{F}_t] &= \mathbb{Q}[\{S_T\geq K\}\mid\mathcal{F}_t] \\ &= \mathbb{Q}[\{\ln(S_T)\geq \ln(K)\}\mid\mathcal{F}_t]. \end{align*} Since $\ln(S_T)\mid\mathcal{F}_t\sim N\left(\ln(S_t)+\left(r-q-\frac{1}{2}\sigma^2\right)(T-t),\sigma^2 (T-t)\right)$, you have for $Z\sim N(0,1)$, \begin{align*} \mathbb{Q}[\{\ln(S_T)\geq \ln(K)\}] &= \mathbb{Q}\left[\left\{\ln(S_t)+\left(r-q-\frac{1}{2}\sigma^2\right)(T-t)+\sigma \sqrt{T-t} Z\geq \ln(K)\right\}\right] \\ &= \mathbb{Q}\left[\left\{Z\geq \frac{\ln(K)-\ln(S_t)-\left(r-q-\frac{1}{2}\sigma^2\right)(T-t)}{\sigma \sqrt{T-t}}\right\}\right] \\ &= \mathbb{Q}\left[\left\{Z\geq -\frac{\ln\left(\frac{S_t}{K}\right)+\left(r-q-\frac{1}{2}\sigma^2\right)(T-t)}{\sigma \sqrt{T-t}}\right\}\right] \\ &= 1-\mathbb{Q}\left[\left\{Z\leq-\frac{\ln\left(\frac{S_t}{K}\right)+\left(r-q-\frac{1}{2}\sigma^2\right)(T-t)}{\sigma \sqrt{T-t}}\right\}\right] \\ &= 1-\Phi\left(-\frac{\ln\left(\frac{S_t}{K}\right)+\left(r-q-\frac{1}{2}\sigma^2\right)(T-t)}{\sigma \sqrt{T-t}}\right) \\ &= \Phi\left(\frac{\ln\left(\frac{S_t}{K}\right)+\left(r-q-\frac{1}{2}\sigma^2\right)(T-t)}{\sigma \sqrt{T-t}}\right) \\ &= \Phi(d_2). \end{align*}

Of course, you can take simply the log-normal density and compute the expectation as integral. There are many more ways to derive the famous Black-Scholes formula...

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