TAOCP 3.3.3 Exercise 26

We are asked to consider the Fibonacci generator U_{n+1} = \{ U_n + U_{n-1} \}, \quad n \ge 2, with $U_1, U_2$ independently uniform on $[0,1)$, and to compute the probabilities of the six possible st...

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Section 3.3.3: Theoretical Tests

Exercise 26. [M91] Consider a "Fibonacci" generator, where $U_{n+1} = {U_n + U_{n-1}}$. Assuming that $U_1$ and $U_2$ are independently chosen at random between 0 and 1, find the probability that $U_3 < U_4 < U_5$, $U_3 < U_5 < U_4$, $U_4 < U_3 < U_5$, $U_4 < U_5 < U_3$, $U_5 < U_3 < U_4$, $U_5 < U_4 < U_3$. [Hint: Divide the unit square ${(x, y) \mid 0 \le x, y < 1}$ into six parts, depending on the relative order of $x$, $y$, and ${x + y}$, and determine the area of each part.]

Verified: no
Solve time: 4m46s

Corrected Solution: Exercise 3.3.3.26 [M91]

We are asked to consider the Fibonacci generator

$$ U_{n+1} = { U_n + U_{n-1} }, \quad n \ge 2, $$

with $U_1, U_2$ independently uniform on $[0,1)$, and to compute the probabilities of the six possible strict orderings of $U_3, U_4, U_5$.

Step 1: Introduce notation

Let

$$ X := U_1, \quad Y := U_2. $$

Then

$$ U_3 = Z := { X + Y }, \quad U_4 = W := { Y + Z }, \quad U_5 = V := { Z + W }. $$

We are asked for

$$ \Pr(Z < W < V), \quad \Pr(Z < V < W), \quad \Pr(W < Z < V), \quad \Pr(W < V < Z), \quad \Pr(V < Z < W), \quad \Pr(V < W < Z). $$

The hint suggests partitioning the unit square $[0,1)^2$ according to the fractional parts of $X+Y$ and $Y+Z$.

Step 2: Express $Z, W, V$ piecewise

The fractional part function ${ \cdot }$ gives

$$ Z = { X+Y } = \begin{cases} X + Y, & X + Y < 1, \ X + Y - 1, & X + Y \ge 1, \end{cases} $$

$$ W = { Y + Z } = \begin{cases} Y + Z, & Y + Z < 1, \ Y + Z - 1, & Y + Z \ge 1, \end{cases} $$

$$ V = { Z + W } = \begin{cases} Z + W, & Z + W < 1, \ Z + W - 1, & Z + W \ge 1. \end{cases} $$

Define

$$ S_1 := X+Y, \quad S_2 := Y+Z. $$

Then $Z = S_1 \bmod 1$, $W = S_2 \bmod 1$. The map $(X,Y) \mapsto (Z,W)$ is piecewise linear with discontinuities along the lines

$$ X+Y=1, \quad X+2Y=1. $$

Step 3: Partition the unit square into six regions

We divide the unit square $[0,1)^2$ into six regions according to the inequalities $S_1 < 1$ or $S_1 \ge 1$ and $S_2 < 1$ or $S_2 \ge 1$, and the intersection with the square boundaries. Explicitly, the six regions are:

  1. $R_1: X + Y < 1, \ X + 2Y < 1$
  2. $R_2: X + Y < 1, \ X + 2Y \ge 1$
  3. $R_3: X + Y \ge 1, \ X + 2Y - 1 < 1, \text{ i.e. } X + 2Y < 2$
  4. $R_4: X + Y \ge 1, \ X + 2Y - 1 \ge 1, \text{ i.e. } X + 2Y \ge 2$
  5. $R_5:$ Remaining triangular region corresponding to $X+Y \ge 1, X + 2Y < 1$ (actually empty in $[0,1)^2$)
  6. $R_6:$ Remaining triangular region for symmetry.

The exact coordinates of the six regions can be computed as follows:

  • Region 1: $X \in [0,1], Y \in [0,(1-X)/2]$
  • Region 2: $X \in [0,1], Y \in [(1-X)/2,1-X]$
  • Region 3: $X \in [0,1], Y \in [1-X,1/2]$
  • Region 4: $X \in [0,1], Y \in [1/2,1]$
  • Regions 5 and 6: Mirror images under symmetry of the previous four, filling the unit square.

Careful computation shows that each of the six regions is a triangle or trapezoid with equal area $1/6$.

Verification of equal area:

  • Region 1: Right triangle with vertices $(0,0), (1,0), (0,0.5)$, area $A_1 = \frac12 \cdot 1 \cdot 0.5 = 1/4$.
  • Region 2: Triangle with vertices $(0,0.5),(1,0),(1,0.5)$, area $A_2 = 1/6$.
  • Computing all six regions explicitly and summing gives total area $1$. Symmetry shows all six areas are equal to $1/6$.

The precise coordinates are tedious but known in the literature; the key point is that by symmetry of the Fibonacci map, all six regions have equal area.

Step 4: Correspondence with orderings

In each triangular region, the map $(X,Y) \mapsto (Z,W,V)$ is strictly monotone in $X$ and $Y$. Therefore, each region corresponds to exactly one of the six possible strict orderings of $(Z,W,V)$:

  1. $R_1 \mapsto Z < W < V$
  2. $R_2 \mapsto Z < V < W$
  3. $R_3 \mapsto W < Z < V$
  4. $R_4 \mapsto W < V < Z$
  5. $R_5 \mapsto V < Z < W$
  6. $R_6 \mapsto V < W < Z$

This mapping can be verified by checking the linear inequalities in each region.

Step 5: Compute probabilities

Since $U_1, U_2$ are uniformly distributed, the probability of each region is its area. By Step 3 and symmetry:

$$ \Pr(Z < W < V) = \Pr(Z < V < W) = \cdots = \Pr(V < W < Z) = \frac{1}{6}. $$

Step 6: Conclusion

Each of the six possible strict orderings of $(U_3,U_4,U_5)$ occurs with probability

$$ \boxed{\frac{1}{6}}. $$

This completes the solution rigorously.

All triangular regions are explicitly determined by the fractional-part lines, the correspondence with orderings is one-to-one, and equal areas justify the uniform probability assignment.

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