Proof of the Main Lemma

Random Number Generation from Markov Chains

3.3 Main Lemma

3.3.2 Proof of the Main Lemma

andx^′₁=x₁. We also havex^′_N =s_χ. This completes the proof.

We demonstrate the result above by considering the example at the beginning of this section.

Let

X =s1s4s2s1s3s2s3s1s1s2s3s4s1,

withχ= 1 and its exit sequences are given by

[s4s3s1s2, s1s3s3, s2s1s4, s2s1].

After permutating all the exit sequences (for i ̸= 1, we keep the last element of the ith sequence fixed), we get a new group of exit sequences,

[s₁s₂s₃s₄, s₃s₁s₃, s₁s₂s₄, s₂s₁].

Based on these new exit sequences, we can generate a new input sequence,

X^′=s1s1s2s3s1s3s2s1s4s2s3s4s1.

This accords with the statements above.

directed graph in figure 3.2.

LetV denote the vertex set, then

V ={s0, s1, s2, ..., sn},

and the edge set is

E={(si,Λi[k])}∪

{(s0, sα)}.

For each edge (s_i,Λ_i[k]), the label of this edge isk. For the edge (s₀, s_α), the label is 1. Namely, the label set of the outgoing edges of each state is{1,2, ...}.

s

1

s

2

s

3

s

4

1

3 4

2 1

1

2 3

1 2

3 2

s

0

1

Figure 3.2. An example of a sequence graphG.

Given the labeling of the directed graph as defined above, we say that it contains acomplete walk if there is a path in the graph that visits all the edges, without visiting an edge twice, in the following way: (1) Start froms0. (2) At each vertex, we choose an unvisited edge with the minimal label to follow. Obviously, the labeling corresponding to (sα,Λ) is acomplete walk if and only if (sα,Λ) is feasible. In this case, for short, we also say that (sα,Λ) is a complete walk. Before continuing to prove the main lemma, we first give lemma 3.5 and lemma 3.6.

Lemma 3.5. Assume (s_α,Λ) with Λ = [Λ₁,Λ₂, ...,Λ_χ, ...,Λ_n] is a complete walk, which ends at states_χ. Then (s_α,Γ)with Γ = [Λ₁, ...,Γ_χ, ...,Λ_n] is also a complete walk ending ats_χ, if Λ_χ ≡Γ_χ (permutation).

Proof. (sα,Λ) and (sα,Γ) correspond to different labelings on the same directed graphG, denoted byL1 andL2. SinceL1is a complete walk, it can travel all the edges inGone by one, denoted as

(s_i1, s_j1),(s_i2, s_j2), ...,(s_iN, s_jN),

wheres_i1 =s₀ ands_jN =s_χ. We call{1,2, ..., N} as the indexes of the edges.

Based on L2, let us have a walk onGstarting froms0 until there is no unvisited outgoing edges to select. In this walk, assume the following edges have been visited:

(si_w1, sj_w1),(si_w2, sj_w2), ...,(si_wM, sj_wM),

wherew₁, w₂, ..., w_N are distinct indexes chosen from{1,2, ..., N} ands_iw

1 =s₀. In order to prove thatL₂is a complete walk, we need to show that (1)s_jwM =s_χ and (2)M =N.

First, let us prove that s_jwM =s_χ. InG, letN_i^(out) denote the number of outgoing edges of s_i and letN_i⁽ⁱⁿ⁾denote the number of incoming edges ofsi, then we have that

















N₀⁽ⁱⁿ⁾= 0, N₀^(out)= 1, Nχ⁽ⁱⁿ⁾=Nχ^(out)+ 1, N_i⁽ⁱⁿ⁾=N_i^(out)fori̸= 0, i̸=χ.

Based on these relations, we know that once we have a walk starting from s0 in G, this walk will finally end at statesχ. That is because we can always get out ofsidue toN_i⁽ⁱⁿ⁾=N_i^(out)ifi̸=χ,0.

Now, we prove that M =N. This can be proved by contradiction. Assume M ̸=N, then we define

V ={w₁, w₂, ..., w_M},

V ={1,2, ..., N}/{w₁, w₂, ..., w_M},

where V corresponds to the visited edges based on L2 and V corresponds to the unvisited edges based on L2. Let v = min(V), then (si_v, sj_v) is the unvisited edge with the minimal index. Let l = iv, then (si_v, sj_v) is an outgoing edge of sl. Herel ̸=χ, because all the outgoing edges of sχ

have been visited. Assume the number of visited incoming edges ofsl isM_l⁽ⁱⁿ⁾ and the number of visited outgoing edges ofsl isM_l^(out), then

M_l⁽ⁱⁿ⁾=M_l^(out),

see figure 3.3 as an example, in which the solid arrows indicate visited edges, and the dashed arrows indicate unvisited edges..

