Pricing a Digital Call Under Linear Gaussian Price Dynamics
Summary
The document corrects a digital-call valuation under the stated risk-neutral dynamics, where the price has drift proportional to its current level and constant diffusion. The original calculation omitted the conditional mean of the terminal price, leading to a probability that did not depend on the current underlying price and failed to approach the option payoff at expiry. Solving the stochastic differential equation gives a Gaussian terminal price whose mean is the current price grown at the drift rate and whose variance comes from the integrated diffusion term.
The call value is the discounted risk-neutral probability that the terminal price exceeds the strike. For a Gaussian terminal distribution, this probability is expressed through the standard normal cumulative distribution function using the mean-minus-strike divided by the terminal standard deviation. As expiry approaches, the standardized distance moves toward positive or negative infinity according to whether the underlying is above or below the strike, recovering the digital payoff away from the strike. The result depends on the document's specific dynamics and payoff assumptions.
Key ideas
- The terminal price under the stated linear Gaussian dynamics has a nonzero conditional mean based on the current price.
- The terminal variance is obtained by integrating the squared diffusion coefficient over the remaining time.
- A digital call is valued as the discounted risk-neutral probability that the terminal price exceeds the strike.
- The expiry limit recovers the indicator payoff when the terminal price is not exactly at the strike.
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# Digital call under Ornstein-Uhlenbeck dynamics
# Digital call under Ornstein-Uhlenbeck dynamics
I am trying to price a digital option with payoff $\mathbb{I}_{S_T>K}$, where $S_t$ follows the Ornstein-Uhlenbeck dynamics $\mathrm{d}S_t=rS_t\mathrm{d}t+\sigma\mathrm{d}W^{\mathbb{Q}}_t$ in the risk-neutral measure $\mathbb{Q}$. I have managed to calculate that $\mathrm{d}(\mathrm{e}^{-rt}S_t)=\sigma\mathrm{e}^{-rt}\mathrm{d}W^{\mathbb{Q}}_t$, so the conditional distribution is
$$\mathrm{e}^{-rT}S_T|\mathrm{e}^{-rt}S_t\sim\mathcal{N}\left(0,\frac{\sigma^2}{2r}(\mathrm{e}^{-2rt}-\mathrm{e}^{-2rT})\right).$$
Therefore, assuming that my calculations make sense, the value of my digital option is
\begin{align*} V(t,S_t) &=\mathrm{e}^{-r(T-t)}\mathbb{E}^{\mathbb{Q}}\mathbb{I}_{S_T>K}\\ &=\mathrm{e}^{-r(T-t)}\mathbb{Q}\left(X_T-Y_T>K|\mathcal{F}_t\right)\\ &=\mathrm{e}^{-r(T-t)}\mathbb{Q}\left(\mathrm{e}^{-rT}(X_T-Y_T)>K\mathrm{e}^{-rT}\Big|\mathcal{F}_t\right)\\ &=\mathrm{e}^{-r(T-t)}\mathbb{Q}\left(Z>\frac{K\mathrm{e}^{-rT}}{\sqrt{\frac{\sigma^2}{2r}(\mathrm{e}^{-2rt}-\mathrm{e}^{-2rT})}}\Big|\mathcal{F}_t\right)\\ &=\mathrm{e}^{-r(T-t)}\Phi\left(\frac{-K\mathrm{e}^{-rT}}{\sqrt{\frac{\sigma^2}{2r}(\mathrm{e}^{-2rt}-\mathrm{e}^{-2rT})}}\right). \end{align*}
However, in the limit $t\to T$, I don't seem to get $V(t,S_t)\to\mathbb{I}_{S_T>K}$ a.s., where have I gone wrong?
## Answer by foreignvol (score 3, accepted)
https://quant.stackexchange.com/a/60175
The problem seems to be that you forgot the mean of the process.
If $ds_t = rs_tdt + \sigma dW_t^\mathbb Q$, then the solution of the SDE is given by $$s_T = s_te^{r(T-t)} + \sigma\int_t^Te^{r(T-u)}dW^\mathbb Q_u.$$ Since the last integral is Gaussian, the distribution of the terminal price is given by $$s_T \sim\mathrm N\left(s_te^{r(T-t)}, \frac{\sigma^2}{2r}\left[e^{2r(T-t)}-1\right]\right).$$
Now, for the digital option, this translates into $$ V(t,s_t) = e^{-r(T-t)}\mathbb Q\left[s_T>k\right] = e^{-r(T-t)}\Phi\left[\frac{s_te^{r(T-t)} - k}{\sqrt{\frac{\sigma^2}{2r}\left[e^{2r(T-t)}-1\right]}}\right], $$ where I used that $\Phi(x) = 1 - \Phi(-x)$.
In particular, when $t\to T$, the argument of $\Phi$ will diverge to $\pm\infty$ depending on the sign of $s_t - k$, which means that $V(t,s_t)\to 1_{\left\{s_t>k\right\}}$.Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)
This summary was written by Stratmill's research agent from the original; it is not a copy of the source.