Spatial Random Processes
Introduction
Literature
-
R. van der Hofstad: Random graphs and complex nerworks.
www.win.tue.nl/~hofstad/NotesRGCN2009.pdf
B.Bollobas: Random graphs
For probability background – R.Durrett, probability theory and examples (2005)
Main subject
We will look at the Erdos-Renyi random graph (1960)
𝑛
Consider n vertices, let each one of the ( ) possible edges appear independently with
2
possibility 𝑝 ∈ [0,1]
Such a graph is referred to as 𝐺(𝑛, 𝑝).
Question: What is the size of the largest component? Denote it as 𝐶𝑚𝑎𝑥 of 𝐺(𝑛, 𝑝).
𝜆
𝑛
We define: 𝑝 = , 𝜆 > 0
Exercise: Prove that 𝐸[𝑑𝑒𝑔𝑟𝑒𝑒(𝑣)] =
𝑛−1
𝜆
𝑛
We will see that:
𝑛→∞
If 𝜆 < 1, than ∃𝐶𝜆 > 0 𝑠. 𝑡. ∀𝜖 > 0. 𝑃[(𝐶𝜆 − 𝜖) log 𝑛 ≤ |𝐶𝑚𝑎𝑥 | ≤ (𝐶𝜆 + 𝜖) log 𝑛] →
𝑛→∞
If 𝜆 > 1, then ∃𝐶′𝜆 > 0 𝑠. 𝑡. ∀𝜖 > 0 𝑃[(𝑐 ′𝜆 − 𝜖)𝑛] ≤ |𝐶𝑚𝑎𝑥 | ≤ (𝐶 ′𝜆 + 𝜖)𝑛] →
1
1
Fix vertex v. Do breadth first exploration of component 𝐶(𝑣) of 𝐺(𝑛, 𝑝) containing v vertices.
Can be either N (neutral), A(Active) or I(Inactive).
At time 𝑡 = 0, 𝑣 ∈ 𝐴 and all other vertices are in N.
At time 𝑡 ≥ 1, if 𝐴 = 𝜙, terminate.
Otherwise, choose 𝑤 ∈ 𝐴.
Put all N neighbors of 𝑊 in 𝐴. Put 𝑊 in 𝐼.
Note: Algorithm terminates after |𝐶(𝑣)| steps.
Suppose at time 𝑡,
|𝑁| = 𝑛 − 𝑘 vertices are left and |𝐴| ≥ 1.
The number of vertices that go from N to A in stem 𝑡 has the distribution:
𝜆
𝜆
𝐵𝑖𝑛(𝑛 − 𝑘, 𝑝) = 𝐵𝑖𝑛 (𝑛 − 𝑘, ) ≈ 𝐵𝑖𝑛(𝑛, )
𝑛
𝑛
𝑛
Recall that 𝑋~𝐵(𝑛, 𝑝), 𝑝 ∈ [0,1] ↔ 𝑃[𝑋 = 𝑘] = ( ) 𝑝𝑘 (1 − 𝑝)𝑛−𝑘 , 0 ≤ 𝑘 ≤ 𝑛
𝑘
Expect similarity with a branching process!
Branching Process
t=0
t=1
𝜆
𝐵𝑖𝑛(𝑛, )
𝑛
𝜆
𝐵𝑖𝑛(𝑛, )
𝑛
t=2
𝑃𝑜𝑖(𝜆)
𝑌~𝑃𝑜𝑖(𝜆), 𝜆 ∈ (0, ∞)𝑖𝑓 𝑃[𝑌 = 𝑘] =
𝜆𝑘 ----−𝜆
𝑒
, 𝑘∈ℤ
𝑘!
𝑛→∞ 𝜆𝑘
𝑒 −𝜆 ,
𝑘!
𝜆
Exercise: If 𝑋~𝐵𝑖𝑛(𝑛, 𝑛), then 𝑃[𝑋 = 𝑘] →
𝑘∈ℤ
We will see that:
If 𝜆 < 1, then branching process (𝑃𝑜𝑠(𝜆)) dies out almost surely.
If 𝜆 > 1, then it lives forever with probability >0.
Further Questions
2
-
𝜆 = 1 → |𝐶𝑚𝑎𝑥 | ≈ 𝑛3
-
If 𝜆 = 1 + 𝛿𝑛 , 𝛿𝑛 → 0
Other properties of 𝐺(𝑛, 𝑝)
Add geometry to the graph
Fix a large graph 𝐺𝑛
Retain (delete) every edge independently with probability 𝑝.
If 𝐺𝑛 = 𝑘𝑛 (the full graph), we get 𝐺(𝑛, 1 − 𝑝)
Consider a random walk on 𝐺𝑛 , study the components 𝐺𝑛 \ {𝑋𝑡1 , … , 𝑋𝑡𝑛 }, 𝑡𝑛 ≥ 0
Study random graphs s.t. 𝑃[𝑑𝑒𝑔𝑟𝑒𝑒(𝑣) ≥ 𝑘] ≈ 𝑘 −𝜏 , 𝜏 > 0
-
𝑛→∞
Branching Processes
Time 𝑛 = 0, 1, …
𝑋1,1
t=0
𝑋2,1
𝑋2,𝑋1,1
t=1
t=2
In generation 𝑛 ≥ 0, individual 𝑖 gives birth to 𝑋𝑛+1,𝑖 children, then dies.
Formally, we define {𝑋𝑛,𝑖 |𝑛 ≥ 1, 𝑖 ≥ 1} as independent, identically distributed (iid) random
variables with values in ℤ > 0
Define 𝑍𝑛 , (𝑛 ≥ 0) recursively by
𝑍0 = 1
⋮
𝑍𝑛=∑𝑍𝑛 −1 𝑋
𝑖=1
𝑛,𝑖
𝑍0 = to the total number of individuals in generation 1.
Notation 𝑋 = 𝑋1,1 (since all distribute identically)
𝑃𝑖 = 𝑃[𝑋 = 𝑖] =probability that individual has 𝑖 children.
𝜂 = 𝑃 [⋃{𝑍𝑛 = 0}] = 𝑃[𝑝𝑟𝑜𝑐𝑒𝑠𝑠 𝑑𝑖𝑒𝑠 𝑜𝑢𝑡]
𝑛≥1
For 𝑠 ∈ [0,1]
𝐺𝑥 (𝑠) = 𝐸[𝑆 𝑥 ]
“offspring generation function”
Theorem 1.1:
- If 𝐸[𝑥] < 1, 𝑡ℎ𝑒𝑛 𝜂 = 1
- If 𝐸[𝑥] > 1, 𝑡ℎ𝑒𝑛 𝜂 < 1
- If 𝐸[𝑥] = 1, 𝑎𝑛𝑑 𝑃[𝑋 = 1] < 1, 𝑡ℎ𝑒𝑛 𝜂 = 1
Moreover, 𝜂 is the smallest solution in [0,1] of 𝜂 = 𝐺𝑥 (𝜂)
Exercise 1.1: Prove 𝐸[𝑍𝑛 ] = 𝜇𝑛 , where 𝜇 = 𝐸[𝑋]
Exercise 1.2: For 𝜇 < 1, prove 𝐸[𝑇] =
--------End of lesson 1
1
1−𝜇
where 𝑇 = ∑∞
𝑛=0 𝑍𝑛
Reminder
-
{𝑥𝑛,𝑖 }𝑛,𝑖≥1 𝑖𝑖𝑑, ℤ ≥ 0 − 𝑣𝑎𝑙𝑢𝑒𝑑
𝑍𝑛−1
- 𝑍0 = 1, 𝑍𝑛 = ∑𝑖=1
𝑋𝑛,𝑖
- 𝑋 = 𝑋1,1
- 𝑃𝑖 = 𝑃[𝑋 = 𝑖], 𝑖 = 𝑖𝑛 ℤ ≥ 0
- 𝜂 = 𝑃[⋃𝑛≥0{𝑍𝑛 = 0}]
- 𝐺𝑥 (𝑠) = 𝐸[𝑠 𝑥 ], 𝑠 ∈ [0,1]
Theorem 1.1 (survival vs. extinction):
if 𝐸[𝑋] < 1 then 𝜂 = 1,
If 𝐸[𝑋] > 1 then 𝜂 < 1,
If 𝐸[𝑋] = 1 and 𝑃[𝑋 = 1] < 1 then 𝜂 = 1
𝜂 is the smallest solution in [0,1] of 𝜂 = 𝐺𝑥 (𝜂)
Example:
𝑋~𝑃𝑜𝑖(𝜆), 𝜆 > 0
∞
𝑖
𝐺𝑥 (𝑠) = ∑∞
𝑖=0 𝑠 𝑝𝑖 = ∑𝑖=0
𝑠𝑖 𝜆𝑖 −𝜆
𝑒
𝑖!
= 𝑒 𝜆(𝑠−1)
Proof: First prove that 𝜂 = 𝐺𝑥 (𝜂)
Set 𝜂𝑛 = 𝑃 [
𝑍𝑛 = 0 ]
⏟
𝑖𝑛𝑐𝑟𝑒𝑎𝑠𝑖𝑛𝑔 𝑖𝑛 𝑛
By 𝜎 additivity, 𝜂𝑛 ↑ 𝜂 as 𝑛 → ∞.
Let 𝐺𝑛 (𝑠) = 𝐸[𝑠 𝑍𝑛 ], 𝑠 ∈ [0,1]
(↑ means that 𝜂𝑛 ≤ 𝜂𝑛+1 , 𝜂𝑛 → 𝜂)
Note that is we set 𝐺𝑛 (0) = 𝜂𝑛
𝑖
𝐺𝑛 (0) = ∑∞
𝑖=1 0 𝑃[𝑍𝑛 = 𝑖] = 𝜂𝑛 + 0 + 0 + ⋯ + 0 = 𝜂𝑛 .
𝑍𝑛
𝐺𝑛 (𝑠) = ∑∞
𝑖=0 𝐸[𝑠 1{𝑍𝑛 =𝑖} ] , where 1𝐴 (𝑤) = {
TODO: Draw the full drawing
𝑡=1
𝑡=𝑛
1 𝑖𝑓 𝑤 ∈ 𝐴
0 𝑖𝑓 𝑤 ∈ Ω\𝐴
(𝑍1 )
(1)
𝑍𝑛 = 𝑍𝑛 + ⋯ + 𝑍𝑛
(𝑖)
where 𝑍𝑛 Is the number of generation n-individuals descending from
(𝑖) (𝑑)
generation 1-individual 𝑖 and 𝑍𝑛 = 𝑍𝑛−1
(𝑖)
(𝑍𝑛 )
𝑖≥0
𝑖
𝑖
𝑍𝑛−1 ]
𝑍𝑛−1 ]
are independent of 𝑍1 hence 𝐺𝑛 (𝑠) = ∑∞
𝑝𝑖 = ∑∞
𝑝𝑖 =
𝑖=0 𝐸[𝑠
𝑖=0 𝐸[𝑠
𝑖
∑∞
𝑖=0 𝐺𝑛−1 (𝑠) 𝑝𝑖 = 𝐺𝑥 (𝐺𝑛−1 (𝑠)).
= 𝐺1 (𝐺𝑛−1 (𝑠))
𝑠 = 0: 𝜂𝑛 = 𝐺1 (𝜂𝑛−1 ), 𝑛 → ∞
𝜂 → 𝐺1 (𝜂)
By the dominated convergence theorem.
𝐺1 (𝜂𝑛−1 ) = 𝐸[(𝜂𝑛−1 )𝑍1 ]
𝑛→∞
(𝜂𝑛−1 ) 𝑧1 →
𝜂 𝑍1
Let 𝜓 ∈ [0,1], 𝜓 = 𝐺𝑥 (𝜓)
Now we need to show that 𝜂 ≤ 𝜓, or equivalently, 𝜂𝑛 ≤ 𝜓 ∀𝑛.