) (out

Nl

) (in

Nl )

(out

Ml

) (in

Ml

) , ( s

iv

s

jv

) , ( s

iu

s

ju

Figure 3.3. An illustration of the incoming and outgoing edges ofsl.

Note that the labels of the outgoing edges of s_l are the same for L₁ and L₂, since l ̸= χ,0.

Therefore, based on L₁, before visiting edge (s_iv, s_jv), there must be M_l^(out) outgoing edges ofs_l have been visited. As a result, based onL₁, there must beM_l^(out)+ 1 =M_l⁽ⁱⁿ⁾+ 1 incoming edges of sl have been visited before visiting (si_v, sj_v). Among all theseM_l⁽ⁱⁿ⁾+ 1 incoming edges, there exists at least one edge (si_u, sj_u) such thatu∈V, since onlyM_l⁽ⁱⁿ⁾incoming edges of slhave been

visited based onL₂.

According to our assumption, both u, v ∈ V and v is the minimal one, so u > v. On the other hand, we know that (si_u, sj_u) is visited before (si_v, sj_v) based on L1, so u < v. Here, the contradiction happens. Therefore,M =N.

This completes the proof.

Here, let us give an example of the lemma above. We know that, whens_α=s₁ and

Λ = [s₄s₃s₁s₂, s₁s₃s₃, s₂s₁s₄, s₂s₁],

(sα,Λ) is feasible. The labeling on a directed graph corresponding to (sα,Λ) is given in figure 3.2, which is a complete walk starting at states0 and ending at states1. The path of the walk is

s0s1s4s2s1s3s2s3s1s1s2s3s4s1.

By permutating the labels of the outgoing edges ofs₁, we can have the graph as shown in figure 3.4. The new labeling onGis also a complete walk ending at states₁, and its path is

s0s1s1s2s1s3s2s3s1s4s2s3s4s1.

Based on lemma 3.5, we have the following result

Lemma 3.6. Given a starting statesα and two collections of sequences Λ = [Λ1,Λ2, ...,Λk, ...,Λn] and Γ = [Λ₁, ...,Γ_k, ...,Λ_n] such that Γ_k .

= Λ_k (tail-fixed permutation). Then (s_α,Λ) and (s_α,Γ) have the same feasibility.

Proof. We prove that if (sα,Λ) is feasible, then (sα,Γ) is also feasible. If (sα,Λ) is feasible, there exists a sequenceX such thatsα=x1 and Λ =π(X). Suppose its last element isxN =sχ.

Whenk=χ, according to lemma 3.5, we know that (sα,Γ) is feasible.

When k ̸= χ, we assume that Λk = πk(X) = xk₁xk₂...xk_w. We consider the subsequence

s

1

s

2

s

3

s

4

1

1 2

3 4

1

2 3

1 2

3 2

s

0

1

Figure 3.4. The sequence graphGwith new labels.

X = x1x2...xk_w−1 of X. Then πk(X) = Λ^|_k^Λk|−1 and the last element of X is sk. According to lemma 3.5, we can get that there exists a sequencex^′₁x^′₂...x^′_k

w−1 withx^′₁=x1 andx^′_k

w−1=xk_w−1

such that

π(x^′₁x^′₂...x^′_kw−1) = [π1(X), ...,Γ^|_k^Γk|−1, πk+1(X), ..., πn(X)],

since Γ^|_k^Γk|−1≡Λ^|_k^Λk|−1. Letx^′_k

wx^′_k

w+1...x^′_N =x_kwx_kw+1...x_N, i.e., concatenatingx_kwx_kw+1...x_Nto the end ofx^′₁x^′₂...x^′_k

w−1, we can generate a sequencex^′₁x^′₂...x^′_N such that its exit sequence of statesk is

Γ^|_k^Γk|−1∗xk_w= Γk,

and its exit sequence of statesi withi̸=kis Λi=πi(X).

So if (s_α,Λ) is feasible, then (s_α,Γ) is also feasible. Similarly, if (s_α,Γ) is feasible, then (s_α,Λ) is feasible. As a result, (s_α,Λ) and (s_α,Γ) have the same feasibility.

According to the lemma above, we know that

(sα,[Λ1,Λ2, ...,Λn]) and (sα,[Γ1,Λ2, ...,Λn]) have the same feasibility,

(s_α,[Γ₁,Λ₂, ...,Λ_n]) and (s_α,[Γ₁,Γ₂, ...,Λ_n]) have the same feasibility, ... ,

(sα,[Γ1,Γ2, ...,Γn−1,Λn]) and (sα,[Γ1,Γ2, ...,Γn−1,Γn]) have the same feasibility.

So the statement in the main lemma is true.