Risk-Neutral Pricing of a Cash-or-Nothing Option in the Bachelier Model

# Valuation of Cash-Or-Nothing option


# Valuation of Cash-Or-Nothing option












Studying options pricing, I'm stuck with the following problem:

> The price of a stock is described by the dynamic: $$dS_t = \mu\, dt + \sigma\,dW_t$$ Compute the fair price of a Cash or Nothing Option with pay-off function $V(S_T) = \mathbb{1}_{S_T<K}$. Hint: Replace $\mu$ such that, the discounted price at maturity $S(T)$ under the risk-free measure is a martingale.

It means that the option can just be exercised at maturity $T$ and has value $1$ when at maturity the underlying price is below the strike.

My thoughts: Use a discretization process like Euler-Maruyama and then compute recursively the value of $S(T)$. Then using the pay-off function, approximate it with a Monte-Carlo simulation.

However, I don't know how to use this Hint. My professor said it could be really useful but I do not know how to use it. Any help with this problem would be really meaningful.

Many thanks.

## Answer by Kevin (score 4, accepted)

https://quant.stackexchange.com/a/46629

You can use such an approximation but there are known analytical prices. You have a special case in which the stock price is normally distributed. See Bachelier Model.

Set $\mu=r-q$ (if you have dividends, or simply $\mu=r$ if there are no dividends). So if you change from the real worl probability measure $\mathbb{P}$ to the risk-neutral measure $\mathbb{Q}$ you get that $\mathrm{d}S_t=(r-q)\mathrm{d}t+\sigma \mathrm{d}W_t$. Then, using risk-neutral pricing, the inital value of your claim is given by \begin{align*} V_0 &= e^{-rT} \mathbb{E}^\mathbb{Q}[{1}_{\{S_T< K\}}] \\ &= e^{-rT} \mathbb{Q}[\{S_T< K\}]. \end{align*}

Thus, all you need to do is to find the probability distribution of $S_T$ under $\mathbb{Q}$. Using again that $\mathrm{d}S_t=(r-q)\mathrm{d}t+\sigma \mathrm{d}W_t$, we see that $(S_t)$ is an arithmetric Brownian motion under $\mathbb{Q}$ and thus normally distributed. Furthermore, \begin{align*} S_T= S_0+(r-q)T + \sigma W_T \sim N\big(S_0+(r-q)T,\sigma^2T\big), \end{align*} since $W_T\sim N(0,T)$. Now, set $m=S_0+(r-q)T$ and $s=\sigma\sqrt{T}$. Then, $S_T=m+sZ$ where $Z\sim N(0,1)$. Thus, \begin{align*} V_0 &= e^{-rT} \mathbb{Q}[\{m+sZ< K\}]\\ &= e^{-rT} \mathbb{Q}\left[\left\{Z< \frac{K-m}{s}\right\}\right]\\ &= e^{-rT} \Phi\left(\frac{K-m}{s}\right)\\ &= e^{-rT} \Phi\left(-\frac{S_0-K+(r-q)T}{\sigma\sqrt{T}}\right) \\ &= e^{-rT} \left(1- \Phi\left(\frac{S_0-K+(r-q)T}{\sigma\sqrt{T}}\right)\right), \end{align*}

where $\Phi$ denotes the cumulative distribution function of a standard normal distribution.

Let me highlight that, of course, you can price such a claim with Euler Maruyama. You can also employ finite differences or Fourier transforms. You could even build a (binomial) tree. But if there is a simple analytical answer available, it is to be preferred.

By the way, in the Black-Scholes model, the price of a Cash-Or-Nothing option is given by $e^{-rT}\Phi(-d_2)=e^{-rT}\big(1-\Phi(d_2)\big)$, see here.