Back to marketplace

Continuous-time Markov Chains

No description yet. This curated deck is ready for daily study.

Total cards: 14•Content version: v4•Public deck•Updated: 8/1/2026

Card preview (14)

#1
Front (Question)

Continuous-time Markov chains ▪ Definition

Back (Answer)

Let (X(t))ₜ˲₀ continuous-time stochastic process with countable state space S if: ▪ for all n⩾0 ▪ all times 0⩽s₁<s₂<…<sₙ<s<t ▪ all states i₁,i₂,…,iₙ,i,t ϵ S it holds that P(X(t+s)=j|X(s)=i, X(sₙ)=iₙ,…, X(s₁)=i₁) = P(X(t+s)=j|X(s)=i) then this stochastic process is a continuous-time Markov chain

#2
Front (Question)

Continuous-time Markov chains ▪ Components

Back (Answer)

▪ Jump Chain A discrete time Markov chains that gives us the transition probabilities ▪ Holding time For each state we have an exponentially distributed holding time that controls the amount of time spent in each state

#3
Front (Question)

Continuous-time Markov chains ▪ Construction

Back (Answer)

Consider ▪ a discrete-time Markov chain (jump chain) ▪ on a countable state space ▪ having transition matrix P ▪ with pᵢᵢ=0 for all non-absorbing state i ϵ S ▪ let (vᵢ)ᵢϵs be a collection of a positive and finite real numbers (holding time parameters) A continuous-time stochastic process (X(t))ₜ˲₀ such that: ① if X(t) = i, the times until the stages changes has distribution Exp(vᵢ) ② if X(t) = i the next state will be j with probability pᵢj is a continuous-time Markov chain (i.e. it satisfies Markov property (MP)) (Holding times relative to different states are independent)

#4
Front (Question)

Continuous-time Markov chains ▪ Transition probability functions

Back (Answer)

Let (X(t))ₜ˲₀ continuous-time Markov chain on a countable state space S Let pᵢj = P(X(t+s)=j | X(s) = i) a transition probability function equal to pᵢj = P(X(t)=j | X(0) = i) for all s,t⩾0 denote the probability that the process presently in state i will be in state j at a time t later

#5
Front (Question)

Continuous-time Markov chains ▪ Transition matrix

Back (Answer)

For any t⩾0 we can define the transition matrix P(t)=(pᵢj(t))ᵢ,jϵs Properties i) for every t⩾0, P(t) is a stochastic matrix ii) P(0) = I (identity matrix) iii) for every s,t⩾0, P(t+s) = P(t)P(s) = P(s)P(t)

#6
Front (Question)

Continuous-time Markov chains ▪ Stationary and limiting distribution

Back (Answer)

(As in discrete-time chains, for “nice” chains a unique stationary distribution exists and it is equal to the limiting distribution) Let (X(t))ₜ˲₀ continuous-time Markov chain on a countable state space S and with transition matrix P(t) A probability distribution π = (πᵢ)ᵢϵs (row vector) is a stationary distribution for (X(t))ₜ˲₀ if π = πP(t) for all t⩾0

#7
Front (Question)

Continuous-time Markov chains ▪ Definition of limiting distribution

Back (Answer)

A probability distribution π = (πᵢ)ᵢϵs is the limiting distribution of the continuous-time Markov chain (X(t))ₜ˲₀ if πj = lim P(X(t)=j | X(o)=i) t → ∞

#8
Front (Question)

Continuous-time Markov chains ▪ The generator matrix Transition rate - Definition

Back (Answer)

Consider a continuous-time Markov chain (X(t))ₜ˲₀ Assume X(o) = i The chain will jump to the next state at time T₁ where: ▪ T₁~Exp(vᵢ) ▪ vᵢ –> holding time parameter at state i For a very small h we have: P(T₁⩽h) = 1 - e^(-vᵢh) = vᵢh + o(h) (Taylor expansion) Thus, in a short time interval of length h, the probability of leaving state 1 is approximately vᵢ*h For this reason vᵢ is often called the transition rate out of state i

#9
Front (Question)

Continuous-time Markov chains ▪ The generator matrix Transition rate - Formally

Back (Answer)

Formally we can write the transition rate as vᵢ = lim P(X(h)!=i | X(0) = i) / h h–>0

#10
Front (Question)

Continuous-time Markov chains ▪ The generator matrix Transition rate from state i to state j

Back (Answer)

Since we go from state i to state j with probability Pᵢj we call the transition rate from state i to state j the quantity gᵢj = vᵢ * Pᵢj

#11
Front (Question)

Continuous-time Markov chains ▪ The generator matrix Definition

Back (Answer)

Let (X(t))ₜ˲₀ - continuous-time Markov chain on a countable state space S - having embedded (jump) chain with transition matrix P - holding time parameters vᵢ → with i ϵ S The generator matrix G = gᵢ,j → with i,j ϵ S has entries: gᵢ,j: ▪ vᵢ * Pᵢj if i!=j ▪ -vᵢ if i=j Remark: The rows of the generator matrix G sum to 0 Generation matrix is useful to analyze continuous-time Markov chains

#12
Front (Question)

Continuous-time Markov chains ▪ The generator matrix Lemma (proposizione preliminare)

Back (Answer)

Let (X(t))ₜ˲₀ - continuous-time Markov chain with transition matrix P(t) - holding time parameters vᵢ –> with i ϵ S Then we have (if i!=j) ▪ lim (Pᵢᵢ(h) - 1) / h = gᵢᵢ h→0 and lim Pᵢj(h) / h = gᵢj h→0 ▪ Equivalently Pᵢᵢ(h) = 1 - vᵢ * h + o(h) and Pᵢj(h) = gᵢj + o(h)

#13
Front (Question)

Continuous-time Markov chains ▪ The generator matrix Find the Stationary distribution using the generation matrix

Back (Answer)

Consider a continuous-time Markov chain (X(t))ₜ˲₀ with - state space S - generator matrix G The probability distribution π on S is a stationary distribution for (X(t))ₜ˲₀ if and only if πG = 0

#14
Front (Question)

Continuous-time Markov chains ▪ The generator matrix Interpretation of πG = 0

Back (Answer)

In the long run the rate at which transitions into state i occur must equal the rate at which transitions out of state j occur Indeed, we have: (sommatoria con k diverso da j) πG = 0 ↔ ∑ gₖj * πₖ - vj * πj = 0 ↔ vj * πj = ∑ gₖj * πₖ Where ▪ vj * πj = rate of which the process leaves state j as: → πj = the amount of time the process spends time in state j → vj = the rate at which the process leaves state j ▪ ∑ gₖj * πₖ = rate at which the process enters state j as: → πₖ = the amount of time the process spends in state k → gₖj = the rate at which transitions from state k to state j occur