Mdthbs

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)

There are several derivations and interpretations for $N(d_1)$ and $N(d_2)$. As you know,
begin{align*}
C(t,S_t)=S_te^{-q(T-t)}N(d_1) -Ke^{-r(T-t)}N(d_2).
end{align*}
We can also show that
begin{align*}
mathbb{Q}_S[{S_Tgeq K}]&=e^{-q(T-t)}N(d_1), \
mathbb{Q}[{S_Tgeq K}] &=e^{-r(T-t)}N(d_2).
end{align*}
Thus, $N(d_i)$ 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.

Alternatively, but a similar idea, we can show that
begin{align*}
Delta = frac{partial C(t,S_t)}{partial S_t} =e^{-q(T-t)}N(d_1), \
kappa = frac{partial C(t,S_t)}{partial K} =e^{-r(T-t)}N(d_2).
end{align*}
and if you recall the idea of a $Delta$ hedge, this interpretation of $N(d_1)$ 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.

You can see $N(d_1)$ and $N(d_2)$ also as prices of binary options ($S_te^{-q(T-t)}N(d_1)$ refers to the price of an asset-or-nothing call and $e^{-r(T-t)}N(d_2)$ to the price of a cash-or-nothing call).

The derivation follows standard arguments, i.e.
begin{align*}
C(t,S_t) &= e^{-r(T-t)}mathbb{E}^mathbb{Q}[max{S_T-K,0}]\
&= e^{-r(T-t)}mathbb{E}^mathbb{Q}[(S_T-K)mathbb{1}_{{S_Tgeq K}}]\
&= e^{-r(T-t)}left(mathbb{E}^mathbb{Q}[S_Tmathbb{1}_{{S_Tgeq K}}] - Kmathbb{E}^mathbb{Q}[mathbb{1}_{{S_Tgeq K}}]right).
end{align*}
From here, you can immediately see the decomposition into exercise probabilities and binary options.

In order to compute the probabilities, simply note that (as an example)
begin{align*}
mathbb{E}^mathbb{Q}[mathbb{1}_{{S_Tgeq K}}] &= mathbb{Q}[{S_Tgeq K}] \
&= mathbb{Q}[{ln(S_T)geq ln(K)}].
end{align*}
Since $ln(S_T)sim Nleft(ln(S_0)+left(r-q-frac{1}{2}sigma^2right)T,sigma^2 Tright)$, you have for $Zsim N(0,1)$,
begin{align*}
mathbb{Q}[{ln(S_T)geq ln(K)}] &= mathbb{Q}[{ln(S_0)+left(r-q-frac{1}{2}sigma^2right)T+sigma^2 T Zgeq ln(K)}] \
&= mathbb{Q}left[left{Zgeq frac{ln(K)-ln(S_0)-left(r-q-frac{1}{2}sigma^2right)T}{sigma^2T}right}right] \
&= mathbb{Q}left[left{Zgeq -frac{lnleft(frac{S_0}{K}right)+left(r-q-frac{1}{2}sigma^2right)T}{sigma^2T}right}right] \
&= 1-mathbb{Q}left[left{Zleq-frac{lnleft(frac{S_0}{K}right)+left(r-q-frac{1}{2}sigma^2right)T}{sigma^2T}right}right] \
&= 1-Phileft(-frac{lnleft(frac{S_0}{K}right)+left(r-q-frac{1}{2}sigma^2right)T}{sigma^2T}right) \
&= Phileft(frac{lnleft(frac{S_0}{K}right)+left(r-q-frac{1}{2}sigma^2right)T}{sigma^2T}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...