Indeed: 𝜂0 = 0 ≤ 𝜓 ∈ [0,1], if 𝜂𝑛−1 ≤ 𝜓, then 𝜂𝑛 = 𝐺𝑥 (𝜂𝑛−1 ) ≤ 𝐺𝑥 (𝜓)
𝑏𝑦 𝑎𝑠𝑠𝑢𝑚𝑝𝑡𝑖𝑜𝑛
=
𝜓.
Case 𝑃[𝑋 ∈ {0,1}] = 1
𝑃[𝑋 = 0] = 𝑝 > 0
𝑛→∞
𝜂𝑛 = 1 − 𝑃[𝑍𝑛 > 0] = 1 − (1 − 𝑝)𝑛 →
1.
Now assume 𝑃[𝑥 ≤ 1] < 1.
𝐺𝑥′ (𝑠) = 𝐸[𝑋𝑠 𝑋−1 ] > 0
𝐺𝑥′′ (𝑠) = 𝐸[𝑋(𝑋 − 1)𝑠 𝑋−2 ] > 0, 𝑠 ∈ (0,1).
(Ex: Prove this by using the mean value theorem and the dominant convergence theorem.)
𝐺𝑥 (1) = 1
TODO: Draw the graph of the function 𝐺𝑥 (𝑠)
Either 𝑠 = 𝐺𝑥 (𝑠) has one or two solutions in [0,1].
One solution ↔ 𝐺𝑥′ (1) ≤ 1
𝐺𝑥′ (1) = 𝐸[𝑋]
Exercise 1.4:
Compute 𝜂
If 𝑝0 = 1 − 𝑝, 𝑝2 = 𝑝, 𝑝 ∈ [0,1]
“Binary branching”.
Exercise 1.5: Prove 𝑃[𝑍𝑛 > 0] ≤ 𝜇𝑛 , 𝜇 = 𝐸[𝑋].
Hint: use the Chebishev inequality and the previous exercise.
Dominated Convergence Theorem
𝑛→∞
Let (𝑋𝑛 )𝑛≥1 be a sequence of random variables, 𝑋𝑛 → 𝑋 a.s.
Assume that ∃ a random variable 𝑌 ≥ 0 almost surely 𝐸[𝑌] < ∞ such that |𝑋𝑛 | ≤ 𝑌 ∀𝑛 almost
𝑛→∞
surely, Then 𝐸[𝑋𝑛 ] →
𝐸[𝑋].
Exercise 1.6:
Use the Dominated Convergence Theorem to fill in the details of the proof of theorem 1.1.
Change of Notation (random walk perspective)
TODO: Draw the drawing
𝑡=0
𝑡 =n
Redenote 𝑋𝑛,𝑖 as they appear in breadth-first search exploration of the tree.
𝑖. 𝑒. (𝑋1,1 , 𝑋2,1 , … , 𝑋2,2 , 𝑋3,1 , … ) =: (𝑋1 , 𝑋2 , 𝑋3 , … )
Define the random variables 𝑁𝑘 , 𝐼𝑘 , 𝑘 ≥ 0 such that 𝑋𝑘 = 𝑋(𝑁𝑘 ,𝐼𝑘)
Observation: (𝑁𝑘 , 𝐼𝑘 ) depend only on (𝑋1 , … , 𝑋𝑘−1 )
Lemma 1.7:
(𝑋𝑘 )𝑘≥1 are iid with the same distribution as 𝑋.
Proof: Let 𝑋1 , … , 𝑋𝑘 ∈ ℤ ≥ 0
𝑃[𝑋1 = 𝑥1, … , 𝑋𝑘 = 𝑥𝑘 ] = ∑ ∑ 𝑃 [(𝑋1 = 𝑥1 , … , 𝑋𝑘−1 = 𝑥𝑘−1 , 𝑁𝑘 = 𝑛, 𝐼𝑘 = 𝑖 ) 𝑋𝑛,𝑖 = 𝑥𝑘 ]
𝑛≥1 𝑖≥1
𝑛−1 ∞
𝐴
𝑖−1
𝐴 depends only on 𝜎 ({𝑋𝑚,𝑗 }𝑚=1,𝑖=1 {𝑋𝑛,𝑗 }𝑗=1 )
∑𝑛 ∑𝑖 𝑃[𝐴]𝑃[𝑋 = 𝑥𝑘 ] = 𝑃[𝑋1 = 𝑥1 , … , 𝑋𝑘−1 = 𝑥𝑘−1 ]𝑃[𝑋 = 𝑥𝑘 ]
𝑃[𝑋 = 𝑥𝑘 ]
𝑖𝑛𝑑𝑢𝑐𝑡𝑖𝑜𝑛
=
𝑃[𝑋 = 𝑥1 ] ∙ … ∙
Independence of 𝑋𝑛,𝑖
Notation (ctd): (𝑆𝑖 )𝑖≥0 equals to the number of active individuals at stage 𝑖.
𝑆0 = 1
𝑆𝑖 = 𝑆𝑖−1 + (𝑋𝑖 − 1), 𝑖 ≥ 1 = (𝑋1 , … , 𝑋𝑖 ) − (𝑖 − 1) = number of active individuals.
𝑇 = (∑𝑛≥0 𝑍𝑛 ) = inf{𝑖 ≥ 0 ∶ 𝑆𝑖 = 0}
𝑆0 = 1
𝑆1 = 1 + (2 − 1)
𝑆2 = 2 + (2 − 1)
𝑆3 = 3 + (0 − 1)
𝑆4 = 2 + (0 − 1) = 1
𝑠5 = 1 + (0 − 1) = 0
------- end of lesson 2
In the proof of theorem 1.1 we showed: 𝐺𝑛 (𝑠) = 𝐸[𝑆𝑍𝑛 ] = 𝐺𝑋 (𝐺𝑛−1 (𝑠)), 𝑠 ∈ [0,1]
Exercise 1.7: Set 𝐺𝑇 (𝑠) = 𝐸[𝑆 𝑇 ], 𝑠 ∈ [0,1]
Prove that 𝐺𝑇 (𝑠) = 𝑠 ∙ 𝐺𝑋 (𝐺𝑇 (𝑠)), where 𝐺𝑋 (𝑠) = 𝐸[𝑆 𝑋 ]
Exercise 1.8: Prove the following :
If 𝜇 = 𝐸[𝑋] = 1, then 𝐸[𝑇] = ∞
(hint: Prove first that 𝐸[𝑇] = 1 + 𝐸[𝑇])
Exercize 1.9: Recall 𝐸[𝑍𝑛 ] = 𝜇𝑛
-
𝑍
Prove that (𝑀𝑛 = 𝜇𝑛𝑛 )
𝑛≥0
is a martingale (w.r.t. natural filtration). Show that:
lim 𝑀𝑛 = 𝑀∞ exists and 0 ≤ 𝑀𝑛 ≤ ∞ almost surely.
𝑛→∞
-
For 𝜇 > 1and 𝐸[𝑋 2 ] < ∞ show that (𝑀𝑛 )𝑛≥0 is bounded in 𝐿2 . Use theorem 1.1 and
onvergence theorem (𝐿2 ) to prove that 𝑃[𝑀∞ = 0] = 𝜂 = 𝑃[𝑇 < ∞]. Deduce that
{𝑀∞ = 0} = {𝑇 < ∞} almost surely.
(⇒ 𝑍𝑛 ~(𝑀∞ ∙ 𝜇𝑛 )𝑜𝑛 {𝑇=∞} )
Theorem 1.10: (Extinction with large total progeny)
𝑒 −𝐼𝑘
If 𝜇 = 𝐸[𝑋] > 1, then 𝑃[𝑘 ≤ 𝑇 < ∞] ≤ 1−𝑒 𝐼 where 𝐼 = sup(𝑡 − log 𝐸[𝑒 𝑡𝑋 ]) > 0
𝜆
Remark: In 𝐺 (𝑛, 𝑛) , 𝜆 > 1, we will see that ∃𝑘 > 0 𝑠. 𝑡. there are no components of size
1
between 𝑘 log 𝑛 and 𝑘 ∙ 𝑛, with high probability as 𝑛 → ∞.
Lemma 1.11: (Cramer/Chernoff bound):
Let {𝑋𝑖 }∞
𝑖=1 be iid.
1. For any 𝑎 ∈ (𝐸[𝑋1 ], ∞), 𝑃[∑𝑛𝑖=1 𝑋𝑖 ≥ 𝑛𝑎] ≤ 𝑒 −𝑛𝐼(𝑎)
2. For any 𝑎 ∈ (−∞, 𝐸[𝑋1 ]), 𝑃[∑𝑛𝑖=1 𝑋𝑖 ≤ 𝑛𝑎] ≤ 𝑒 −𝑛𝐼(𝑎)
Where 𝐼(𝑎) = sup(𝑡𝑎 − log 𝐸[𝑒 𝑡𝑋1 ]) in (1)
t≥0
and 𝐼(𝑎) = sup(𝑡𝑎 − log 𝐸[𝑒 𝑡𝑋1 ]) in (2)
t≤0
Proof of Lemma 1.11:
𝑛
𝑛
𝑛
𝑃 [∑ 𝑋𝑖 ≥ 𝑛𝑎] = 𝑃 [exp (−𝑡𝑛𝑎 + 𝑡 ∑ 𝑋𝑖 ) ≥ 1] = 𝐸[1𝐴 ] ≤ 𝐸 1𝐴 ∙ exp (−𝑡𝑛𝑎 + 𝑡 ∑ 𝑋𝑖 )
⏟
𝑖=1
𝑖=1
𝑖=1
∀𝑡≥0
[
]
≥1
𝑛
𝑛
≤ 𝐸 exp (−𝑡𝑛𝑎 + 𝑡 ∑ 𝑋𝑖 ) = 𝑒
⏟
𝑖=1
[
]
≥0
𝑡𝑋𝑖 ])
= exp(−𝑛(𝑡𝑎 − log 𝐸[𝑒
)
Now optimize over 𝑡 ≥ 0 ⇒(1).
−𝑡𝑛𝑎
𝐸 [∏ 𝑒
𝑛
𝑡𝑋𝑖
]=𝑒
−𝑡𝑛𝑎
𝑖=1
∏ 𝐸[𝑒 𝑡𝑋𝑖 ]
𝑖=1
(2) is left as an exercise!
Proof of theorem 1.10:
∞
∞
∞
𝑃[𝑘 ≤ 𝑇 < ∞] = ∑ 𝑃 [ 𝑇
⏟= 𝑛 ] ≤ ∑ 𝑃[𝑋1 + ⋯ + 𝑋𝑛 = 𝑛 − 1] = ∑ 𝑃[𝑋1 + ⋯ + 𝑋𝑛 ≤ 𝑛]
𝑛=𝑘
⊆{𝑆𝑛 =0}
𝑛=𝑘
𝑛=𝑘
𝑆𝑛 = (𝑋1 + ⋯ + 𝑋𝑛 ) − (𝑛 − 1)
𝑙𝑒𝑚𝑚𝑎 1.11,(2),𝑎=1
𝑃[𝑘 ≤ 𝑇 < ∞]
≤
∞
∑ 𝑒 −𝑛𝐼(1)
𝑛=𝑘
Define – 𝑓(𝑡) = 𝑡 − log 𝐸[𝑒 𝑡𝑋 ] , 𝑡 ≤ 0
𝑓(0) = 0 − log 1 = 0
1
𝑓 ′ (𝑡) | = 1 −
∙ 𝐸[𝑋𝑒 0𝑋 ] = 1 − 𝐸[𝑋] < 0
0𝑋 ]
𝐸[𝑒
𝑡=0
Since the derivative is negative, there is some negative value where 𝑓(𝑡) approaches from
above.
So ∃𝑡 < 0: 𝑓(𝑡) > 0 ⇒ 𝐼(1) > 0
Duality Principle
Assume 𝜂 ∈ (0,1)
(𝜂 = 𝑃[𝑇 < ∞])
Condition the branching process to die out. (only consider the probability 𝑃[𝐴|𝑇 < ∞] =
𝑃[𝐴∩{𝑇<∞}]
𝑃[𝑇<∞]
(A event)
Question: How does this change the branching process?
