Prove, using Problems 64 and 66, that for a linearly ordered…

Prove, using Problems 64 and 66, that for a linearly ordered sequence of sets $L^*$ and $i\lt j\lt k$ \[ \varrho(L_i,L_j)+\varrho(L_j,L_k)=\varrho(L_i,L_k). \]

Step 1: Property $b_{ijk}$

By Problem 66, for a linearly ordered sequence $L_i=A_i+B_{n-i+1}+C$ and any $i\lt j\lt k$,

\[ L_i\overline{L_j}L_k=V, \qquad \overline{L_i}L_j\overline{L_k}=V. \]

Step 2: Apply Problem 64

By the identity of Problems 63–64,

\[ \varrho(L_i,L_j)+\varrho(L_j,L_k)-\varrho(L_i,L_k)=2\Big[P\big(L_i\overline{L_j}L_k\big)+P\big(\overline{L_i}L_j\overline{L_k}\big)\Big] = 2[P(V)+P(V)] = 0. \]

\[ \boxed{\varrho(L_i,L_j)+\varrho(L_j,L_k)=\varrho(L_i,L_k)\qquad(i\lt j\lt k).} \quad\blacksquare \]

Step 3: A direct formula for the distances

Using the four cases of Problem 66, $L_i$ and $L_j$ ($i\lt j$) differ exactly on $(A_j\setminus A_i)+(B_{n-i+1}\setminus B_{n-j+1})$, two disjoint pieces. So

\[ \varrho(L_i,L_j) = \big[P(A_j)-P(A_i)\big] + \big[P(B_{n-i+1})-P(B_{n-j+1})\big]. \]

Define $a_t=P(A_t)-P(B_{n-t+1})$. Then $\varrho(L_i,L_j)=a_j-a_i$ for $i\lt j$, and the additivity is obvious:

\[ (a_j-a_i)+(a_k-a_j)=a_k-a_i. \ \checkmark \]

So the events $L_1,\dots,L_n$ behave like the points $a_1\le a_2\le\dots\le a_n$ on the real line — this is the reason for the name "linearly ordered".

Step 4: Example (Problem 66)

With a fair die and $L_1=\{3,4,5\}$, $L_2=\{1,3,5\}$, $L_3=\{1,2,3,5\}$:

\[ \varrho(L_1,L_2)=P(\{1,4\})=\tfrac26,\quad \varrho(L_2,L_3)=P(\{2\})=\tfrac16,\quad \varrho(L_1,L_3)=P(\{1,2,4\})=\tfrac36=\tfrac26+\tfrac16. \ \checkmark \]