Stochastic Calculus

Lesson 8 of 11

Radon-Nikodym on a Coin

Two coins, one tree: the ratio Z of their path probabilities turns real-world expectations into prices.

Two coins on the same tree

Take a stock at 100 that moves to 120 or 90 each step, with r=0r = 0. Say the real coin is fair, p=12p = \tfrac12. For pricing we use a second coin, the one that makes the stock a fair game: 120p~+90(1−p~)=100120\tilde p + 90(1 - \tilde p) = 100 gives p~=13\tilde p = \tfrac13.

Toss twice. Both coins see the same four paths and weight them differently:

ωHHHTTHTTS214410810881P(ω)14141414P~(ω)19292949\begin{array}{c|cccc} \omega & HH & HT & TH & TT \\ \hline S_2 & 144 & 108 & 108 & 81 \\ \mathbb P(\omega) & \tfrac14 & \tfrac14 & \tfrac14 & \tfrac14 \\ \tilde{\mathbb P}(\omega) & \tfrac19 & \tfrac29 & \tfrac29 & \tfrac49 \end{array}

Neither measure calls any path impossible. Two measures that agree on which events are impossible are equivalent, and that is what lets you divide one by the other. The path-by-path ratio is the Radon-Nikodym derivative of P~\tilde{\mathbb P} with respect to P\mathbb P:

Z(ω)=P~(ω)P(ω),Z(HH),Z(HT),Z(TH),Z(TT)=49, 89, 89, 169.Z(\omega) = \frac{\tilde{\mathbb P}(\omega)}{\mathbb P(\omega)}, \qquad Z(HH), Z(HT), Z(TH), Z(TT) = \tfrac49,\ \tfrac89,\ \tfrac89,\ \tfrac{16}9.

On a finite tree it is just a quotient. The word "derivative" comes from the general case, such as Brownian paths, where every single path has probability zero. There ZZ is pinned down by events instead of single paths, through P~(A)=E[Z 1A]\tilde{\mathbb P}(A) = E[Z\,\mathbf 1_A].

Two facts follow at once. Z>0Z > 0 on every path, and

E[Z]=∑ωP~(ω)P(ω) P(ω)=∑ωP~(ω)=1.E[Z] = \sum_\omega \frac{\tilde{\mathbb P}(\omega)}{\mathbb P(\omega)}\,\mathbb P(\omega) = \sum_\omega \tilde{\mathbb P}(\omega) = 1.

Here: 14(49+89+89+169)=14⋅369=1\tfrac14(\tfrac49 + \tfrac89 + \tfrac89 + \tfrac{16}9) = \tfrac14 \cdot \tfrac{36}{9} = 1. Throughout, EE with no tilde means the real measure P\mathbb P, and E~\tilde E means P~\tilde{\mathbb P}.