Definition 1.12: (Conjugate Distribution)
Let 𝑝𝑥 = 𝑃[𝑋 = 𝑥], 𝑥 ∈ ℤ≥0 and assume 𝑝0 > 0 (⇔ 𝜂 > 0).
Then the conjugate distribution (𝑝𝑥′ )𝑥∈ℤ≥0 associated to 𝑝 is defined as 𝑝𝑥′ = 𝜂 𝑥−1 𝑝𝑥 , 𝑥 ∈ ℤ≥0
Lemma 1.13:
∑ 𝑝𝑥′ = 1
𝑥≥0
Proof:
∑ 𝑝𝑥′ = ∑ 𝜂 𝑥−1 𝑝𝑥 =
𝑥≥0
𝑥≥0
1
1
1
∑ 𝜂 𝑥 𝑝𝑥 = 𝐺𝑋 (𝜂) = ∙ 𝜂 = 1
𝜂
𝜂
𝜂
𝑥≥0
Denote the branching process with offspring distribution 𝑝′ (𝑝′ branching process) by
′
{𝑋𝑛,𝑖
}𝑛≥1,𝑖≥1
Lemma 1.14 (𝜂 > 0) 𝑃[𝑇 ′ < ∞] = 1
Proof: Let 𝑠 ∈ [0,1], assume that 𝐺𝑋 ′ (𝑠) = 𝑠. Want to show ⇒ 𝑠 = 1, use theorem 1.1.
∞
∞
1
𝐺𝑋 ′ (𝑠) = ∑ 𝑠 𝑃[𝑋 = 𝑖] = ∑ 𝑠 𝑖 (𝜂 𝑖−1 𝑝𝑖 ) = 𝐺𝑋 (𝑠𝜂) = 𝑠
𝜂
𝑖
′
𝑖=0
𝑖=0
So
𝑠𝜂≥𝜂
𝐺𝑋 (𝑠𝜂) = 𝑠𝜂 ⇒
∎
------ End of lesson 3
𝑠𝜂 ≥ 𝜂 ⇒ 𝑠 = 1
Duality Principle
Assume that 𝜂𝑃[𝑇 < ∞] ∈ (0,1)
Given 𝑝𝑥 = 𝑃[𝑋 = 𝑥], 𝑥 ≥ 0, 𝑝0 > 0
Define conjugate distribution 𝑝′ by 𝑝𝑥′ = 𝜂 𝑥−1 𝑝𝑥 , 𝑥 ∈ ℤ≥0
′
{𝑋𝑛,𝑖
}𝑛≥1,𝑖≥1 𝑝′ - A branching process
(𝑇 ′ , 𝑍𝑛′ , 𝑋1′ , 𝑋2′ … ) then 𝑃[𝑇 ′ < ∞] = 1
Theorem 1.15: (duality princinple)
Let 𝑝 = (𝑝𝑥 )𝑥∈ℤ≥0 be a distribution on ℤ≥0 with 𝑝0 > 0 and let 𝑝′ be its conjugate distribution.
Let (𝑋𝑘 )𝑘≥1 be a branching process.
Let (𝑋𝑘′ )𝑘≥1 be a branching process
Then 𝑃[(𝑋1 , … , 𝑋 𝑇 ′ ) = (𝑥1 , … , 𝑥𝑡 )|𝑇 < ∞] = 𝑃[(𝑋1′ , … , 𝑋𝑇′ ) = (𝑥1 , … , 𝑥𝑡 )|𝑇 < ∞]
For 𝑥1 , … , 𝑥𝑛 ∈ ℤ≥0 .
In words, 𝑝-branching process conditioned to die out is distributed as a 𝑝′ branching process.
Proof: Denote 𝐻 = (𝑋1 , … , 𝑋𝑇 ), 𝐻 ′ = (𝑋1′ , … , 𝑋𝑇′ ′ ), ℎ = (𝑥1 , … , 𝑥𝑡 )
Recall:
𝑆𝑖 = (𝑋1 + ⋯ + 𝑋𝑖 ) − (𝑖 − 1)
𝑇 = inf{𝑖 ≥ 1|𝑆𝑖 = 0}
𝑠𝑖 = (𝑥1 , … , 𝑥𝑖 ) − (𝑖 − 1)
Assume that 𝑆𝑖 > 0 for 1 ≤ 𝑖 < 𝑡 and 𝑆𝑡 = 0.
Otherwise (1) otherwise the equation is true trivially (0 = 0)
𝑡
𝑡
𝑖=1
𝑖=1
𝑃[𝐻 = ℎ, 𝑇 < ∞] 𝐵𝑦 𝑙𝑒𝑚𝑚𝑎 1.7 1
1
𝑃[𝐻 = ℎ|𝑇 < ∞] =
=
∏ 𝑝𝑥𝑖 = ∏ 𝑝𝑥′ 𝑖 𝜂1−𝑥𝑖
𝑃[𝑇 < ∞]
𝜂
𝜂
𝑡
=
1 𝑡−∑𝑡 𝑥
∙ 𝜂 𝑖=1 𝑖 ∏ 𝑝𝑥′ 𝑖
𝜂
⏟
𝑖=1
𝑃[𝐻 ′ =ℎ]
∑𝑡𝑖=1 𝑥𝑖
But 𝑡 −
= 𝑡 − (𝑠𝑡 + (𝑡 − 1)) = 𝑡 − 𝑡 + 1 = 1
′
So it equals: 𝑃[𝐻 = ℎ]
Which is what we needed to prove.
Exercise 1.16:
Let 𝑋 be 𝑃𝑜𝑖(𝜆) - distributed, 𝜆 > 0.
Recall from Lemma 1.11
𝐼(𝑎) = sup(𝑡𝑎 − log 𝐸[𝑒 𝑡𝑥 ]) , 𝑎 > 0
(𝑡 ≥ 0 if 𝑎 > 𝜆)
(𝑡 ≤ 0 if 𝑎 < 𝜆)
Show that in both cases (𝑎 > 𝜆, 𝑎 < 𝜆)
𝑎
𝐼(𝑎) = sup (𝑡𝑎 − log 𝐸[𝑒 𝑡𝑥 ]) = 𝜆 − 𝑎 + 𝑎 log ( )
𝜆
t∈Rℝ
Exercise 1.17 (coupling)
Let (𝑝𝑥 )𝑥∈ℤ , (𝑞𝑥 )𝑥∈ℤ be different probability distributions on ℤ.
We want a random variable (𝑋, 𝑌), ℤ2 valued, such that
𝑃[𝑋 = 𝑥] = 𝑝𝑥 , 𝑥 ∈ ℤ≥0
𝑃[𝑌 = 𝑥] = 𝑞𝑥 , 𝑥 ∈ ℤ≥0
𝑃[𝑋 ≠ 𝑌] as small as possible.
Prove that:
1
2
(𝑝𝑥 −min{𝑝𝑥 ,𝑞𝑥 })(𝑞𝑥 −min{𝑝𝑥 ,𝑞𝑥 })
𝑑𝑇𝑉 (𝑝,𝑞)
(i)
∑𝑥∈ℤ(𝑝𝑥 − min{𝑝𝑥 , 𝑞𝑥 }) = ∑𝑥∈ℤ(𝑞𝑥 − min{𝑝𝑥 , 𝑞𝑥 }) = ∑𝑥∈ℤ|𝑝𝑥 − 𝑞𝑥 | = 𝑑𝑇𝑉 (𝑝, 𝑞)
(ii)
𝑐𝑥,𝑦 = min{𝑝𝑥 , 𝑞𝑥 } if 𝑥 = 𝑦, and
(iii)
otherwise. Is a
distribution on ℤ2
With 𝑃[(𝑋, 𝑌) = (𝑥, 𝑦)] = 𝑐𝑥,𝑦 , (𝑥, 𝑦) ∈ ℤ2
get 𝑃[𝑋 ≠ 𝑌] = 𝑑𝑇𝑉 (𝑝, 𝑞) and this is the best.
if 𝑃[(𝑋, 𝑌) = (𝑥, 𝑦)] = 𝑐′𝑥,𝑦 , 𝑃[𝑋 = 𝑥] = 𝑝𝑥 , 𝑃[𝑌 = 𝑥] = 𝑞𝑥 ⇒ 𝑃[𝑋 ≠ 𝑌] ≥
𝑑𝑇𝑉 (𝑝, 𝑞)
Possible for any random variable (𝑋 ′ , 𝑌′) with values in ℤ2 , with 𝑃[𝑋 ′ = 𝑥] =
𝑝𝑥 , 𝑃[𝑌 ′ = 𝑥] = 𝑞𝑥 ⇒ 𝑃[𝑋 ′ ≠ 𝑌 ′ ] > 𝑑𝑇𝑉 (𝑝, 𝑞)
Poisson branching processes
Notation: Use superscript “*” for Poisson branching processes.
(𝑋𝑛∗ , 𝑋𝑖∗ , 𝑇 ∗ , … )
Recall:
𝜆𝑘 −𝜆
𝑒 ,
𝑘!
-
𝑃[𝑋 ∗ = 𝑘] =
-
𝐸[𝑋 ∗ ] = 𝜆
𝐺𝑋 ∗ (𝑠) = 𝑒 𝜆(𝑠−1) , 𝑠 ≥ 0
𝑘 ∈ ℤ≥0
Definition 2.1: Call 0 < 𝜇 < 1 < 𝜆 a conjugate pair if 𝜇𝑒 −𝜇 = 𝜆𝑒 −𝜆
Note:
𝑓: 𝑥 → 𝑥𝑒 −𝑥 is strictly increasing on [0,1]
It is strictly decreasing on [1, ∞]
𝑓(0) = 0, 𝑓(∞) = 0
TODO: Draw function…
So for any 𝜆 > 1, there is a unique conjugate 𝜇 ∈ [0,1) and vice-versa.
Theorem 2.2: (Poisson duality principle):
Let 𝜇 < 1 < 𝜆 be a conjugate pair
The 𝑃𝑜𝑖(𝜆) - a branching process conditioned on extinction is distributed as 𝑃𝑜𝑖(𝜇) - a
branching process.
Proof: By theorem 1.15, The conditioned process is a branching process with offspring
distribution 𝑝𝑥′ = 𝜂𝜆𝑥−1
𝜆𝑥 −𝜆
𝑒 ,
𝑥!
𝑥 > 0 where 𝜂𝜆 = extinction probability for 𝑃𝑜𝑖(𝜆)- branching
process. 𝜂𝜆 ∈ (0,1)
i.e.
(𝜆𝜂𝜆 )𝑥 −𝜆 1
𝑝𝑥′ =
𝑒 ( )
𝑥!
𝜂𝜆
By theorem 1.1:
𝜂𝜆 = 𝐺𝑋 ∗ (𝜂𝜆 ) = 𝑒 𝜆(𝜂𝜆 −1)
So 𝑝𝑥′ =
(𝜆𝜂𝜆 )𝑥
𝑥!
𝑒 −𝜆 ∙
1
𝑒
𝜆(𝜂𝜆 −1)
=
(𝜆𝜂𝜆 )𝑥
𝑥!
𝑒 −𝜆𝜂𝜆 , 𝑥 ≥ 0
Hence, 𝑝′ = 𝑃𝑜𝑖(𝜆𝜂𝜆 )
?
