Computing Risk-Neutral Probabilities in Binomial Trees

From Branch Probabilities to Risk-Neutral Pricing

In a binomial model, a stock has only two possible values at the next step. This makes the risk-neutral probability a direct solution of one linear equation: choose the branch weights so that the discounted stock price has the martingale property. The resulting probabilities are not generally the probabilities observed under the physical measure \(\mathbb{P}\). They are probabilities under a pricing measure \(\mathbb{Q}\), and they are useful only when they are positive and sum to one.

Work on a finite filtered probability space \((\Omega,\mathcal{F},(\mathcal{F}_n)_{n=0}^N,\mathbb{P})\), where \(\mathcal{F}_n\) records the market history through date \(n\). Suppose the bank account grows by a constant gross factor \(R>0\) each period, so its value at date \(n\) is \(R^n\). At a node where the stock price is \(S>0\), the next stock price is either \(Su\) or \(Sd\), with \(u>d>0\). A one-period portfolio holding \(\Delta\) shares and \(b\) units of the bank account costs \(\Delta S+b\) at the node and is worth \(\Delta S_{\text{next}}+Rb\) at the next date.

The martingale equation is \(qSu+(1-q)Sd=RS\). Since \(S>0\) and \(u\ne d\), it has the unique algebraic solution \(q=(R-d)/(u-d)\). It gives a valid equivalent probability precisely when \(d<R<u\). If either inequality fails, the algebraic value is not strictly between zero and one, so it cannot be the probability of an equivalent measure. At \(R=d\) or \(R=u\), one branch receives probability zero; that is not equivalent to a physical measure assigning positive probability to both branches.

One-Step Pricing and Replication

Proof. The martingale equation has the unique solution \(q=(R-d)/(u-d)\), and \(d<R<u\) implies \(0<q<1\). The two replication equations are \(\Delta Su+Rb=X_u\) and \(\Delta Sd+Rb=X_d\). Subtracting them gives \(\Delta=(X_u-X_d)/(S(u-d))\); substituting into the down equation gives \(b=(uX_d-dX_u)/(R(u-d))\). These values solve both equations, and the same subtraction and substitution show that no other holdings can solve them. Their initial cost is $$ \Delta S+b =\frac{(R-d)X_u+(u-R)X_d}{R(u-d)} =\frac{qX_u+(1-q)X_d}{R}, $$ because \(1-q=(u-R)/(u-d)\). Thus the replicating cost equals the discounted risk-neutral expectation. \(\square\)

The bank-account growth factor matters whenever a holding is carried to a later date: \(b\) is its current value, while \(Rb\) is its next-period value. Risk-neutral probabilities, by contrast, are determined by the discounted-stock martingale equation and do not depend on a particular claim's payoff.

Extending the Calculation Through a Tree

Now suppose there are \(N\) periods, the same \(u,d,R\) apply at each node, and \(d<R<u\). At each step the risk-neutral up probability is \(q=(R-d)/(u-d)\), and the down probability is \(1-q\). Assume the physical probability of each branch is strictly between zero and one at every node. Assigning these risk-neutral branch probabilities successively defines a measure \(\mathbb{Q}\) on the finite set of paths. Every path then has positive probability under both \(\mathbb{P}\) and \(\mathbb{Q}\), so the two measures are equivalent.

For a terminal claim \(H\), define \(V_N=H\). At a node at date \(n\), write \(V_{n+1}^{u}\) and \(V_{n+1}^{d}\) for the values at its up and down successor nodes. The value at the node is calculated by backward induction: take the conditional risk-neutral expectation of the next value and discount one period. This is a finite sum, so all the conditional expectations below are integrable.

Proof. At any node, the proposed holdings have next-period values $$ \Delta_n S_nu+Rb_n =\frac{u(V_{n+1}^{u}-V_{n+1}^{d})+uV_{n+1}^{d}-dV_{n+1}^{u}}{u-d} =V_{n+1}^{u}, $$ and $$ \Delta_n S_nd+Rb_n =\frac{d(V_{n+1}^{u}-V_{n+1}^{d})+uV_{n+1}^{d}-dV_{n+1}^{u}}{u-d} =V_{n+1}^{d}. $$ Their current cost is $$ \Delta_n S_n+b_n =\frac{(R-d)V_{n+1}^{u}+(u-R)V_{n+1}^{d}}{R(u-d)} =\frac{qV_{n+1}^{u}+(1-q)V_{n+1}^{d}}{R} =V_n. $$ Thus the holdings at each node cost exactly the value required there and produce the recursively specified successor values. Starting at date \(0\), choose these holdings and rebalance at each successor node. The portfolio value equals \(V_n\) at each date by the displayed equations, and at \(N\) it equals \(H\). Rebalancing costs no additional cash because the value of the portfolio before and after choosing the prescribed holdings at a node is \(V_n\). This constructs the claimed self-financing replication. \(\square\)

Martingale Check and a Common Pitfall

At every node, the conditional expectation of the next discounted stock is $$ \mathbb{E}^{\mathbb{Q}}\left[\frac{S_{n+1}}{R^{n+1}}\middle|\mathcal{F}_n\right] =\frac{S_n(qu+(1-q)d)}{R^{n+1}} =\frac{S_n}{R^n}, $$ because \(qu+(1-q)d=R\). This verifies the martingale property: on the finite tree the discounted stock is integrable, and its conditional expectation at each date equals its current value. The same finite-tree argument applies to the discounted value process \(V_n/R^n\): the backward recursion gives \(\mathbb{E}^{\mathbb{Q}}[V_{n+1}/R^{n+1}\mid\mathcal{F}_n]=V_n/R^n\).

A common error is to use a physical up probability \(p\) in the pricing expectation without checking the martingale condition. In general \(p\) need not satisfy \(pu+(1-p)d=R\), so discounting the physical expectation does not produce the replicating cost. Another frequent error is to carry a bank-account position forward as \(b\) rather than \(Rb\). The first worked example shows why: using \(-80\) instead of \(-88\) as the next-period bank contribution would incorrectly give terminal values \(52\) and \(8\), not \(44\) and \(0\). The stock holding \(\Delta=11/12\) is correct; the error is omitting bank-account growth.