Foundations of Data Science

The Gaussian distribution on the interval [-1,1]
Consider the interval [−1, 1]. Let δ be a “grid size” specified later and let G be the
graph consisting of a path on the 2 + 1 vertices {−1, −1 + δ, −1 + 2δ, . . . , 1 − δ, 1} having
δ
self loops at the two ends. Let π x = ce −αx2 for x ∈ {−1, −1 + δ, −1 + 2δ, . . . , 1 − δ, 1}
where α > 1 and c has been adjusted so that ∑ x π x = 1.
We now describe a simple Markov chain with the π x as its stationary probability and
argue its fast convergence. With the Metropolis-Hastings’ construction, the transition
probabilities are
( )
( )
p x,x+δ = 1 2 min 1, e−α(x+δ)2
and p x,x−δ = 1 e −αx2 2 min 1, e−α(x−δ)2
.
e −αx2
Let S be any subset of states with π(S) ≤ 1 . First consider the case when S is an interval
2
[kδ, 1] for k ≥ 2. It is easy to see that
π(S) ≤
≤
∫ ∞
x=(k−1)δ
∫ ∞
= O
(k−1)δ
(
ce −αx2 dx
x
(k − 1)δ ce−αx2 dx
)
ce −α((k−1)δ)2
α(k − 1)δ
Now there is only one edge from S to ¯S and total conductance of edges out of S is
∑ ∑
π i p ij = π kδ p kδ,(k−1)δ = min(ce −αk2 δ 2 , ce −α(k−1)2 δ 2 ) = ce −αk2 δ 2 .
i∈S
j /∈S
Using 2 ≤ k ≤ 1/δ, α ≥ 1, and π( ¯S) ≤ 1, we have
Φ(S) =
flow(S, ¯S)
π(S)π( ¯S) ≥ δ2 α(k − 1)δ
ce−αk2
ce −α((k−1)δ)2
≥ Ω(α(k − 1)δe −αδ2 (2k−1) ) ≥ Ω(αδe −O(αδ) ).
.
For δ < 1 α , we have αδ < 1, so e−O(αδ) = Ω(1), thus, Φ(S) ≥ Ω(αδ). Now, π min ≥ ce −α ≥
e −1/δ , so ln(1/π min ) ≤ 1/δ.
If S is not an interval of the form [k, 1] or [−1, k], then the situation is only better
since there is more than one “boundary” point which contributes to flow(S, ¯S). We do
not present this argument here. By Theorem 5.5 in Ω(1/α 2 δ 3 ε 3 ) steps, a walk gets within
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