Computing Density Processes on Finite-State Trees

From Terminal Weights to Densities at Every Node

A finite probability tree makes the conditional-expectation definition of a density process concrete. At the terminal time, the density compares the probability of each outcome under two measures. At an earlier time, the density at a node compares the probabilities assigned to that whole node. These node values must fit together: the density at a parent is the conditional average, under the original measure, of the densities at its children.

We work on a finite probability space \((\Omega,\mathcal{F},\mathbb{P})\) equipped with the filtration \((\mathcal{F}_t)_{t=0}^N\) generated by a finite tree. Here \(\mathcal{F}_t\) records which node has been reached by time \(t\), and \(\mathcal{F}_N=\mathcal{F}\) distinguishes the terminal leaves. Assume each terminal leaf has positive \(\mathbb{P}\)-probability. A Brownian motion and stochastic integrals are not needed on this finite-tree model.

For any event \(A\), the defining property of \(L\) is \(\mathbb{Q}(A)=\mathbb{E}[\mathbf{1}_A L]\). In a finite tree, conditional expectation on a node is an ordinary weighted average over the leaves below that node. This makes both the terminal derivative and all earlier values directly computable.

The Node-Ratio Formula and Its Consistency

Proof. Fix a node \(v\) at time \(t\). Its event consists of the terminal leaves \(\omega\) below \(v\). Conditional expectation on this event is the \(\mathbb{P}\)-weighted average of \(L\) over those leaves. Therefore $$ Z_t(v)=\frac{1}{\mathbb{P}(v)} \sum_{\omega\text{ below }v}\mathbb{P}(\{\omega\})L(\omega) =\frac{1}{\mathbb{P}(v)} \sum_{\omega\text{ below }v}\mathbb{Q}(\{\omega\}) =\frac{\mathbb{Q}(v)}{\mathbb{P}(v)}. $$ The denominator is positive because \(v\) contains at least one leaf and every leaf has positive \(\mathbb{P}\)-probability. Equivalence gives \(\mathbb{Q}(v)>0\), so \(Z_t(v)>0\). At time zero the only node is the root, whose probability is one under both measures; hence \(Z_0=1\).

To verify the conditional-expectation identity, fix a time-\(t\) node \(v\) and sum over its children \(u\). Since \(Z_{t+1}(u)=\mathbb{Q}(u)/\mathbb{P}(u)\), the conditional average under \(\mathbb{P}\) is $$ \mathbb{E}[Z_{t+1}\mid\mathcal{F}_t](v) =\sum_{u\text{ child of }v}\frac{\mathbb{P}(u)}{\mathbb{P}(v)} \frac{\mathbb{Q}(u)}{\mathbb{P}(u)} =\frac{\sum_u\mathbb{Q}(u)}{\mathbb{P}(v)} =\frac{\mathbb{Q}(v)}{\mathbb{P}(v)} =Z_t(v). $$ The children partition the event \(v\), which justifies the penultimate equality. Each \(Z_t\) is integrable because the space is finite and all its values are finite. Thus the conditional-expectation identity proves that \((Z_t)\) is a \(\mathbb{P}\)-martingale. \(\square\)

The identity also gives a practical rule when conditional branch probabilities are supplied. If the children of \(v\) have \(\mathbb{P}\)-transition probabilities \(p(u\mid v)\) and \(\mathbb{Q}\)-transition probabilities \(q(u\mid v)\), then $$ Z_{t+1}(u)=Z_t(v)\frac{q(u\mid v)}{p(u\mid v)}. $$ This follows by writing each node probability as its parent probability times its transition probability. The formula requires \(p(u\mid v)>0\); equivalence ensures \(q(u\mid v)>0\) as well.

Worked Calculations on Two-Step Trees

At any time, a density is a value attached to the currently observed node—not an average over possibilities under \(\mathbb{Q}\). The averaging that gives the next density is under \(\mathbb{P}\), because \(Z_t\) is defined by \(\mathbb{P}\)-conditional expectation.

Constructing a Measure from Terminal Weights

Proof. Each assigned leaf mass is positive because both \(L(\omega)\) and \(\mathbb{P}(\{\omega\})\) are positive. The masses sum to \(\sum_\omega L(\omega)\mathbb{P}(\{\omega\})=\mathbb{E}_{\mathbb{P}}[L]=1\), so they define a probability measure. Since every leaf has positive mass under both measures, the measures are equivalent. For any event \(A\), summing the assigned masses over its leaves gives \(\mathbb{Q}(A)=\sum_{\omega\in A}L(\omega)\mathbb{P}(\{\omega\})=\mathbb{E}_{\mathbb{P}}[\mathbf{1}_A L]\). This identifies \(L\) as the Radon–Nikodym derivative. The conditional-expectation definition then gives the stated node values. \(\square\)

Common Mistake: Using a Leaf Density at an Earlier Node

A terminal density is generally not the density at an earlier time. For example, in the first worked example, the density at leaf \(A_1\) is \(4/15\), but the density at node \(A\) is \(1/2\). Replacing \(Z_1(A)\) by \(4/15\) is incorrect: the node contains both \(A_1\) and \(A_2\), so its density must average over both leaves with their conditional \(\mathbb{P}\)-probabilities.

The wrong calculation would set \(Z_1(A)=4/15\), then claim the root average is \(0.6(4/15)+0.4(7/4)=4/25+7/10=43/50\), not \(1\). The correction is to use \(Z_1(A)=0.75(4/15)+0.25(6/5)=1/2\). Then \(0.6(1/2)+0.4(7/4)=1\). A failed normalization or parent-child average is a useful warning that a value has been attached to the wrong node or averaged using the wrong probabilities.

Finite trees are useful because they turn an abstract change of measure into a sequence of auditable arithmetic checks. To verify a proposed density process, check that terminal values are positive, their \(\mathbb{P}\)-expectation is one, and each parent value is the conditional \(\mathbb{P}\)-average of its children. In a finite filtration these checks establish the required martingale consistency at every time.