𝜆𝜂𝜆 𝑒 −𝜆𝜂𝜆 = 𝜇𝑒 −𝜇
Have to prove 𝜆𝜂𝜆 = 𝜇
𝜆𝜂𝜆 𝑒 −𝜆𝜂𝜆 = 𝜆𝑒 𝜆(𝜂𝜆 −1) 𝑒 −𝜆𝜂𝜆 = 𝜆𝑒 −𝜆
So, 𝜆𝜂𝜆 ≠ 𝜆 (since 𝜆 > 𝑞 (theorem 1.1) ⇒ 𝐸[𝑋 ∗ ] > 1 ⇒ 𝜂𝜆 > 1)
And (𝜆𝜂𝜆 )𝑒 −(𝜆𝜂𝜆 ) = 𝜆𝑒 −𝜆
Since 𝑥𝑒 −𝑥 ↑ on [0,1], ↓ on [1, ∞]
There is at most 1 𝑥 diferent from 𝜆 such that 𝑥𝑒 −𝑥 = 𝜆𝑒 −𝜆 = 𝜇
Hence, 𝜆𝜂𝜆 = 𝜇 ∎
Theorem (Cayley’s formula):
The number of labeled trees on 𝑛 vertices is equal to 𝑛𝑛−2
TODO: Draw a tree
Theorem 3.15: In reference
---- End of lesson 4
∞
∗
𝑇 = ∑ 𝑍𝑛∗ = inf{𝑖 ≥ 0: (𝑋1 + ⋯ + 𝑋𝑖 ) − (𝑖 − 1) = 0}
𝑛=1
Theorem 2.3 (Total Progeny of Poisson Branching Process):
For 𝑃𝑜𝑖(𝜆) - branching process. 𝜆 > 0
(𝜆𝑛 )𝑛−1 −𝜆𝑛
𝑃[𝑇 ∗ = 𝑛] =
𝑒
,
𝑛!
𝑛≥1
Proof: By Induction on 𝑛.
∗
Check for 𝑛 = 1: 𝑃[𝑇 ∗ = 1] = 𝑃[𝑋1,1
= 0] = 𝑒 −𝜆 which fully agrees with the formula.
𝑛 ≥ 2, assume the result holds for all 𝑛′ < 𝑛.
𝑛−1
∗
𝑃[𝑇 = 𝑛] = ∑ 𝑃[𝑇 ∗ = 𝑛, 𝑋1,1
= 𝑖]
∗
𝑖=1
If we observe the children that “sprung” from all children of the first one, the sum of all vertices
is their sum + 1 (for the root) equals the sum of all vertices in the graph.
∗
𝑇 ∗ is distributed as 1 + 𝑇 (1) + ⋯ + 𝑇 (𝑋1,1 ) , where {𝑇 (2) }𝑖≥1 are iid with the same distribution as
𝑇∗
𝑛−1
∗
𝑃[𝑇 = 𝑛] = ∑ 𝑃[𝑇 (1) + ⋯ + 𝑇 (𝑖) = 𝑛 − 1, 𝑋1,1
= 𝑖]
∗
𝑖=1
But these random variables are independent (by our assumption)
𝑛−1
=∑
∗
𝑃[𝑇 (1) = 𝑛1 , … , 𝑇 (𝑖) = 𝑛𝑖 , 𝑋1,1
= 𝑖] =
∑
𝑖=1 (𝑛1 ,…,𝑛𝑖 )
𝑛𝑘 ≥1,
∑𝑘 𝑛𝑘 =𝑛−1
𝑛−1
∑
∑
∗
𝑃[𝑇 (1) = 𝑛1 ] ∙ … ∙ 𝑃[𝑇 (𝑖) = 𝑛𝑖 ] ∙ 𝑃[𝑋1,1
= 𝑖]
𝑖=1 (𝑛1 ,…,𝑛𝑖 )
𝑛𝑘 ≥1,
∑𝑘 𝑛𝑘 =𝑛−1
Now we can use our induction hypothesis!
Each of the 𝑛𝑖 ’s is smaller than 𝑛.
𝑛−1
∑
𝑖
∑
𝑛−1−𝑖 −𝜆(𝑛−1)
𝜆
𝑒
∙ ∏(
𝑖=1 (𝑛1 ,…,𝑛𝑖 )
𝑛𝑘 ≥1,
∑𝑘 𝑛𝑘 =𝑛−1
So, so far we have shown that:
𝑘=1
(𝑛𝑘 )𝑛𝑘−1 𝜆𝑖 −𝜆
)∙ 𝑒
𝑛𝑘 !
𝑖!
𝑛−1
∗
𝑛−1 −𝜆𝑛
𝑃[𝑇 = 𝑛] = 𝜆
𝑒
1
∑
𝑖!
𝑖=1
⏟
𝑖
∑
∏(
(𝑛1 ,…,𝑛𝑖 ) 𝑘=1
𝑛𝑘 ≥1
∑𝑘 𝑛𝑘 =𝑛−1
(𝑛𝑘 )𝑛𝑘 −1
) = 𝜆𝑛−1 𝑒 −𝜆𝑛 ∙ 𝑎𝑛
𝑛𝑘 !
=:𝑎𝑛
Lemma 2.4: Let 𝐿𝑛 be the number of labeled trees on 𝑛 vertices.
Then, 𝐿𝑛 = 𝑛𝑛−2 = (𝑛 − 1)! 𝑎𝑛
𝑛𝑛−2
With the lemma 2.4, we get 𝑃[𝑇 ∗ = 𝑛] = 𝜆𝑛−1 𝑒 −𝜆𝑛 ∙ (𝑛−1)! =
(𝜆𝑛)𝑛−1
𝑛!
𝑒 −𝜆𝑛 ∎ (theorem 2.3).
How do we prove lemma 2.4? We use lemma 2.5.
Lemma 2.5 (Cayley’s Formula):
𝐿𝑛 = 𝑛𝑛−2
Example:
𝑛=5
TODO: Draw graph
Proof that lemma 2.5 ⇒ Lemma 2.4:
For 𝑖 ≥ 1, 𝑛1 , … , 𝑛𝑖 , 𝑛𝑘 ≥ 1,
∑𝑘 𝑛𝑘 = 𝑛 − 1, define 𝑡(𝑛1 ,…,𝑛𝑖) =number of labeled trees of 𝑛 vertices such that 𝑣1 has 𝑖
neighbors, (tree\{𝑣1 }) – has components with 𝑛1 , … , 𝑛𝑖 vertices.
To choose such a tree:
1. Split {𝑣2 , … , 𝑣𝑛 } into sets of size 𝑛1 , … , 𝑛𝑖
2. Choose 𝑖 labeled trees of size 𝑛1 , … , 𝑛𝑖
3. Choose a vertex in each of the 𝑖 trees, and connect it to 𝑣1
𝑖
𝑡(𝑛1 ,…,𝑛𝑖)
(𝑛𝑘 )𝑛𝑘−1
𝑛−1
𝑛 −2
𝑛 −2
=(
) ∙ (𝑛1 1 ∙ … ∙ 𝑛𝑖 𝑖 ) ∙ (𝑛1 ∙ … ∙ 𝑛𝑖 ) = (𝑛 − 1)! ∏ (
)
𝑛1 , … , 𝑛𝑖
𝑛𝑘 !
𝑘=1
Hence,
𝑛−1
𝐿𝑛 = ∑
𝑛−1
∑
𝑛𝑘 ≥1
𝑖=1
∑𝑘 𝑛𝑘 =𝑛−1
(𝑈𝑛𝑜𝑟𝑑𝑒𝑟𝑒𝑑!)
Now we’re done!
⇒ 𝐿𝑛 = (𝑛 − 1)! ∙ 𝑎𝑛
𝑡(𝑛1 ,…,𝑛𝑖) = ∑
𝑖=1
1
𝑖!
∑
𝑛𝑘 ≥1
∑𝑘 𝑛𝑘 =𝑛−1
𝑂𝑟𝑑𝑒𝑟𝑒𝑑!
𝑡(𝑛1 ,…,𝑛𝑖)
Lemma 2.5 (𝐿𝑛 = 𝑛𝑛−2 , 𝑛 ≥ 1):
Definition: A rooted tree is a tree with one distinguished vertex called “the root” and the edges
are all oriented, and oriented away from the root.
Example:
TODO: Draw an oriented tree
It’s not possible that a vertex will have 2 incoming edges. So a vertex has 1 incoming degree.
For labeled vertices 𝑣1 , … , 𝑣𝑛 , 𝐸 = {〈𝑣𝑖 , 𝑣𝑗 〉, 𝑖 ≠ 𝑗}
𝑠𝑛 = |{(𝑒1 , … , 𝑒𝑛 ) ∈ 𝐸 𝑛−1 |({𝑣1 , … , 𝑣𝑛 }, {𝑒1 , … , 𝑒𝑛−1 }) − 𝑖𝑠 𝑎 𝑟𝑜𝑜𝑡𝑒𝑑 𝑡𝑟𝑒𝑒}|
Example:
TODO: Draw vertices example
To choose such a sequence as (was supposed to be drawn) above:
1. Choose a labeled tree on {𝑣1 , … , 𝑣𝑛 }
2. Choose a root
3. Choose order in which to add the 𝑛 − 1 edges.
Hence, 𝑠𝑛 = 𝐿𝑛 ∙ 𝑛 ∙ (𝑛 − 1)! = 𝐿𝑛 ∙ 𝑛!
Alternatively, 𝑠𝑛 = 𝑠𝑛′ , where 𝑆𝑛′ is the number of rooted trees after you remove the last 𝑘
edges.
𝑠𝑛′
= |{(𝑒1 , … , 𝑒𝑛 ) ∈ 𝐸 𝑛−1 |({𝑣1 , … , 𝑣𝑛 }, {𝑒1 , … , 𝑒𝑛−1−𝑘 }) − 𝑖𝑠 𝑎 𝑟𝑜𝑜𝑡𝑒𝑑 𝑓𝑜𝑟𝑒𝑠𝑡 𝑤𝑖𝑡ℎ 𝑘 + 1 𝑐𝑜𝑚𝑝𝑜𝑛𝑒𝑛𝑡𝑠 ∀𝑘 = 0, … , 𝑛 − 1}|
𝑠𝑛 ≤ 𝑠𝑛′ obviously (every sequence in a set of 𝑠𝑛 can generate a sequence in 𝑠𝑛′ )
But also 𝑠𝑛′ ≤ 𝑠𝑛 , since for 𝑘 = 1, every forest is a rooted tree.
Suppose edges 𝑒1 , … , 𝑒𝑛−1−𝑘 have been added.
Number of choices for 𝑒𝑛−𝑘 = (𝑢, 𝑣)
⏟
𝑛
∙
⏟
𝑘
𝑢 𝑎𝑟𝑏𝑖𝑡𝑟𝑎𝑟𝑦 𝑣 𝑎𝑛𝑦 𝑟𝑜𝑜𝑡 𝑜𝑡ℎ𝑒𝑟 𝑡ℎ𝑎𝑛 𝑡ℎ𝑒 𝑜𝑛𝑒 𝑜𝑓 𝐶(𝑢)
Hence
𝑛−1
𝑆𝑛 =
𝑠𝑛′
= ∏(𝑛𝑘) = 𝑛𝑛−1 ∙ (𝑛 − 1)! = 𝑛𝑛−2 𝑛! = 𝐿𝑛 (𝑛!)
𝑘=1
--- end of lesson 5
Theorem 2.3: For 𝑃𝑜𝑖(𝜆) = 𝐵. 𝑃.
𝑃[𝑇 ∗ = 𝑛] =
(𝜆 > 0),
(𝜆𝑛 )𝑛−1 −𝜆𝑛
𝑒
,
𝑛!
𝑛≥1
Proof of Lemma 2.4:
(Correction):
𝑖 ≥ 1,
𝑛1 , … , 𝑛𝑖 ,
𝑛𝑘 ≥ 1,
∑ 𝑛𝑘 = 𝑛 − 1
𝑘
TODO: Draw correction drawing
𝑡(𝑛1 , … , 𝑛𝑖 ) = number of labeled trees on 𝑛 vertices {𝑣1 , … , 𝑣𝑛 }, where 𝑣1 belongs to labeled
edges 1, … , 𝑖.
