probability of wiring sequence and conditional expectations in random configuration graph
$begingroup$
I am reading about the Bollobas Configuration random graph and try to understand the proof of following Proposition:
(see Prop 2.1 in https://arxiv.org/pdf/1512.03084.pdf, page 7-8)
Proposition:
Considering the last step of the ACG construction for finite $N$ with probabilities $P,Q$ conditioned on the $X$ from the first step, one has
The conditional probability of any wiring sequence $W=(l in [E])$ is:
$$ mathbb{P}(W mid X)= C^{-1} prod_{kj} (Q_{kj})^{e_{kj}(W)},$$
$$C = E! sum_e prod_{kj} frac{(Qkj)^{e_{kj}}}{e_{kj}!} prod_j ({e_j}^-!) prod_k ({e_k}^+ !)$$
and where the sum above runs over all collections $e=(e_{kj})$ satisfying
$${e_k}^+ = sum_{j} e_{kj}, quad {e_j}^- = sum_{k}e_{kj}, quad E= sum_{kj}e_{kj}$$The conditional probability of any edge of the wiring sequence $W$ having type $(k, j)$ is
$$p = mathbb{E}[e_{kj} mid X]/E$$
I don't understand how they prove the very first formula as they simply just state that
The denominator of it is $C = sum_{σ,σ ̃in S(E)} prod_{ l∈[E]} Q_{k_{σ(l)} j_{σ ̃(l)}}$ , from which the formula of $C$ can be verified by induction on $E$.
How do they get to this expression?
probability graph-theory random-graphs
asked Dec 23 '18 at 14:18
AlisatAlisat
448
$endgroup$
add a comment |
$begingroup$
I am reading about the Bollobas Configuration random graph and try to understand the proof of following Proposition:
(see Prop 2.1 in https://arxiv.org/pdf/1512.03084.pdf, page 7-8)
Proposition:
Considering the last step of the ACG construction for finite $N$ with probabilities $P,Q$ conditioned on the $X$ from the first step, one has
The conditional probability of any wiring sequence $W=(l in [E])$ is:
$$ mathbb{P}(W mid X)= C^{-1} prod_{kj} (Q_{kj})^{e_{kj}(W)},$$
$$C = E! sum_e prod_{kj} frac{(Qkj)^{e_{kj}}}{e_{kj}!} prod_j ({e_j}^-!) prod_k ({e_k}^+ !)$$
and where the sum above runs over all collections $e=(e_{kj})$ satisfying
$${e_k}^+ = sum_{j} e_{kj}, quad {e_j}^- = sum_{k}e_{kj}, quad E= sum_{kj}e_{kj}$$The conditional probability of any edge of the wiring sequence $W$ having type $(k, j)$ is
$$p = mathbb{E}[e_{kj} mid X]/E$$
I don't understand how they prove the very first formula as they simply just state that
The denominator of it is $C = sum_{σ,σ ̃in S(E)} prod_{ l∈[E]} Q_{k_{σ(l)} j_{σ ̃(l)}}$ , from which the formula of $C$ can be verified by induction on $E$.
How do they get to this expression?
probability graph-theory random-graphs
asked Dec 23 '18 at 14:18
AlisatAlisat
448
$endgroup$
add a comment |
$begingroup$
I am reading about the Bollobas Configuration random graph and try to understand the proof of following Proposition:
(see Prop 2.1 in https://arxiv.org/pdf/1512.03084.pdf, page 7-8)
Proposition:
Considering the last step of the ACG construction for finite $N$ with probabilities $P,Q$ conditioned on the $X$ from the first step, one has
The conditional probability of any wiring sequence $W=(l in [E])$ is:
$$ mathbb{P}(W mid X)= C^{-1} prod_{kj} (Q_{kj})^{e_{kj}(W)},$$
$$C = E! sum_e prod_{kj} frac{(Qkj)^{e_{kj}}}{e_{kj}!} prod_j ({e_j}^-!) prod_k ({e_k}^+ !)$$
and where the sum above runs over all collections $e=(e_{kj})$ satisfying
$${e_k}^+ = sum_{j} e_{kj}, quad {e_j}^- = sum_{k}e_{kj}, quad E= sum_{kj}e_{kj}$$The conditional probability of any edge of the wiring sequence $W$ having type $(k, j)$ is
$$p = mathbb{E}[e_{kj} mid X]/E$$
I don't understand how they prove the very first formula as they simply just state that
The denominator of it is $C = sum_{σ,σ ̃in S(E)} prod_{ l∈[E]} Q_{k_{σ(l)} j_{σ ̃(l)}}$ , from which the formula of $C$ can be verified by induction on $E$.
How do they get to this expression?
probability graph-theory random-graphs
asked Dec 23 '18 at 14:18
AlisatAlisat
448
$endgroup$
I am reading about the Bollobas Configuration random graph and try to understand the proof of following Proposition:
(see Prop 2.1 in https://arxiv.org/pdf/1512.03084.pdf, page 7-8)
Proposition:
Considering the last step of the ACG construction for finite $N$ with probabilities $P,Q$ conditioned on the $X$ from the first step, one has
The conditional probability of any wiring sequence $W=(l in [E])$ is:
$$ mathbb{P}(W mid X)= C^{-1} prod_{kj} (Q_{kj})^{e_{kj}(W)},$$
$$C = E! sum_e prod_{kj} frac{(Qkj)^{e_{kj}}}{e_{kj}!} prod_j ({e_j}^-!) prod_k ({e_k}^+ !)$$
and where the sum above runs over all collections $e=(e_{kj})$ satisfying
$${e_k}^+ = sum_{j} e_{kj}, quad {e_j}^- = sum_{k}e_{kj}, quad E= sum_{kj}e_{kj}$$The conditional probability of any edge of the wiring sequence $W$ having type $(k, j)$ is
$$p = mathbb{E}[e_{kj} mid X]/E$$
I don't understand how they prove the very first formula as they simply just state that
The denominator of it is $C = sum_{σ,σ ̃in S(E)} prod_{ l∈[E]} Q_{k_{σ(l)} j_{σ ̃(l)}}$ , from which the formula of $C$ can be verified by induction on $E$.
How do they get to this expression?
probability graph-theory random-graphs
probability graph-theory random-graphs
asked Dec 23 '18 at 14:18
AlisatAlisat
448
asked Dec 23 '18 at 14:18
AlisatAlisat
448
asked Dec 23 '18 at 14:18
AlisatAlisat
448
asked Dec 23 '18 at 14:18
AlisatAlisat
448
asked Dec 23 '18 at 14:18
AlisatAlisat
448
448
add a comment |
add a comment |
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