Solution: Radon-Nikodym in a Binomial Model — Pricing via Change of Measure

Exercise: Radon-Nikodym in a Binomial Model: Pricing via Change of Measure

Part 1

Part 2

Outcomes in $\Omega = \{uu, ud, du, dd\}$:

$\mathbb{P}(uu) = p^2 = 0.36$, $\mathbb{P}(ud) = p(1 - p) = 0.24$, etc. $\mathbb{Q}(uu) = q^2 = 0.25$, etc.

Part 3

$Z(\omega) = \mathbb{Q}(\omega)/\mathbb{P}(\omega)$:

Check $\mathbb{E}^{\mathbb{P}}[Z] = \sum_\omega Z(\omega)\mathbb{P}(\omega) = \sum_\omega \mathbb{Q}(\omega) = 1$. ✓ (Trivially — the sum of $\mathbb{Q}$-probabilities is $1$.)

Part 4

Payoffs: $(S_2 - 100)_+$ is $21$ for $S_2 = 121$ and $0$ otherwise.

Both methods agree. The option fair price is \5.25$, and it equals the$\mathbb{Q}$-expectation of the payoff or equivalently the$\mathbb{P}$-expectation of the payoff weighted by the Radon-Nikodym derivative$Z$.

Takeaways

• $\mathbb{Q}$ is discrete here; Radon-Nikodym derivatives are ratios of probabilities. This is the simplest non-trivial instance of the theorem: in a finite probability space, $Z(\omega) = \mathbb{Q}(\omega)/\mathbb{P}(\omega)$.
• Risk-neutral probability $q$ is the one that makes $\mathbb{E}^{\mathbb{Q}}[S_{t+1}/S_t] = 1 + r$ — i.e. the stock's expected return equals the risk-free rate. Here $0.5\cdot 1.1 + 0.5\cdot 0.9 = 1$. ✓
• Pricing via $\mathbb{Q}$ is equivalent to pricing via $Z$-weighted $\mathbb{P}$. Same answer, two viewpoints. In continuous-time models the $Z$-weighted viewpoint (importance sampling) is often more tractable.
• This generalises to $n$ periods. $Z_n = \prod_{t=1}^n Z_t$ with $Z_t$ the per-period Radon-Nikodym derivative. In the limit $n \to \infty$ with $\Delta t \to 0$, $Z_n$ becomes the Girsanov-type exponential for Brownian motion.