And 𝑣1 has 𝑛𝑘 descendants attached to edge 𝑘, 𝑘 ∈ {1, … , 𝑖}
Reminder of what we did last time:
𝑛−1
(
)
⏟𝑛1 , … , 𝑛𝑖
𝑡(𝑛1 , … , 𝑛𝑖 ) =
𝑜𝑓 𝑤𝑎𝑦𝑠 𝑡𝑜
𝑝𝑎𝑟𝑡𝑖𝑡𝑖𝑜𝑛 {𝑣2 ,…,𝑣𝑛 }
𝑖𝑛𝑡𝑜 𝑔𝑟𝑜𝑢𝑝𝑠 𝑜𝑓
𝑠𝑖𝑧𝑒 𝑛1 ,…,𝑛𝑖
𝑛1 −2
𝑛 −2
=
𝑛
× … × 𝑛𝑖 𝑖
⏟1
𝑜𝑓 𝑐ℎ𝑜𝑖𝑐𝑒𝑠 𝑜𝑓
𝑙𝑎𝑏𝑒𝑙𝑒𝑑 𝑠𝑢𝑏𝑡𝑟𝑒𝑒𝑠
×
(𝑛1 × … × 𝑛𝑖 )
⏟
𝑜𝑓 𝑐ℎ𝑜𝑖𝑐𝑒𝑠 𝑜𝑓
𝑛𝑒𝑖𝑔ℎ𝑏𝑜𝑟𝑠 𝑜𝑓 𝑣1
𝑎
Remark: (𝑏 , … , 𝑏 ) - is the number of ways to partition 𝑎 objects into 𝑘 labeled bins of sizes
1
𝑘
𝑏1 , … , 𝑏𝑘
And the number of labeled trees on 𝑛 vertices equals
𝑛−1
∑
𝑖=1
1
𝑖!
∑
𝑡(𝑛1 ,…,𝑛𝑖) ∎
(𝑛1 ,…,𝑛𝑖 )
⏟𝑛𝑘≥1,∑𝑘 𝑛−1
# 𝑜𝑓 𝑡𝑟𝑒𝑒𝑠 𝑤𝑖𝑡ℎ deg(𝑣1 )=𝑖
Corollary 2.6 (differentiability of the extinction probability):
Let 𝜂(𝜆) = 𝑃𝜆 [𝑇 ∗ < ∞], 𝜆 > 1
And let 0 < 𝜇 < 1 be the conjugate of 𝜆 (cf. definition 2.1)
So 𝜇𝑒 −𝜇 = 𝜆𝑒 −𝜆
Then
−𝜂′ (𝜆) =
𝜂(𝜆)(𝜆 − 𝜇)
<∞
𝜆(1 − 𝜇)
Proof: Assume 𝜆 > 1 + 𝜖, 𝜖 > 0
By theorem 2.3:
∞
𝜂(𝜆) = ∑
𝑛=1
(𝜆𝑛)𝑛−1 −𝜆𝑛
𝑒
𝑛!
By using the mean-value theorem, and the dominant convergence theorem, we can check that
we can differentiate term-by-term.
So:
∞
𝜂
′ (𝜆)
=∑
𝑛=1
(𝜆𝑛)𝑛−1 −𝜆𝑛 (𝑛 − 1)𝜆𝑛−2 𝑛𝑛−1 −𝜆𝑛
𝑛𝑒
−
𝑒
=
𝑛!
𝑛!
𝑏𝑦 𝑡ℎ𝑒𝑜𝑟𝑒𝑚
1.15 (𝑑𝑢𝑎𝑙𝑖𝑡𝑦)
1
1
1
1
𝐸𝜆 [𝑇 1{𝑇 ∗ <∞} ] (1 − ) + 𝜂(𝜆) = 𝐸𝜆 [𝑇 ∗ |𝑇 ∗ < ∞] ∙ 𝜂(𝜆) (1 − ) + 𝜂(𝜆)
=
𝜆
𝜆
𝜆
𝜆
1
1
𝜂(𝜆) 𝑏𝑦 𝐸𝑥.1.3(?) 𝜇(𝜆) (1 − 𝜆) 𝜂(𝜆) 𝜂(𝜆)(𝜆 − 1) + 𝜂(𝜆)(1 − 𝜇)
∗
𝐸𝜇 [𝑇 ] ∙ 𝜂(𝜆) (1 − ) +
=
+
=
𝜆
𝜆
1−𝜇
𝜆
𝜆(1 − 𝜇)
𝜂(𝜆)(𝜆 − 𝜇)
=
∎
𝜆(1 − 𝜇)
∗
Recall exercise 1.17:
𝑝, 𝑞 be different distributions on ℤ, then there exists a distribution 𝑐 on ℤ2 such that if
(𝑑)
(𝑑)
(𝑑)
(𝑋, 𝑌) = 𝑐, then 𝑋 = 𝑝, 𝑌 = 𝑞 and 𝑃[𝑋 ≠ 𝑌] = 𝑑 𝑇𝑉 (𝑝, 𝑞) = 1 − ∑𝑥∈ℤ min{𝑝𝑥 , 𝑞𝑥 }
Exercise 2.7: Let (𝑋𝑖 )𝑛𝑖=1 be iid 𝑥𝑖 ~𝐵𝑒(𝑝), i.e. 𝑃[𝑋𝑖 = 1] = 𝑝, 𝑃[𝑋𝑖 = 0] = 1 − 𝑝, 𝑝 ∈ [0,1],
then show that ∑𝑛𝑖=1 𝑥𝑖 ~𝐵𝑖𝑛(𝑛, 𝑝)
Exercise 2.8: Let (𝑋𝑖 )𝑛𝑖=1 be independent, 𝑥𝑖 ~𝑃𝑜𝑖 (𝜆𝑖 ), 𝜆𝑖 > 0
Show that ∑𝑛𝑖=1 𝑋𝑖 ~𝑃𝑜𝑖(∑𝑛𝑖=1 𝜆𝑖 )
Theorem 2.9: (coupling of Binomial and Poisson random variables)
For 𝜆 > 0, there exists a ℤ2 values random variable (𝑋, 𝑌) such that:
𝜆
-
𝑋~𝐵𝑖𝑛 (𝑛, 𝑛)
-
𝑌~𝑃𝑜𝑖(𝜆)
-
𝑃[𝑋 ≠ 𝑌] ≤
𝜆2
𝑛
Proof: By Ex.1.17, there are iid random variables {(𝐼𝑘 , 𝐽𝑘 )}𝑛𝑘=1 such that:
𝜆
-
𝐼𝑘 ~𝐵𝑒 (𝑛)
-
𝐽𝑘 ~𝑃𝑜𝑖 (𝑛)
-
𝑃[𝐼𝑘 = 𝐽𝑘 ] = ∑𝑥∈ℤ≥0 min{𝑃[𝐼1 = 𝑥], 𝑃[𝐽1 = 𝑥]} =
𝜆
𝜆
𝜆 𝜆 𝜆 1−𝑥≤𝑒 −𝑥
𝜆 𝜆 𝜆
min {1 − , 𝑒 −𝑛 } + min { , 𝑒 −𝑛 } = 1 − + 𝑒 −𝑛
𝑛
𝑛 𝑛
𝑛 𝑛
Define:
𝜆
𝑛
(𝑋, 𝑌) = ∑(𝐼𝑘 , 𝐽𝑘 )
𝑘=1
𝜆
By exercise 2.7, 𝑋~𝐵𝑖𝑛 (𝑛, 𝑛)
By exercise 2.8: 𝑌~𝑃𝑜𝑖(𝜆)
Finally:
𝑛
𝜆 𝜆 𝜆
𝜆
𝑃[𝑋 ≠ 𝑌] ≤ ∑ 𝑃[𝐼𝑘 ≠ 𝐽𝑘 ] = 𝑛 ∙ ( − 𝑒 −𝑛 ) ≤ 𝜆 ∙
𝑛 𝑛
𝑛
𝑘=1
𝜆
𝑛
Exercise 2.10: Let (𝑋𝑛 )𝑛≥1 , 𝑋𝑛 ~𝐵𝑖𝑛 (𝑛, ) , 𝑌~𝑃𝑜𝑖(𝜆), 𝜆 ≥ 0.
(𝑑)
𝑛→∞
Prove that 𝑋𝑛 → 𝑌, meaning: For any 𝐴 ⊆ ℤ, 𝑃[𝑋𝑛 ∈ 𝐴] →
𝑃[𝑌 ∈ 𝐴]
Theorem 2.11 (Poisson and Binomial branching processes):
𝜆
𝑛
Let 𝑇 and 𝑇 ∗ be the total progenies of a 𝐵𝑖𝑛 (𝑛, ) - branching process 𝑇 and of a 𝑃𝑜𝑖(𝜆) branching process 𝑇 ∗, 𝜆 > 0. Then for 𝑘 ≥ 1 𝑃[𝑇 ≥ 𝑘] = 𝑃[𝑇 ∗ ≥ 𝑘] + 𝑒(𝑘, 𝑛)
Where |𝑒(𝑘, 𝑛)| ≤
2𝜆2 𝑘−1
∑𝑠=1 𝑃[𝑇 ∗
𝑛
> 𝑠] ≤
2𝜆2 𝑘
𝑛
Proof: By theorem 2.9, we can define the branching processes:
(𝑋1 , 𝑋2 , … ) (Bin)
(𝑋1∗ , 𝑋2∗ , … ) (Poi)
(random walk perspective)
Such that 𝑃[𝑋𝑖 ≠ 𝑋𝑖∗ ] ≤
𝜆2
,
𝑛
𝑖≥1
𝑘 ≥ 1:
{𝑇 ≥ 𝐾} depends only on {𝑋1 , … , 𝑋𝑘−1 } - A key observation!
Now:
𝑘−1
𝑃[𝑇 ≥ 𝑘, 𝑇 < 𝑘] ≤ ∑ 𝑃[𝑋𝑖 = 𝑋𝑖∗ 𝑓𝑜𝑟 1 ≤ 𝑖 ≤ 𝑠 − 1, 𝑋𝑠 ≠ 𝑋𝑠∗ 𝑇 ≥ 𝑘 ≥ 𝑠]
∗
𝑠=1
We can deduce from 𝑇 ≥ 𝑘 ≥ 𝑠 that 𝑇 ∗ ≥ 𝑠!
