Understanding $N(d_1)$ and $N(d_2)$

Question

Firstly, if the solution to geometric Brownian motion is $S_t = S_0 \exp((r-\sigma^2)t + \sigma W_t$ then if I have a payment that is not necessarily a full call option e.g. if the exercise price $K$ is crossed then a payment of 5000 shares at the time is released if the stock price is greater than its initial value by 20% after 6 months, how does that arrive at $N(d_1)$ or $N(d_2)$? The connection I'm making here is that $d_1 = \dfrac{\ln(S_t/K) + (r - \sigma^2)\sqrt{t}}{\sigma \sqrt{t}}$ and so if I was to solve it as per a method I had been taught it would go as follows $V_0 = \exp(-rt)E[X|\mathscr{F}_t]$ where X is the payment specified.

Then: \begin{align*} V_0 &= \exp(-rt)\cdot E[10,000\cdot\mathbb{I}{(S_{0.5} > 1.2S_0)}|\mathscr{F}_t]\\ &=10,000\cdot\exp(-rt)\cdot P\left((r-\dfrac{1}{2}\sigma^2)\cdot t+ \sigma W_t > \ln(1.2)\right)\\ &= 10,000\cdot\exp(-rt)\cdot P\left(W_t > \dfrac{\ln(1.2) - (r-\dfrac{1}{2}\sigma^2)t}{\sigma}\right). \end{align*} This is not similar to $N(d_1)$.

Is the initial formula $S_t = S_0\cdot\exp((r-\sigma^2)t + \sigma W_{t}\sqrt{t} )$ rather?

Answer 1

The Black Scholes (1973) model assumes that $\mathrm{d}S_t=rS_t\mathrm{d}t+\sigma S_t\mathrm{d}W_t$. Thus, $$S_t=S_0\exp\left(\left(r-\frac{1}{2}\sigma^2\right)t+\sigma W_t\right).$$ Please note the factor $-\frac{1}{2}\sigma^2t$ in the exponential. If you incorporate dividends, replace $r$ by $r-q$. You do not need an extra term $\sqrt{t}$ in front of the Brownian motion since, by definition, $W_t\sim N(0,t)$.

You can then indeed use risk-neutral pricing which expresses the time $t$ price of a claim $X$ as $$\mathbb{E}^\mathbb{Q}\left[\frac{B_t}{B_T} X\bigg|\mathcal{F}_t\right].$$ Here, $(B_t)$ is a locally risk-free bank account governed by $\mathrm{d}B_t=r_tB_t\mathrm{d}t$. In the Black-Scholes world, $r_t\equiv r$ such that $\frac{B_t}{B_T}=e^{-r(T-t)}$.

You describe a European-style option which pays one unit of the stock if the terminal stock price exceeds the strike price. These are known as asset-or-nothing calls. Their price simply equals $$S_te^{-q(T-t)}N(d_1)$$ with the standard $$ d_1=\frac{\ln\left(\frac{S_0}{K}\right)+\left(r-q-\frac{1}{2}\sigma^2\right)(T-t)}{\sigma\sqrt{T-t}}.$$ All what you need to do in your special application is to set $K= 1.2S_0$ and $T=\frac{1}{2}$ and multiply the option price by 5,000.

Answer 2

In its simplest form, an option is a combination of two binary options.

The buyer of a call option is long of an "asset-or-nothing" binary call. I.E. if Spot>Strike, it is worth Spot; else 0. To fund that, he is selling a "cash-or-nothing" call: worth Strike if Spot>Strike, else 0.

The positive value of the option obviously derives from the fact that as Spot moves beyond Strike, Spot moves but Strike does not.

By definition, the two components have to have the same probability of being worth zero or something, because they share the same Strike. The value of of the cash-or-nothing is the present value of Strike * probability of this-versus-0. This is N(d2): the probability of the options having value.

N(d1) is the change in the value of the asset-or-nothing per change in Spot. This is higher because Spot as well as probability of Spot>Strike both move in tandem. N(d1) captures the likelihood of a bigger payoff should the probability of any payoff rise.

Your formulation looks like N(d2) to me; albeit if you are exercising on shares rather than cash, the exchange is not your 20% strike but the value of the shares being exchanged, if above . So should be N(d1). Try E(10,000 * St, P(St > 1.2 * S0))!