Mixing Time – General Chains
Seminar on Random Walks on Graphs
2009/2010
Ilan Ben Bassat
Omri Weinstein
Mixing Time - Lazy Chains
Given that:
Then:
 max
2

8
1
2 t
| Pt (w)   (w) |
(1  )
2
 min
Proof Sketch
• Bound distance to stationary distribution by some function h:
| pt ( S )   ( S ) | ht ( x)
• Bound the value of ht(x) inductively.
• Derive a bound on the mixing time.
Proof Sketch
• Bound distance to stationary distribution by some function h:
| pt ( S )   ( S ) | ht ( x)
• Bound the value of ht(x) inductively.
• Derive a bound on the mixing time.
Bounding Distributions Distance
For every
0  x 1
ht ( x)  max{  ( p ( w)   ( w)) ( w)}
t
w
  [0,1] ,   ( w) ( w)  x}
n
w
ht ( x)  max{  ( p t ( w)   ( w)) ( w)}   ( w) ( w)  x}
w
w
So, choosing   1
wS
for every S yields:
| pt ( S )   ( S ) | ht ( ( S ))
And for every vertex w we can get:
| pt (w)   (w) | min{ ht ( (w)), ht (1   (w))}
H(x) Values
Order the graph vertices:
And define
k
pt ( w1 )
p ( w2 )
p ( wn )
 t
 ...  t
 ( w1 )
 ( w2 )
 ( wn )
 k    ( wi ). Find k
i 1
such that:
Then the value of ht(x) is obtained by:
And we get:
 k 1  x   k
Function H - Analysis
• Piece-wise linear function.
• Concave
• Breakpoints at 0  0  1   ...   n  1
• On the interval [0,1]: ht (0)  ht (1)  0 and 0  ht ( x)  1
Proof Sketch
• Bound distance to stationary distribution by some function h:
| pt ( S )   ( S ) | ht ( x)
• Bound the value of ht(x) inductively.
• Derive a bound on the mixing time.
Proof Sketch and Intuition
We will prove for every x and t:
1
ht ( x)  (ht 1 ( x   )  ht 1 ( x   ))
2
Intuition: Bound the value h(x) in time t by going one step
backwards, and a bit towards the endpoints.
Prove will be in three parts:
 For a finite group of discrete x values x   i .
 For all x values satisfying   x    12 (and symmetric case).
 For all x values   x   and   1  
k 1
k 1
k
k
k 1
2
k
Proof – Breakpoints
1
ht ( x)  (ht 1 ( x  2 min{ x,1  x})  ht 1 ( x  2 min{ x,1  x}))
2
Fix k and let ui   Pi , j
j k
n
x  k
ht ( x)   ( pt 1 ( wi )   ( wi ))ui
i 1
n
  ( wi )ui  x
i 1
x {1... n }
t 1
Proof – cont’d
1
1
ht ( x)  ht 1 ( x' )  ht 1 ( x' ' )
2
2
x'  x  2 min{ x,1  x}
x' '  x  2 min{ x,1  x}
1
ht ( x)  (ht 1 ( x  2 min{ x,1  x})  ht 1 ( x  2 min{ x,1  x}))
2
Proof – second case
Proof – Third Part
l 1  x  l
ht ( x) 
 l 1 
1
 l
2
1
(ht 1 ( x  2( x  (1   )( 2 l  1)))  ht 1 ( x  2( x  (1   )( 2 l  1)))
2
x  2
(1   )( 2 l  1)
 2  4( l   l 1 )  2  4 max
x
So, for every x we get:
Base Case
Induction
For t=0 trivial.
For t  1 and
0 x
1
2
1
ht ( x)  (ht 1 ( x   x x)  ht 1 ( x   x x))
2
1
 2 t 1
 2 t 1
 (C x   x x (1 
)  C x   x x (1 
) )
2
2
2
( 1   x  1   x )
 2 t 1
 C (1 
)
x
2
2
Induction
( 1   x  1   x )
2
 (1 
)
2
2
2
1   x  1   x  2 
4
1   x  2  1  2 2
2
1
 x2
4
 2
8 max   2
Proof Sketch
• Bound distance to stationary distribution by some function h:
| pt ( S )   ( S ) | ht ( x)
• Bound the value of ht(x) inductively.
• Derive a bound on the mixing time.
Mixing Time Proof
2 t
| Pt ( w)   ( w) | ht ( ( w))  C min{  ( w) , 1   ( w) }(1 
)
2
1
2 t

(1 
)
2
 min
Given  max
2

8