𝑘−1
≤ ∑ 𝑃[𝑋𝑠 ≠ 𝑋𝑠∗ , 𝑇 ∗ ≥ 𝑠]
𝑠=1
But these two events are independent
𝑘−1
= ∑ 𝑃[𝑋𝑠 ≠
𝑠=1
Similarly:
𝑘−1
𝑋𝑠∗ ]𝑃[𝑇 ∗
𝜆2
≥ 𝑠] ≤ ∑ 𝑃[𝑇 ∗ ≥ 𝑠]
𝑛
𝑠=1
𝑘−1
𝑃[𝑇 ≥ 𝑘, 𝑇 < 𝑘] ≤ ∑ 𝑃[𝑋𝑖 = 𝑋𝑖∗ , 𝑓𝑜𝑟 𝑖 ≤ 𝑠 − 1, 𝑋𝑖 ≠ 𝑋𝑖∗ , 𝑇 ∗ ≥ 𝑘 ≥ 𝑠] ≤
∗
𝑠=1
𝑘−1
∑⏟
𝑃[𝑋𝑠 ≠ 𝑋𝑠∗ ] 𝑃[𝑇 ∗ ≥ 𝑠]
𝑠=1
≤
𝜆2
𝑛
We are interested in: |𝑃[𝑇 ≥ 𝑘] − 𝑃[𝑇 ∗ ≥ 𝑘]| ≤
|𝑃[𝑇 ≥ 𝑘, 𝑇 ∗ ≥ 𝑘] + 𝑃[𝑇 ≥ 𝑘, 𝑇 ∗ < 𝑘] − 𝑃[𝑇 ∗ ≥ 𝑘, 𝑇 ≥ 𝑘] − 𝑃[𝑇 ∗ ≥ 𝑘, 𝑇 < 𝑘]| ≤
𝑘−1
2𝜆2
∑ 𝑃[𝑇 ∗ ≥ 𝑠] ∎
𝑛
𝑠=1
--- end of lesson 6
Exercise 1.17 𝑝, 𝑞 different distributions on ℤ
(𝑝𝑥 − min{𝑝𝑥 , 𝑞𝑥 })(𝑞𝑦 − min{𝑝𝑦 , 𝑞𝑦 })
𝑜𝑓 𝑥 = 𝑦
𝑐𝑥,𝑦 = {
𝑑 𝑇𝑉 (𝑝, 𝑞)
min{𝑝𝑥 , 𝑞𝑥 } 𝑖𝑓 𝑥 = 𝑦
(𝑥, 𝑦) ∈ ℤ2
1
1
𝑑𝑇𝑉 (𝑝, 𝑞) = ∑|𝑝𝑥 − 𝑞𝑥 | = ( ∑ (𝑝𝑥 − 𝑞𝑥 ) + ∑ (𝑞𝑥 − 𝑝𝑥 )) =
2
2
𝑥∈ℤ
𝑥:𝑝𝑥 ≥𝑞𝑥
𝑥:𝑞𝑥 <𝑝𝑥
1
(∑ 𝑝𝑥 − ∑ 𝑝𝑥 − ∑ 𝑞𝑥 + ∑ 𝑞𝑥 − ∑ 𝑝𝑥 ) =
2
𝑥∈ℤ
𝑥:𝑝𝑥 <𝑞𝑥
𝑥:𝑝𝑥 ≥𝑞𝑥
𝑥:𝑝𝑥 <𝑞𝑥
𝑥:𝑝𝑥 <𝑝𝑥
1
(1 − ∑ 𝑝𝑥 − ∑ 𝑞𝑥 + ∑ 𝑞𝑥 − ∑ 𝑞𝑥 − ∑ 𝑝𝑥 ) =
2
𝑥:𝑝𝑥 <𝑝𝑥
𝑥:𝑝𝑥 ≥𝑞𝑥
𝑥∈ℤ
𝑥:𝑞𝑥 ≤𝑝𝑥
𝑥:𝑝𝑥 <𝑞𝑥
1
(2 − 2 ∑ min{𝑝𝑥 − 𝑞𝑥 }) = 1 − ∑ min{𝑝𝑥 , 𝑞𝑥 }
2
𝑥∈ℤ
𝑥∈ℤ
𝑐 is a distribution on ℤ2 :
∑ ∑ 𝑐𝑥,𝑦 = ∑ 𝑐𝑧,𝑧 + ∑ ∑ 𝑐𝑥,𝑦 = ∑ min{𝑝𝑧 , 𝑝𝑧 } + ∑(𝑝𝑥 − min{𝑝𝑥 , 𝑞𝑥 }) = 1
𝑥
𝑦
𝑧∈ℤ
𝑃[𝑋 = 𝑥] = ∑ 𝑐𝑥,𝑦
𝑦∈ℤ
𝑥∈ℤ 𝑦≠𝑥
𝑧∈ℤ
𝑥∈ℤ
(𝑋, 𝑌)~𝑐
(𝑝𝑥 − min{𝑝𝑥 , 𝑞𝑥 })𝑑𝑇𝑉 (𝑝, 𝑞)
= min{𝑝𝑥 , 𝑞𝑥 } +
= 𝑝𝑥
𝑑𝑇𝑉 (𝑝, 𝑞)
Similarly, 𝑃[𝑌 = 𝑥] = 𝑞𝑥
𝑃[𝑋 ≠ 𝑌] = 1 − 𝑃[𝑋 = 𝑌] = 1 − ∑ 𝑐𝑥,𝑥 = 1 − ∑ min{𝑝𝑥 , 𝑞𝑥 } = 𝑑𝑇𝑉 (𝑝, 𝑞)
𝑥∈ℤ
𝑥∈ℤ
Cannot do better!
Assume (𝑋 ′ , 𝑌 ′ ): 𝑋 ′ ~𝑝, 𝑌 ′ ~𝑞
𝑃[𝑋 ′ = 𝑌 ′ ] = ∑ 𝑃[𝑋 ′ = 𝑌 ′ = 𝑥]
𝑥∈ℤ
But 𝑃[𝑋 ′ = 𝑌 ′ = 𝑥] ≤ 𝑃[𝑋 ′ = 𝑥] = 𝑝𝑥
And also 𝑃[𝑋 ′ = 𝑌 ′ = 𝑥] ≤ 𝑃[𝑌 ′ = 𝑥] = 𝑞𝑥
And both are smaller or equal to min{𝑝𝑥 , 𝑞𝑥 } ⇒
𝑃[𝑋 ′ ≠ 𝑌 ′ ] = 1 − 𝑃[𝑋 ′ = 𝑌 ′ ] ≥ 1 − ∑ min{𝑝𝑥 , 𝑞𝑥 } = 𝑑𝑇𝑉 (𝑝, 𝑞)
𝑥∈ℤ
3. The Eedos-Renyi random graph
Definition 3.1: E.R. random graphs, 𝐺(𝑛, 𝑝) for 𝑛 ≥ 1, 𝑝 ∈ [0,1].
Let (𝐼{𝑖,𝑗} )1≤𝑖<𝑗≤𝑛 be iid random variables with distribution 𝐵𝑒(𝑝).
Then 𝐺(𝑛, 𝑝) = ([𝑛], 𝐸)
Where [𝑛] = {1, … , 𝑛}, 𝐸 = {{𝑖, 𝑗} ≤ [𝑛]: 𝑖 ≠ 𝑗, 𝐼{𝑖,𝑗} = 1}
Exercise 3.2:
Let 𝑋 ≔ number of triangles in 𝐺(𝑛, 𝑝)
𝑛
𝐸[𝑋] = ( ) 𝑝3
3
1
Let 𝑝 = 𝑛𝛼 , 𝛼 > 0
Show that:
-
𝑛→∞
For 𝛼 > 1, 𝑃[𝑋 > 0] →
𝑛→∞
For 𝛼 < 1, 𝑃[𝑋 > 0] →
𝛼 = 1?
What about 𝐾𝑙 , 𝑙 ≥ 4?
(hint, prove that
𝑣𝑎𝑟(𝑋) 𝑛→∞
→ 0
𝐸[𝑋]2
0 (can be proven using a chebishev inequality)
1 (should use a second moment method! – the variant of 𝑋)
and use chebishev).
Notation:
- For 𝑢, 𝑣 ∈ [𝑛], "𝑢 ↔ 𝑣” means “There is a nearest-neighbor path from 𝑢 to 𝑣 in
𝐺(𝑛, 𝑝)”
- For 𝑣 ∈ [𝑛], 𝐶(𝑣) = {𝑤 ∈ [𝑛]: 𝑣 ↔ 𝑤}
Consider the following exploration algorithm for 𝐶(1) set 𝐴0 = {1}, 𝑁1 = [𝑛]\{1}
-
For 𝑡 > 0, if 𝐴𝑡−1 = ∅, then 𝐴𝑡 = ∅, 𝑁𝑡 = ∅ (terminate)
If 𝐴𝑡−1 ≠ ∅, let 𝑤𝑡−1 be the smallest vertex in 𝐴𝑡−1 and define:
𝑣𝑡 = {𝑤 ∈ 𝑁𝑡−1 |{𝑤𝑡−1 , 𝑤} ∈ 𝐸}
𝐴𝑡 ≔ (𝐴𝑡−1 \{𝑤𝑡−1 }) ∪ 𝑉𝑡
𝑁𝑡 ≔ 𝑁𝑡−1 \𝑉𝑡
Example:
Draw graphical example.
Note that:
- |𝐴0 | = 1, |𝐴𝑡 | = |𝐴𝑡−1 | + |𝑉𝑡 | − 1 for 𝑡 ≥ 1
- |𝑇𝑡 | = 𝑛 − |𝐴𝑡 | − 𝑡, 𝑡 ≥ 0
- |𝐶(1)| = inf{𝑡 ≥ 1 | |𝐴𝑡 | = 0}
Lemma 3.3: For 𝑥1 , … , 𝑥𝑘 ∈ ℤ≥0 , let:
- 𝑠0 = 1, 𝑠𝑡 = 𝑠𝑡−1 + 𝑥𝑡 − 1 for 𝑡 ≥ 1
- 𝑛𝑡 = 𝑛 − 𝑠𝑡 − 𝑡
If 𝑠1 , 𝑠2 , … , 𝑠𝑘 ≥ 1, 𝑛0 , 𝑛1 , … , 𝑛𝑘−1 ≥ 0
Then
𝑘
𝑃[|𝑉1 | = 𝑥1 , … , |𝑉𝑘 | = 𝑥𝑘 ] = ∏(𝑏(𝑛𝑡−1 , 𝑝, 𝑥𝑡 ))
𝑛
Where 𝑏(𝑛, 𝑝, 𝑘) = ( ) 𝑝𝑘 (1 − 𝑝)𝑛−𝑘
𝑘
𝑡=1
Proof: By picture.
TODO: Draw picture proof.
𝑃[|𝑉1 | = 𝑥1 , … , |𝑉𝑘 | = 𝑥𝑘 ] =
∑ ∑ 𝑃 [|𝑉1 | = 𝑥1 , … , |𝑉𝑘−1 | = 𝑥𝑘−1 , 𝐴𝑘−1 = 𝐴, 𝑁𝑘−1 = 𝑁, ∑ 𝐼{𝑎,𝑤} = 𝑥𝑘 ]
𝐴⊆[𝑛] 𝑁⊆[𝑛]
𝐴≠∅
𝑤∈𝑁
𝑎 is the smallest vertex in 𝐴.
{|𝑉1 } = 𝑥1 , … , 𝑁𝑡−1 = 𝑁} ∈ 𝜎(𝐼𝑒 : 𝑒 𝑛𝑜𝑡 𝑎𝑛 𝑒𝑑𝑔𝑒 𝑓𝑟𝑜𝑚 𝑎 𝑡𝑜 𝑁)
{ ∑ 𝐼{𝑎,𝑤} = 𝑥𝑘 } ∈ 𝜎(𝐼𝑒 : 𝑒 𝑒𝑑𝑔𝑒 𝑓𝑟𝑜𝑚 {𝑎} 𝑡𝑜 𝑁)
𝑤∈𝑁
So these two events must be independent!
So:
𝑃[|𝑉1 | = 𝑥1 , … , |𝑉𝑘 | = 𝑥𝑘 ] = ∑ ∑ 𝑃[|𝑉1 | = 𝑥1 , … , 𝑁𝑘−1 = 𝑁] ∙ 𝑏(𝑛𝑘−1 , 𝑝, 𝑥𝑘 )
If the probability of 𝑃[|𝑉1 | = 𝑥1 , … , 𝑁𝑘−1 = 𝑁] is positive then necessarily |𝑁| = |𝑁𝑘−1 | = 𝑛𝑘−1
⇒ 𝑃[|𝑉1 | = 𝑥1 , … , |𝑉𝑘 | = 𝑥𝑘 ] = 𝑃[𝑉1 = 𝑥! , … , |𝑉𝑘−1 | = 𝑥𝑘−1 ] × 𝑏(𝑛𝑘−1 , 𝑝, 𝑥𝑘 )
So now we can use induction to complete the proof.
