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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?

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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 $V(t,s_t)\to 1_{\left\{s_t>k\right\}}$.

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  • $\begingroup$ ah!! don’t know how I missed this, thank you! $\endgroup$
    – user107224
    Dec 27 '20 at 11:23

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