# Formula for the discounted payoff of a digital option

In "Heard on the Street" it states that the expected discounted payoff of a digital option is $$H\exp^{-r(T-t)}N(d_2)$$

where $$H$$ is the payoff of the option, the exponential is the discounting.

Why do we have the $$N(d_2)$$, what does it represent and why is it there?

The digital option pays $$H$$ at time $$T$$ if $$S_T \geq K$$ , so its option time at time $$t$$ is given by

$$V_t=E_t\left[e^{-r(T-t)}H 1_{\{S_T \geq K\}}\right]=e^{-r(T-t)}H* P_t(S_T \geq K)$$

The model used is Black-model, that $$dS_t=rS_tdt+\sigma dW_t$$

or $$S_T=S_te^{\left(r-\frac12 \sigma^2\right)(T-t)+\sigma (W_T-W_t)}{}$$

Calculate $$P_t(S_T \geq K)$$

$$P_t(S_T \geq K)=P_t(S_te^{\left(r-\frac12 \sigma^2\right)(T-t)+\sigma (W_T-W_t)}{} \geq K)=P_t(W_T-W_t \geq\frac{log\frac{K}{S_0}-\left(r-\frac12 \sigma^2\right)(T-t)}{\sigma})$$

$$W_T-W_t |W_t$$ is centered and normally distributed with variance $$T-t$$
$$P_t(W_T-W_t \geq\frac{log\frac{K}{S_0}-\left(r-\frac12 \sigma^2\right)(T-t)}{\sigma})=P(Y \geq\frac{log\frac{K}{S_t}-\left(r-\frac12 \sigma^2\right)(T-t)}{\sigma \sqrt{T-t}})$$

where $$Y \sim \mathcal{N}(0,1)$$

Using the symmetry of the normal distribution,

$$P(Y \geq\frac{log\frac{K}{S_t}-\left(r-\frac12 \sigma^2\right)(T-t)}{\sigma \sqrt{T-t}})=P(Y \leq -\frac{log\frac{K}{S_t}-\left(r-\frac12 \sigma^2\right)(T-t)}{\sigma \sqrt{T-t}})$$

Define $$d_2=-\frac{log\frac{K}{S_t}-\left(r-\frac12 \sigma^2\right)(T-t)}{\sigma \sqrt{T-t}}=\frac{log\frac{S_t}{K}+\left(r-\frac12 \sigma^2\right)(T-t)}{\sigma \sqrt{T-t}}$$

$$P(Y \leq -\frac{log\frac{K}{S_t}-\left(r-\frac12 \sigma^2\right)(T-t)}{\sigma \sqrt{T-t}})=P(Y \leq d_2)=N(d_2)$$

where $$N$$ is the cdf of a standard normal variable.

Finally,

$$V_t=E_t\left[e^{-r(T-t)}H 1_{\{S_T \geq K\}}\right]=e^{-r(T-t)}H*N(d_2)$$

$$N\left(d_2\right)$$ is the risk-neutral probability that the spot is greater than the strike at maturity, therefore the RN probability that you get your payoff.

• Why is it the risk netural probability? this is what i dont get – Trajan Dec 30 '19 at 18:01

Recall that the price of your contract is \begin{align*} V_t = e^{-r(T-t)} \mathbb{E}^\mathbb{Q} [H1_{\{S_T>K\}}|\mathcal{F}_t] \end{align*} because your option always pays $$H$$ if $$S_T>K$$. Next, \begin{align*} V_t &=He^{-r(T-t)} \mathbb{E}^\mathbb{Q} [1_{\{S_T>K\}}|\mathcal{F}_t] \\ &= He^{-r(T-t)} \mathbb{Q} [{\{S_T>K\}}|\mathcal{F}_t] \\ &= He^{-r(T-t)} N(d_2) \end{align*} The fact that the option price equals the discounted (conditional) expectation of the payoff is linked with no-arbitrage via the fundamental theorem of asset pricing.

You can compute the probability of $$\{S_T>K\}$$ under the Black-Scholes model to obtain $$N(d_2)$$.

• How does $\mathbb{Q} [{\{S_T>K\}}|\mathcal{F}_t] =N(d_2) ??$ – Trajan Dec 30 '19 at 18:28