# Pricing Option with Payoff of $S_T$ units of a European Digital Call [duplicate]

I am in a Black Scholes market with the usual riskless asset $$B$$ and risky asset $$S$$ with dynamics given by \begin{align*} dB_t &= rB_t dt, B_0 = 1, \\ dS_t &= rS_t dt + \sigma S_t dW_t^{\mathbb{Q}}. \end{align*}

I am asked to use the probabilistic risk-neutral pricing formula to calculate the price of an option with payoff \begin{align*} S_T 1_{S_T > K} \end{align*} for a constant $$K$$. Clearly, if $$S_T$$ were replaced by a constant, say $$A$$, this would be straightforward: it would just be $$A$$ units of a European digital call and we would have the option value $$V_t$$ at time $$t\in[0,T]$$, given by \begin{align*} V_t &= e^{-r(T-t)} A \mathbb{E}^\mathbb{Q}[1_{S_T > K} | S_t] \\ &= e^{-r(T-t)} A \mathbb{Q}(S_T > K | S_t = K). \end{align*}

However I am unsure how to deal with the inclusion of the random variable $$S_T$$ inside the expectation. Writing $$S_T = S_t e^{(r-\frac{\sigma^2}{2})(T-t) + \sigma(\hat W_T - \hat W_t)}$$ doesn't seem to help since $$(\hat W_T - \hat W_t)$$ is not independent of the indicator function. It seems like it is not feasible to use the tower property and nest any expectations.

Any hints would be much appreciated. Thanks for the help.

• The most elegant trick is a change of numeraire, see these answers: 1, 2, and 3. Your question is very similar to those in the links. Alternatively, there's the tedious way of brute-force integration of probability density functions. Jan 2 at 20:54
• @Kevin thanks very much, I had not learned about change of numeraire before but having learned about it, it definitely is a elegant trick! Jan 2 at 22:35