Finance & Options

Lesson 5 of 12

Fair Games and the Markov Property

Under the risk-neutral odds every discounted price is a fair game, and carrying the running maximum in the state turns a path-dependent lookback into a small backward recursion.

The discounted stock is a fair game

Take the tree used all through this lesson: S0=100S_0 = 100, each step multiplies the price by u=1.25u = 1.25 or d=0.8d = 0.8, and cash grows by 1+r=1.051 + r = 1.05 per step. Write qq for the risk-neutral up-probability from the lesson on Risk-Neutral Pricing, with the per-step growth erΔte^{r\Delta t} written as 1+r1 + r:

q=1+r−du−d=1.05−0.80.45=59,1−q=49.q = \frac{1 + r - d}{u - d} = \frac{1.05 - 0.8}{0.45} = \frac59, \qquad 1 - q = \frac49.

Write E~n\tilde E_n for the expectation under these odds given the first nn tosses. After one step the stock is 125 or 80, so E~0[S1]=59⋅125+49⋅80=105\tilde E_0[S_1] = \tfrac59 \cdot 125 + \tfrac49 \cdot 80 = 105. Divide by 1.05 and you are back at 100. The same algebra works at every node:

E~n ⁣[Sn+1(1+r)n+1]=Sn(1+r)n⋅qu+(1−q)d1+r=Sn(1+r)n,\tilde E_n\!\left[\frac{S_{n+1}}{(1+r)^{n+1}}\right] = \frac{S_n}{(1+r)^n}\cdot\frac{qu + (1-q)d}{1 + r} = \frac{S_n}{(1+r)^n},

because qq was chosen to make qu+(1−q)d=1+rqu + (1-q)d = 1 + r. A process whose best forecast of the next value, given everything so far, is its current value is a martingale: a fair game with no drift.

The raw price SnS_n is not a martingale under qq. It grows at the risk-free rate. Under a realistic up-probability such as p=2/3p = 2/3, even the discounted price drifts upward: (23⋅125+13⋅80)/1.05≈104.76>100(\tfrac23 \cdot 125 + \tfrac13 \cdot 80)/1.05 \approx 104.76 > 100. A process that rises on average like this is a submartingale, and the tilt is the risk premium investors earn for holding the stock. The risk-neutral odds are the ones that remove that tilt once prices are measured in units of the bank account.