Exercise: Numéraire change for digital options

In Black-Scholes, the price of a European call is

This decomposition has a beautiful probabilistic interpretation in terms of two different numéraires: the bank account $B_t$ and the stock $S_t$.

Tasks

1. Bank-account measure $\mathbb{Q}$. Under $\mathbb{Q}$ (numéraire $B_t = e^{rt}$), $\Phi(d_2) = \mathbb{Q}(S_T > K)$. State the risk-neutral valuation formula for the digital cash-or-nothing call paying $\mathbb{1}_{\{S_T > K\}}$ and verify the formula gives $e^{-rT}\Phi(d_2)$.
2. Stock measure $\mathbb{Q}^S$. When the stock is the numéraire, define $\mathbb{Q}^S$ by

Show that $\mathbb{E}^{\mathbb{Q}^S}[1] = 1$ (so $\mathbb{Q}^S$ is a probability measure) and that it's equivalent to $\mathbb{Q}$.

1. Asset-or-nothing call. Price the asset-or-nothing call paying $S_T \mathbb{1}_{\{S_T > K\}}$ under both numéraires. Show that under $\mathbb{Q}^S$, the price is $S_0 \mathbb{Q}^S(S_T > K)$.
2. Identify $\Phi(d_1)$. Show that $\Phi(d_1) = \mathbb{Q}^S(S_T > K)$.
3. Reinterpret Black-Scholes. Combine the previous parts to give the probabilistic interpretation: the call price decomposes as the expected stock received (weighted by ITM probability under stock measure) minus the strike paid (weighted by ITM probability under bond measure).

Hint

For part 4, under $\mathbb{Q}^S$, $S_t/B_t$ has a different drift than under $\mathbb{Q}$. Specifically, $W^{\mathbb{Q}^S}_t = W^{\mathbb{Q}}_t - \sigma t$ is a Brownian motion under $\mathbb{Q}^S$.