UPT Example: 2-State Synchrony Markov Chain

UPT Example: 2-State Synchrony Markov Chain

This page gives a concrete toy example of Unit Perception Tests (UPTs) and the resulting 2×2 empirical transition matrix.

1. Setup

We consider a binary synchrony state at each timeline index t:

• State S: synchrony detected at time t (the UPT returns 1).
• State N: no synchrony detected at time t (the UPT returns 0).

We write this as a state variable X_t with two possible values:

• <math>X_t \in \{S, N\}</math>.

The underlying data (streams A_t and B_t) and the Unit Perception Test S(t) are abstracted away here; we focus only on the resulting sequence of states.

2. Example Sequence of UPT Outcomes

Suppose that, after running UPTs over 10 consecutive timeline indices, we observe the following sequence of synchrony states:

• <math>(X_1, X_2, X_3, X_4, X_5, X_6, X_7, X_8, X_9, X_{10}) = (S, N, N, S, S, N, S, N, N, S)</math>.

For convenience, rewrite this as:

• Time 1: S
• Time 2: N
• Time 3: N
• Time 4: S
• Time 5: S
• Time 6: N
• Time 7: S
• Time 8: N
• Time 9: N
• Time 10: S

We will use this sequence to estimate the transition probabilities between S and N.

3. Transition Counts

We look at **one-step transitions** X_t → X_{t+1} for t = 1, …, 9. The possible ordered pairs are:

• S → S
• S → N
• N → S
• N → N

From the example sequence:

1. Time 1 → 2: S → N
2. Time 2 → 3: N → N
3. Time 3 → 4: N → S
4. Time 4 → 5: S → S
5. Time 5 → 6: S → N
6. Time 6 → 7: N → S
7. Time 7 → 8: S → N
8. Time 8 → 9: N → N
9. Time 9 → 10: N → S

Counting each transition type:

• S → S: occurs 1 time (at 4 → 5)
• S → N: occurs 3 times (at 1 → 2, 5 → 6, 7 → 8)
• N → S: occurs 3 times (at 3 → 4, 6 → 7, 9 → 10)
• N → N: occurs 2 times (at 2 → 3, 8 → 9)

We can summarize these counts as N_{ij}, where i and j are in {S, N}:

• <math>N_{SS} = 1</math>
• <math>N_{SN} = 3</math>
• <math>N_{NS} = 3</math>
• <math>N_{NN} = 2</math>

4. Empirical 2×2 Transition Matrix

The empirical transition probability from state i to state j is defined as:

• <math>\hat{P}_{ij} = \dfrac{N_{ij}}{\sum_k N_{ik}}</math>

For our 2-state case {S, N}:

• Total departures from S:
  • <math>\sum_k N_{Sk} = N_{SS} + N_{SN} = 1 + 3 = 4</math>
• Total departures from N:
  • <math>\sum_k N_{Nk} = N_{NS} + N_{NN} = 3 + 2 = 5</math>

So the empirical transition probabilities are:

• From S:
  • <math>\hat{P}_{SS} = \dfrac{N_{SS}}{N_{SS} + N_{SN}} = \dfrac{1}{4} = 0.25</math>
  • <math>\hat{P}_{SN} = \dfrac{N_{SN}}{N_{SS} + N_{SN}} = \dfrac{3}{4} = 0.75</math>

• From N:
  • <math>\hat{P}_{NS} = \dfrac{N_{NS}}{N_{NS} + N_{NN}} = \dfrac{3}{5} = 0.6</math>
  • <math>\hat{P}_{NN} = \dfrac{N_{NN}}{N_{NS} + N_{NN}} = \dfrac{2}{5} = 0.4</math>

We can write the empirical 2×2 transition matrix \hat{P} as:

<math> \hat{P} = �egin{pmatrix} \hat{P}_{SS} & \hat{P}_{SN} \ \hat{P}_{NS} & \hat{P}_{NN} \end{pmatrix} = �egin{pmatrix} 0.25 & 0.75 \ 0.60 & 0.40 \end{pmatrix}. </math>

Here the first row/column corresponds to state S, and the second row/column to state N.

5. Interpretation

From this small example, we can already see qualitative behavior:

• When the system is in synchrony (S), it tends to move to no-synchrony (N) on the next step with probability 0.75.
• When the system is in no-synchrony (N), it tends to move back to synchrony (S) with probability 0.6.
• Both states are unstable in the sense that the most likely transition is to the *other* state.

With longer sequences (more UPTs), these empirical probabilities stabilize and define a perceptive envelope over synchrony dynamics: a concise description of how likely synchrony is to persist, dissolve, or re-emerge over time.