Reading a Dendrogram as a Record of Merges
A dendrogram is a tree-shaped record of an agglomerative clustering procedure. Its leaves represent the original observations. Each internal node records a merge: two existing clusters join to form a new cluster. Reading the tree means tracing which observations belong to each branch and interpreting the height at which its branches meet.
The horizontal arrangement of leaves is often chosen for display and can vary without changing the clustering. Membership is encoded by the branching structure, not by how close two labels appear on the page. The vertical coordinate, called the merge height, reports the dissimilarity associated with the merge under the selected linkage rule and dissimilarity convention.
A height is not a cluster label, a count of observations, or necessarily the dissimilarity between a particular pair of original observations. Its interpretation depends on the method. For single linkage it represents the minimum pairwise dissimilarity between the joined clusters; for complete linkage it represents the maximum; for average linkage it is the mean pairwise dissimilarity. Other linkage rules define the relevant merge dissimilarity differently. Therefore, read a height together with the stated linkage rule and distance convention.
The merge height answers a different question from membership. The branches meeting at a node tell you which two clusters were joined; the height tells you the dissimilarity reported for that merge. A cluster can contain several observations even though its branch begins at a single earlier node.
Track Membership Through Successive Merges
To read a dendrogram reliably, start at the leaves and work upward. At each internal node, write down the two clusters represented by its child branches, then take their union. Keep every cluster not involved in that merge unchanged. This gives the full partition at each stage, not merely the membership of the newest cluster.
Trace each branch downward to its leaf labels or to an earlier merge node. Those are the two clusters being joined.
Combine all observations in the two child clusters. Do not replace a multi-observation branch with only the label nearest the merge.
Read the vertical coordinate of the node, then state what dissimilarity that height represents for the selected linkage rule.
Confirm that each original observation appears exactly once, and that the number of clusters has decreased by one for this merge.
In a dendrogram with \(n\) observations, the starting partition has \(n\) singleton clusters. Each binary merge reduces the number of current clusters by exactly one. After \(r\) merges, the number of clusters is \(n-r\), provided each merge joins two distinct current clusters. This count is a useful check, but it does not identify the members of those clusters; membership must still be traced through the branches.
Worked Example: Reading a Complete-Linkage Merge
A hypothetical insurer groups five claim portfolios, labelled A, B, C, D, and E, using complete linkage. The reported early merges are A with B at height 1.4, followed by D with E at height 2.1. The next merge joins the cluster {A, B} with C at height 4.6. Reconstruct the partition after each merge and interpret the height 4.6.
Method. Follow the merge sequence from the singleton leaves. At each step, replace the two joined clusters with their union. Complete linkage defines the distance between two clusters as the largest pairwise dissimilarity between an observation in one cluster and an observation in the other.
Initially: the partition is {A}, {B}, {C}, {D}, {E}. It contains five clusters.
First merge, height 1.4: {A} and {B} join, so their union is {A, B}. The partition becomes {A, B}, {C}, {D}, {E}, with four clusters. The height 1.4 is the complete-linkage dissimilarity reported for the A–B merge.
Second merge, height 2.1: {D} and {E} join to form {D, E}. The partition becomes {A, B}, {C}, {D, E}, with three clusters. The earlier cluster {A, B} remains intact; it is not split when a different pair merges.
Third merge, height 4.6: the child branches represent {A, B} and {C}, so the new cluster is {A, B, C}. The full partition is now {A, B, C}, {D, E}, with two clusters. Under complete linkage, 4.6 represents the largest of the dissimilarities from A to C and from B to C. It is not necessarily the dissimilarity between A and C alone.
Check: three merges have occurred among five observations, and \(5-3=2\) clusters remain. Every label A through E appears once in the stated partition. The common mistake would be to read 4.6 as the distance between the labels visually nearest the node, rather than as the complete-linkage distance between the two child clusters.
Why the Partitions Are Nested
Agglomerative clustering is nested because a merge only combines two existing clusters; it never divides a cluster or moves one of its members elsewhere. More precisely, consider two successive partitions. If the merge joins clusters \(C_1\) and \(C_2\), the next partition contains \(C_1 \cup C_2\) in their place, while every other cluster is unchanged. Thus each cluster in the new partition is either an unchanged cluster from before or a union of two previous clusters.
This property extends across multiple steps. Every later cluster is built by taking unions of clusters that existed earlier. Therefore, two clusters from different stages either remain separate or an earlier cluster is entirely contained in a later one; the process does not produce crossing memberships. This is what “nested partitions” means: the tree’s branches preserve the history of membership as clusters grow.
Here \(\mathcal{P}_r\) is the partition after \(r\) merges, and \(C_1\) and \(C_2\) are two distinct clusters in that partition. The set difference removes those two clusters, and the union adds their combined cluster. Since \(C_1\) and \(C_2\) were disjoint members of a partition, their union contains exactly their previous observations, with no duplication or loss.
The cluster count also follows directly. If \(\mathcal{P}_r\) has \(k\) clusters, removing two and adding one gives \(k-2+1=k-1\). Consequently, after \(r\) merges from \(n\) singleton clusters, the count is \(n-r\). This justifies the count check without relying on the visual spacing of branches.
Worked Example: Recovering Partitions from a Merge Sequence
A hypothetical portfolio segmentation starts with six observations, P, Q, R, S, T, and U. A dendrogram reports these successive merges: P with Q at height 0.8; T with U at height 1.5; {P, Q} with R at height 2.7; and {T, U} with S at height 3.4. State the partition after the second merge and after the fourth merge. Explain how the sequence demonstrates nesting.
Method. Begin with six singleton clusters, then apply the listed unions in order. The heights give the dissimilarities for those merges; they do not change the membership arithmetic.