--- end of lesson
Definition: 3.4:
𝑌
𝑡−1
Let (𝐼𝑡,𝑖 )𝑡,𝑖≥1 be iid 𝐵𝑒(𝑝). Define 𝑌0 = 𝑛 − 1, for 𝑡 ≥ 1 : ∑𝑖=1
𝐼𝑡,𝑖 , 𝑌𝑡 = 𝑌𝑡 − 𝑋𝑡 , 𝑆𝑡 =
(𝑋1 + ⋯ + 𝑋𝑡 ) − (𝑡 − 1)
𝑋𝑡 ~|𝑉𝑡 |,
𝑌𝑡 ~|𝑁𝑡 |,
𝑆𝑡 ~|𝐴𝑡 |
Lemma 3.5 For 𝑥𝑖 , 𝑠𝑖 , 𝑛𝑖 as in lemma 3.3,
𝑘
∏ 𝑏(𝑛𝑡−1 , 𝑝, 𝑥𝑡 ) ,
𝑃[𝑋1 = 𝑥1 , … , 𝑋𝑘 = 𝑥𝑘 ] = {
𝑖𝑓 𝑛1 , … , 𝑛𝑡−1 ≥ 0
𝑡=0
0,
𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒
Proof: {𝑋1 = 𝑥1 , … , 𝑋𝑘 = 𝑥𝑘 } ⊆ {𝑌1 = 𝑛1 , … , 𝑌𝑘−1 = 𝑛𝑘−1 }
So if 𝑛𝑖 < 0, then 𝑃[… ] = 0. Otherwise:
𝑃[𝑋1 = 𝑥1 , … , 𝑋𝑘−1 = 𝑥𝑘−1 , 𝑋𝑘 = 𝑥𝑘 ] = 𝑃[𝑋1 = 𝑥1 , … , 𝑋𝑘−1 = 𝑥𝑘−1 ] ∙ 𝑏(𝑛𝑘−1 , 𝑝, 𝑥𝑘 )
𝑏𝑦 𝑑𝑒𝑓
𝑛𝑘−1
Because 𝑋𝑘 = ∑𝑖=1
𝐼𝑘,𝑖
We can continue by induction. And that finishes the proof. ∎
Corollary 3.6: For 𝑘 ≥ 1, and any 𝑓: ℤ𝑘≥0 → ℝ
𝐸[𝑓(|𝑉1 |, … , |𝑉𝑘 |)1{|𝐴1 |>0,…,|𝐴𝑘−1 |>0} ] = 𝐸[𝑓(𝑋1 , … , 𝑋𝑘 )1{𝑆1 >0,…,𝑆𝑘−1 >0} ]
Proof: Combine Lemma 3.3 and Lemma 3.5. ∎
Theorem 3.7: Let 𝑇 − and 𝑇 + be the total progenies of a 𝐵𝑖𝑛(𝑛 − 𝑘, 𝑝) and 𝐵𝑖𝑛(𝑛, 𝑝) branching
processes.
Where 1 ≤ 𝑘 ≤ 𝑛 and 𝑝 ∈ [0,1] then 𝑃[𝑇 − ≥ 𝑘] ≤ 𝑃[|𝐶(1)| ≥ 𝑘] ≤ 𝑃[𝑇 + ≥ 𝑘].
+/−
Proof: For (𝐼𝑡,𝑖 ) as in Definition 3.4, define 𝑋𝑡
+/−
Then (𝑋𝑡 ) are iid~𝐵𝑖𝑛(𝑛/𝑛 − 𝑘, 𝑝).
𝑡≥1
+/−
+/−
+/−
𝑆𝑡 : = (𝑋1 + ⋯ + 𝑋𝑡 ) − (𝑡 − 1),
(𝑑)
+/−
Then 𝑇 +/− = inf {𝑡 ≥ 1|𝑆𝑡 = 0}
𝑛/𝑛−𝑘
= ∑𝑖=1
𝑡 ≥ 1,
𝐼𝑡,𝑖 , 𝑡 ≥ 1.
+/−
𝑆0
=1
𝑃[|𝐶(1)| ≥ 𝑘] = 𝑃[|𝐴1 | > 0, … , |𝐴𝑘−1 | > 0] = 𝑃[𝑆1 > 0, … , 𝑆𝑘−1 > 0] ≤
+
𝑃[𝑆1+ > 0, … , 𝑆𝑘−1
> 0] = 𝑃[𝑇 + ≥ 𝑘]
(by cor 3.6 if 𝑓 ≡ 1) 𝑆𝑡 ≤ 𝑆𝑡+ , 𝑡 ≤ 𝑘 − 1
On the other hand, 𝑃[|𝐶(1)| ≤ 𝑘] = 𝑃[⋃𝑘𝑡=1{|𝑉𝑡 | = 0, |𝐴𝑡−1 | = 1, |𝐴𝑡−2 | > 0, … , |𝐴1 | > 0}]
(exploration stops at 𝑡)
𝑘
= ∑ 𝑃[𝑋𝑡 = 0, 𝑆𝑡−1 = 1, 𝑆𝑡−2 > 0, … , 𝑆1 > 0]
𝑡=1
(by Cor 3.6)
= 𝑃[𝑇 ≤ 𝑘] where 𝑇 = inf{𝑡 ≥ 1|𝑆𝑡 = 0}
Claim: If {𝑇 ≤ 𝑘} ⊆ {𝑇 − ≤ 𝑘}
Indeed, if 𝑇 ≤ 𝑘 𝑌𝑇 = 𝑛 − 1 − 𝑋1 − ⋯ − 𝑋𝑇 = 𝑛 − 𝑇 − 𝑆𝑇 ≥ 𝑛 − 𝑘
So 𝑌1 ≥ 𝑌2 ≥ ⋯ ≥ 𝑌𝑇 ≥ 𝑛 − 𝑘
Hence, 𝑋𝑡− ≤ 𝑋𝑡 , 𝑡 ≤ 𝑇
𝑆𝑡− ≤ 𝑆𝑡 ,
𝑡 ≤ 𝑇 ⇒ 𝑇− ≤ 𝑘
Which proves this claim. ∎
Exercise 3.8:
Consider random variables 𝑁, 𝑀 such that 𝑁~𝐵𝑖𝑛(𝑛, 𝑝) and conditionally on 𝑁,
𝑀~𝐵𝑖𝑛(𝑁, 𝑞), 𝑛 ≥ 1, 𝑝, 𝑞 ∈ [0,1]
𝑘
I.e. 𝑃[𝑀 = 𝑙|𝑁 = 𝑘] = ( ) 𝑞 𝑙 (1 − 𝑞)𝑘−𝑙
𝑙
(0 ≤ 𝑙 ≤ 𝑘 ≤ 𝑛)
Prove that 𝑀~𝐵𝑖𝑛(𝑛, 𝑝𝑞)
It would be probable to assume: |𝑁𝑡 |~𝐵𝑖𝑛(𝑛 − 1, (1 − 𝑝)𝑡 )
But this is not correct…
Proposition 3.9:
For 0 ≤ 𝑡 ≤ 𝑛, 𝑌𝑡 ~𝐵𝑖𝑛(𝑛 − 1, (1 − 𝑝)𝑡 )
Proof: By induction.
𝑌0 = 𝑛 − 1. This is true.
Assume that 𝑌𝑡 ~𝐵𝑖𝑛(𝑛 − 1, (1 − 𝑝)𝑡 )
Need to check that conditionally on 𝑌𝑡 , 𝑌𝑡+1 ~𝐵𝑖𝑛(𝑌𝑡 , 1 − 𝑝)
(By Ex 3.6) 0 ≤ 𝑙 ≤ 𝑘 ≤ 𝑛 − 1
𝑃[𝑌𝑡+1 = 𝑙. 𝑌𝑡 = 𝑘],
𝑌𝑡+1 = 𝑌𝑡 − 𝑋𝑡+1
= 𝑃[ 𝑋
⏟
𝑡+1
= 𝑘 − 𝑙, 𝑌𝑡 = 𝑘]
𝑖𝑛𝑑𝑒𝑝𝑒𝑛𝑑𝑒𝑛𝑐𝑒
=
𝑏(𝑘, 𝑝, 𝑘 − 𝑙)𝑃[𝑌𝑡 = 𝑘] =
=∑𝑘
𝑖=1 𝐼𝑡+1,𝑖
𝑏(𝑘, (1 − 𝑝), 𝑙)𝑃[𝑌𝑡 = 𝑘] ∎
𝑘
(
) 𝑝𝑘−𝑙 (1 − 𝑝)𝑙
𝑘−𝑙
𝐼𝜆 = 𝜆 − 1 − log 𝜆 ,
(see Ex. 1.16, 𝐼(𝜆))
𝜆>0
𝜆
Lemma 3.8: For 𝑝 = 𝑛 , 𝜆 ∈ (0,1)
𝑘 ≥ 1,
𝑃[|𝐶(1)| > 𝑘] ≤ 𝑒 −𝐼𝜆 𝑘
Proof: Let 𝑟 > 0, 𝑟 = − log 𝜆 > 0
𝑃[|𝐶(1)| > 𝑘] = 𝑃[|𝐴1 | > 0, … , |𝐴𝑘 | > 0]
𝐶𝑜𝑟 3.6
= ≤ 𝑃[𝑆1 > 0, … , 𝑆𝑘 > 0] ≤ 𝑃[𝑆𝑘 > 0] =
𝑃 𝑆
⏟𝑘 + 𝑘 − 1 ≥ 𝑘
𝑒𝑞𝑢𝑖𝑣𝑎𝑙𝑒𝑛𝑡 𝑡𝑜:
[
]
𝑆𝑘 + 𝑘 − 1 = 𝑋1 + ⋯ + 𝑋𝑘 = 𝑛 − 1 − 𝑌𝑘 ~𝐵𝑖𝑛(𝑛 − 1,1 − (1 − 𝑝)𝑘 )
𝑒 𝑟(𝑆𝑘 +𝑘−1) ≥𝑒 𝑟𝑘
So 𝑃[|𝐶(1)| > 𝑘] ≤ 𝑃[𝑒 𝑟(𝑆𝑘 +𝑘−1) ≥ 𝑒 𝑟𝑘 ]
𝑏𝑦 𝑀𝑎𝑟𝑘𝑜𝑣
≤
𝑒 −𝑟𝑘 𝐸[𝑒 𝑟(𝑆𝑘 +𝑘−1) ]
𝑛−1
𝐸[𝑒 𝑟(𝑆𝑘 +𝑘−1) ] = ((1 − 𝑞) + 𝑞𝑒 𝑟 )
(𝑑)
𝑆𝑘 + 𝑘 − 1 = 𝜉1 + ⋯ + 𝜉𝑛−1 , 𝜉𝑖 iid ~𝐵𝑒(𝑞)
𝑛−1
⇒ 𝐸[𝑒
𝑟(𝑆𝑘 +𝑘−1)
] = 𝐸 [𝑒
𝑟(∑𝑛−1
𝑖=1 𝜉𝑖 )
𝑛−1
] = ∏(𝐸[𝑒 𝑟𝜉𝑖 ]) = (𝐸[𝑒 𝑟𝜉1 ])
𝑛−1
= ((1 − 𝑞) + 𝑞𝑒 𝑟 )
𝑖=1
So
𝑃[|𝐶(1)| > 𝑘] ≤ 𝑒
−𝑟𝑘
𝑟 𝑛−1
((1 − 𝑞) + 𝑞𝑒 )
(1−𝑥)≤𝑒 −𝑥
≤
𝑟
exp(−𝑟𝑘 + 𝑞(𝑒 − 1)(𝑛 − 1)) ≤ exp(−𝑟𝑘 + 𝑞(𝑒 𝑟 − 1)𝑛)
𝑞 = 1 − (1 − 𝑝)𝑘 ≤ 𝑝𝑘 (∗)
(comment: Seems like 𝑡 turned into 𝑘)
⇒ 𝑃[|𝐶(1)| > 𝑘] ≤ exp(−𝑟𝑘 + 𝑞(𝑒 𝑟 − 1)𝑛) ≤ exp(−𝑟𝑘 + 𝑘𝑝(𝑒 𝑟 − 1)𝑛) =
𝑟
𝑝=
𝜆
𝑛
exp (−𝑘(𝑟 − 𝑝𝑛(𝑒 − 1))) = exp (−𝑘(𝑟 − 𝜆(𝑒 𝑟 − 1)))
If we try to minimize the expression, we get that 𝑟 = − log 𝜆
So let’s just set it in:
𝑒 −𝑘𝐼𝜆
Why is (*) true? Show that at zero they are equal, and the derivative of the right hand side is
always larger then the derivative of the left hand side.
