A Statistical View of Entropy — Question 9

Section 20-7 · Physics pages 3368–3369

Question 9 Correlations reduce joint entropy while leaving individual entropies unchanged Two binary variables A and B have joint probabilities P(00) = P(11) = 1/2 and P(01) = P(10) = 0. Use S = −kB ∑p ln p. Compare this ensemble with two independent fair binary variables. 00: 1/2 01: 0 10: 0 11: 1/2 Schematic; labels, not drawing lengths, specify the data. (a) Find the marginal probabilities and entropy of each variable, then the joint entropy. (b) Calculate the correlation entropy difference SA + SB −SAB. Explain why knowledge of A determines B despite each variable being individually uncertain.
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Question 9 - Solution Correlations reduce joint entropy while leaving individual entropies unchanged 00: 1/2 01: 0 10: 0 11: 1/2 (a) Method and calculation PA(0) = PA(1) = PB(0) = PB(1) = 1/2. SA = SB = kB ln 2, SAB = −2kB(1 2 ln 1 2) = kB ln 2. (b) Interpretation and further calculation SA + SB −SAB = kB ln 2. SAB,independent = kB ln 4 = 2kB ln 2, P(B = A | A) = 1. Only the joint probabilities change between the ensembles. Correlations remove one binary choice without changing either marginal distribution. Physical check Entropy additivity applies to independent variables. Physically creating correlations requires accounting for the preparation process and environment.

Question 9

Correlations reduce joint entropy while leaving individual entropies unchanged

Two binary variables A and B have joint probabilities P(00)=P(11)=1/2P(00)=P(11)=1/2 and P(01)=P(10)=0P(01)=P(10)=0. Use S=−kB∑pln⁡pS=-k_B\sum p\ln p. Compare this ensemble with two independent fair binary variables.

Diagram shown in the original SVG worksheet.

  1. Find the marginal probabilities and entropy of each variable, then the joint entropy.

  2. Calculate the correlation entropy difference SA+SB−SABS_A+S_B-S_{AB}. Explain why knowledge of A determines B despite each variable being individually uncertain.

Question 9 - Solution

Correlations reduce joint entropy while leaving individual entropies unchanged

Diagram shown in the original SVG worksheet.

(a) Method and calculation

PA(0)=PA(1)=PB(0)=PB(1)=1/2.P_A(0)=P_A(1)=P_B(0)=P_B(1)=1/2.SA=SB=kBln⁡2,SAB=−2kB(12ln⁡12)=kBln⁡2.S_A=S_B=k_B\ln2,\quad S_{AB}=-2k_B(\tfrac12\ln\tfrac12)=k_B\ln2.

(b) Interpretation and further calculation

SA+SB−SAB=kBln⁡2.S_A+S_B-S_{AB}=k_B\ln2.SAB,independent=kBln⁡4=2kBln⁡2,P(B=A∣A)=1.S_{AB,\mathrm{independent}}=k_B\ln4=2k_B\ln2,\quad P(B=A\mid A)=1.Only the joint probabilities change between the ensembles. Correlations remove one binary choice without changing either marginal distribution.

Physical check

Entropy additivity applies to independent variables. Physically creating correlations requires accounting for the preparation process and environment.