Initially, the partition is {P}, {Q}, {R}, {S}, {T}, {U}, containing six clusters.
After merge 1, P and Q form {P, Q}. The partition is {P, Q}, {R}, {S}, {T}, {U}, containing five clusters.
After merge 2, T and U form {T, U}. The partition is {P, Q}, {R}, {S}, {T, U}, containing four clusters. This is the requested partition after the second merge.
After merge 3, {P, Q} joins R to form {P, Q, R}. The partition becomes {P, Q, R}, {S}, {T, U}, containing three clusters.
After merge 4, {T, U} joins S to form {S, T, U}. The partition is {P, Q, R}, {S, T, U}, containing two clusters. This is the requested partition after the fourth merge.
The memberships are nested: {P, Q} was not broken apart when R joined it, and {T, U} was not broken apart when S joined it. Each new group contains whole earlier clusters. The count check gives \(6-2=4\) clusters after merge 2 and \(6-4=2\) after merge 4, matching the explicit lists. A common error is to report only the newest cluster, such as {S, T, U}, as the entire partition after merge 4; the other cluster {P, Q, R} must also be included.
What a Merge Height Does—and Does Not—Tell You
A dendrogram’s heights summarize the sequence of merge dissimilarities. For example, a higher merge height indicates that the corresponding merge was made at a larger reported dissimilarity than a lower merge, when the heights use the same scale and convention. It does not, by itself, say how many observations are in either child cluster or how many clusters exist at that stage.
The precise interpretation depends on the linkage rule. Under average linkage, the distance between clusters \(C\) and \(D\) is the mean of the pairwise dissimilarities between their members:
The denominator counts all cross-cluster pairs. If one child cluster has two observations and the other has three, there are six cross-cluster pairwise dissimilarities in the average. The node height records the value used for that merge under the method and convention stated for the analysis. The individual pairwise values need not equal the height.
Worked Example: Computing the Height for an Average-Linkage Merge
Four hypothetical regional risk groups, P, Q, R, and S, have the following pairwise dissimilarities: \(d(P,Q)=2\), \(d(R,S)=3\), \(d(P,R)=6\), \(d(P,S)=8\), \(d(Q,R)=4\), and \(d(Q,S)=6\). Using average linkage, verify the successive merges and find the final merge height.
Method. Initially, all clusters are singletons, so each cluster distance is the corresponding pairwise dissimilarity. After a merge, calculate the average of all cross-cluster pairwise dissimilarities for each relevant pair of clusters.
First merge: the six initial distances are 2, 3, 6, 8, 4, and 6. The smallest is \(d(P,Q)=2\), so {P} and {Q} merge to form {P, Q} at height 2.
Update distances: the distance from {P, Q} to R is \((d(P,R)+d(Q,R))/2=(6+4)/2=5\). The distance from {P, Q} to S is \((d(P,S)+d(Q,S))/2=(8+6)/2=7\). The remaining singleton distance is \(d(R,S)=3\).
Second merge: among 5, 7, and 3, the smallest is 3. Thus {R} and {S} merge at height 3. The partition is now {P, Q}, {R, S}.
Final merge: the four cross-cluster distances between {P, Q} and {R, S} are 6, 8, 4, and 6. Their sum is \(6+8+4+6=24\), and there are \(2 \times 2=4\) such pairs. Therefore, the average-linkage distance is \(24/4=6\), so the final merge height is 6.
Check: there are four observations and three merges, so one cluster remains: \(4-3=1\). The final cluster is {P, Q, R, S}. The common mistake is to use only one cross-cluster distance, such as \(d(P,R)=6\), and treat it as the height. In this example that happens to equal the average by coincidence, but the correct calculation uses all four cross-cluster pairs; their average is 6.
Common Interpretation Traps
- Reading leaf order as membership. Nearby labels along the bottom are not necessarily a cluster. Trace their branches to a shared merge node.
- Confusing height with cluster size. A node at height 4.6 does not mean a cluster contains 4.6 observations or that 4.6 clusters remain.
- Treating a cluster merge as an observation-pair merge. Once a branch contains multiple observations, its next merge joins that whole cluster to another cluster.
- Ignoring the linkage rule. The same child clusters can have different dissimilarities under different linkage definitions, so a height must be interpreted in method context.
- Listing only the newest cluster. A partition includes all current clusters, including those not involved in the latest merge.
For an exam-style interpretation, give three pieces of information together: the two child memberships, the resulting union, and the merge height with its linkage meaning. Then check that the complete partition covers every observation exactly once and has one fewer cluster than before. This approach is aligned with the SOA Exam SRM (Statistics for Risk Modeling) syllabus and avoids relying on visual guesses.
Check Your Understanding
For each problem, track memberships as well as merge heights; report the complete partition whenever asked.
- A five-observation dendrogram has merges A–C at height 1.1 and B–D at height 1.6. What is the complete partition after the second merge? (Answer: {A, C}, {B, D}, {E})
- Starting with seven singleton observations, how many clusters remain after four agglomerative merges, assuming each merge joins two distinct current clusters? (Answer: 3 clusters)
- Under average linkage, clusters {J, K} and {L} have pairwise dissimilarities \(d(J,L)=5\) and \(d(K,L)=9\). What is their merge dissimilarity? (Answer: 7)
- A dendrogram first joins {M, N} with O, then joins the resulting cluster with {P, Q}. State the membership of the final cluster and explain whether {M, N} remains an intact subset. (Answer: The final cluster is {M, N, O, P, Q}; yes, {M, N} remains an intact subset.)
- Six observations undergo merges A–B, C–D, {A, B}–E, and {C, D}–F, in that order. Give the partition after the fourth merge and verify the cluster count. (Answer: {A, B, E}, {C, D, F}; 2 clusters, consistent with \(6-4=2\).)