--- end of lesson
Theorem 3.9 (Upper bound on the largest sub-critical component):
𝜆
1
Fix 𝜆 ∈ (0,1), 𝑝 = 𝑛 , 𝑎 > 𝐼
𝜆
Then there is a 𝛿 = 𝛿(𝑎, 𝜆) such that 𝑃[|𝐶𝑚𝑎𝑥 | > 𝑎 log 𝑛] < 𝑛−𝛿
Proof:
𝑃[|𝐶𝑚𝑎𝑥 | > 𝑎 log 𝑛] = 𝑃 [ ⋃ {|𝐶(𝑣)| > 𝑎 log 𝑛}] ≤ ∑ 𝑃[|𝐶(𝑣)| > 𝑎 log 𝑛] =
𝑣∈[𝑛]
𝑏𝑦 𝐿𝑒𝑚𝑚𝑎 3.8
𝑛𝑃[|𝐶(1)| > 𝑎 log 𝑛]
≤
𝑣∈[𝑛]
𝑛 ∙ 𝑛−𝐼𝜆𝑎 ,
𝐼𝜆 𝑎 > 1 = ∎
Next, lower bound:
Lemma 3.10:
Let 𝑍≥𝑘 = ∑𝑛𝑣=1 1{|𝐶(𝑣)|≥𝑘}
𝜒≥𝑘 = 𝐸[|𝐶(1)|1{|𝐶(1)|≥𝑘} ]
Then 𝑣𝑎𝑟(𝑍≥𝑘 ) ≤ 𝑛𝜒≥𝑘
2
2
Proof: 𝑣𝑎𝑟(𝑍≥𝑘 ) = 𝐸 [(∑𝑛𝑣=1 1{|𝐶(𝑣)|≥𝑘} ) ] − 𝐸[∑𝑛𝑣=1 1{|𝐶(𝑣)|≥𝑘} ] =
𝑛
𝑛
∑ ∑ 𝑃[|𝐶(𝑖)| ≥ 𝑘, |𝐶(𝑗)| ≥ 𝑘] − 𝑃[|𝐶(𝑖)| ≥ 𝑘]𝑃[|𝐶(𝑗)| ≥ 𝑘]
𝑖=1 𝑗=1
Split according to 𝑖 ↔ 𝑗 and 𝑖 ↔ 𝑗
𝑖 connected to 𝑗 and 𝑖 not connected to 𝑗
𝑛
𝑛
𝑃[|𝐶(𝑖)| ≥ 𝑘, |𝐶(𝑗)| ≥ 𝑘, 𝑖 ↔ 𝑗] = ∑ ∑ 𝑃[|𝐶(𝑖)| = 𝑙, 𝑖 ↔ 𝑗, |𝐶(𝑗)| = 𝑚] =
𝑙=𝑘 𝑚=𝑘
𝑛
∑ 𝑃[|𝐶(𝑖) = 𝑙|, 𝑖 ↔ 𝑗] ∙ 𝑃 [|𝐶(𝑗)| = 𝑚] = ∑ 𝑃[|𝐶(𝑖)| = 𝑙, 𝑖 ↔ 𝑗] ∙ 𝑃 [|𝐶(𝑗)| ≥ 𝑘] ≤
(𝑛−𝑙)
𝑙,𝑚
𝑛
(𝑛−𝑙)
𝑙=𝑘
𝑛
∑ 𝑃[|𝐶(𝑖)| = 𝑙, 𝑖 ↔ 𝑗] ∙ 𝑃 [|𝐶(𝑗)| ≥ 𝑘] = ∑ 𝑃[|𝐶(𝑖)| = 𝑙] ∙ 𝑃[|𝐶(𝑗)| ≥ 𝑘]
(𝑛−𝑙)
𝑙=𝑘
𝑙=𝑘
(we can strike these out since it only increases the probability of the event)
So let’s go back to the entire variance:
The term we just calculated cancels the second term! So we are only left with the case where
they are connected:
𝑛
𝑛
𝑣𝑎𝑟(𝑍≥𝑘 ) ≤ ∑ ∑ 𝑃[|𝐶(𝑖)| ≥ 𝑘, 𝐶(𝑗) ≥ 𝑘, 𝑖 ↔ 𝑗]
𝑖=1 𝑗=1
𝑖↔𝑗⇒
𝑇ℎ𝑒𝑠𝑒 𝑠𝑝𝑒𝑐𝑖𝑓𝑦 𝑛
𝑡ℎ𝑒 𝑠𝑎𝑚𝑒 𝑒𝑣𝑒𝑛𝑡
=
𝑛
∑ ∑ 𝑃[|𝐶(𝑖)| ≥ 𝑘] =
𝑖=1 𝑗=1
𝑛
𝑛
∑ 𝐸 1{|𝐶(𝑖)|≥𝑘} ∙ ∑ 1{𝑗∈𝐶(𝑖)} = 𝑛𝜒≥𝑘
𝑖=1
𝑗=1
⏟
[
]
=|𝐶(𝑖)|
Theorem 3.11 (lower bound on largest subcritical component):
𝜆
Fix 𝜆 ∈ (0,1), 𝑜 = 𝑛
1
Then for every 𝑎 < 𝐼 there exists 𝛿 = 𝛿(𝑎, 𝜆), and 𝑐 = 𝑐(𝑎, 𝜆) such that
𝜆
𝑃[|𝐶𝑚𝑎𝑥 | ≤ 𝑎 log 𝑛] ≤ 𝑐𝑛−𝛿
Proof: Let 𝑘 = ⌊𝑎 log 𝑛⌋, 𝑐 denotes a constant depending only on 𝜆, 𝑎 changing from place to
place.
Claim 1: For any 0 < 𝛼 < 1 − 𝐼𝜆 𝑎
𝐸[𝑍≥𝑘 ] ≥ 𝑛𝛼 , for 𝑛 ≥ 𝑐
Claim 2: 𝑣𝑎𝑟(𝑍≥𝑘 ) ≤ 𝑐𝑘𝑛1−𝑎𝐼𝜆
From these claims, it follows that we can prove this theorem.
Why?
𝑃[|𝐶𝑚𝑎𝑥 | ≤ 𝑎 log 𝑛] = 𝑃[𝑍≥𝑘 = 0] ≤ 𝑃[|𝑍𝑘≥0 − 𝐸[𝑍≥𝑘 ]| ≥ 𝐸[𝑍≥𝑘 ]] =
𝑏𝑦 𝑡ℎ𝑒
𝑃[|𝑍𝑘≥0 − 𝐸[𝑍≥𝑘
]|2
≥ 𝐸[𝑍≥𝑘
]2 ]
𝑣𝑎𝑟(𝑍≥𝑘 ) 𝑐𝑙𝑎𝑖𝑚𝑠 𝑐𝑘𝑛1−𝑎𝐼𝜆
≤
≤
𝐸[𝑍≥𝑘 ]2
𝑛2𝛼
2
(For instance, if 𝛼 = 3 (1 − 𝑎𝐼𝜆 ) then it’s ≤
𝑐 log 𝑛
𝑛
(1−𝑎𝐼𝜆 )
3
)
Let’s prove claim 1:
𝜆
Let 𝑇, 𝑇 ∗ be the total progenies of 𝐵𝑖𝑛 (𝑛 − 𝑘, 𝑛) branching process, and 𝑃𝑜𝑖(𝜆𝑛 ) branching
process wher 𝜆𝑛 =
𝜆(𝑛−𝑘)
𝑛
By theorem 3.7, 𝑃[|𝐶(1)| ≥ 𝑘] ≥ 𝑃[𝑇 ≥ 𝑘]
𝑏𝑦
𝑡ℎ𝑒𝑜𝑟𝑒𝑚 2.11
≥
𝜆
1
𝜆(𝑛 − 𝑘)
=
(
)
𝑛 𝑛−𝑘⏟ 𝑛
𝜆𝑛
∗
∗
𝑃[𝑇 ≥ 𝑘] ≥ 𝑃[𝑇 = 𝑘]
𝑏𝑦
𝑡ℎ𝑒𝑜𝑟𝑒𝑚 2.3
=
We can use sterling’s formula:
(𝜆𝑛 𝑘)𝑘−1 −𝜆 𝑘
𝑒 𝑛
𝑘!
𝑃[𝑇 ∗ ≥ 𝑘] −
2𝜆2𝑛 𝑘
𝑛
𝑘 𝑘
𝑘! = ( ) ∙ √2𝜋𝑘(1 + 𝑜(1))
𝑒
And then:
≥
𝜆𝑘𝑛
𝑒
−𝜆𝑛 𝑘 𝑘
𝑒 (1 + 𝑜(1)) ≥ 𝑒
−𝐼𝜆𝑛 𝑘(1+𝜖)
𝑓𝑜𝑟 𝑠𝑢𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡𝑙𝑦
𝑙𝑎𝑟𝑔𝑒 𝑛
𝑘√2𝜋𝑘𝜆𝑛
(𝐼𝜆 = 𝜆 − 1 − log 𝜆)
=
𝑒 −𝐼𝜆 𝑘(1+𝜖)
For 𝜖 ≥ 0: 1 − 𝑎𝐼𝜆 (1 + 𝜖) > 𝛼
for 𝑛 ≥ 𝑐(𝑎, 𝜆, 𝜖)
>𝛼
∗
𝐸[𝑍≥𝑘 ] = 𝑛𝑃[|𝐶(1)| ≥ 𝑘] ≥ 𝑛𝑃[𝑇 ≥ 𝑘] −
𝑐𝑘
𝑛
≥ 𝑛 (𝑒
−𝐼𝜆 𝑘(1+𝜖)
−
𝑐𝑘
)
𝑛
=𝑛
⏞
1−𝐼𝜆 𝑎(1+𝜖)
− 𝑐 log 𝑛 ≥
𝑛𝛼 for 𝑛 ≥ 𝑐.
(For 𝜖 chosen sufficiently small)
Now let’s prove claim 2:
Proof of claim 2:
|𝐶(1)|
We can write |𝐶(1)| = ∑𝑖=1 1{|1|>𝑖} = ∑𝑛𝑖=1 1{|1|>𝑖}
So:
𝑛
𝑛
𝜒≥𝑘 = ∑ 𝑃[|𝐶(1)| ≥ 1, |𝐶(1)| ≥ 𝑘] = 𝑘𝑃[|𝐶(1)| ≥ 𝑘] + ∑ 𝑃[|𝐶(1)| ≥ 𝑖]
𝑖=1
𝑖=𝑘+1
∞
𝑘𝑒 −𝐼𝜆 (𝑘−1) + ∑ 𝑒 −𝐼𝜆 (𝑖−1) ≤ 𝑐𝑘𝑛−𝑎𝐼𝜆 + 𝑐𝑒
⏟ −𝐼𝜆 𝑘 ≤ 𝑐𝑘𝑛−𝑎𝐼𝜆
≤𝑛−𝑎𝐼𝜆
𝑖=𝑘+1
So by lemma 3.10 get claim 2. ∎
Exercise 3.12 (second moment method):
Let 𝑋 be a random variable such that 0 ≤ 𝐸[𝑋] < ∞ and 0 < 𝐸[𝑋 2 ] < ∞
Let 0 ≤ 𝜆 < 1.
𝐸[𝑥]2
Prove that: 𝑃[𝑋 > 𝜆𝐸[𝑋]] ≥ (1 − 𝜆)2 𝐸[𝑥 2 ]
In our proof, we used it with 𝜆 = 0.
Hint: Prove first that (1 − 𝜆)𝐸[𝑋] ≤ 𝐸[𝑋 ∙ 1{𝑋>𝜆𝐸[𝑋]} ]
𝑏𝑦
𝑙𝑒𝑚𝑚𝑎 3.8
≤