## What is Markov Chain ?

Markov Chain is a Stochastic Model in which Future is dependent only on Present not on Past , What I mean to say that is

\[P(X^{t+1}|X^t,X^{t-1},...X^2,X^1) = P(X^{t+1}| X^t)\]#### Transition Probability Matrix

Let us denote

\[p_{ij} = P(X^{n+1} = i | X^n = j)\]Where $[Math Processing Error]p_{ij}$ denotes the probability of going from state “j” to state “i” in one step, similarly we can define $[Math Processing Error]p_{ij}^n$ as the probability of going from state “j” to state “i” in n steps, we can create Transition Probability Matrix as

\[TPM = \begin{bmatrix} p_{11} \ p_{12} \ p_{13} \ . \ .\ \\ p_{21} \ p_{22} \ p_{23} \ . \ .\ \\ \ . \ .\ . \ . \ .\ . \ . \ .\ .\ \\ \ . \ .\ . \ . \ .\ . \ . \ .\ .\ \\ \end{bmatrix}\]## Why Convergence of Markov Chain Important ?

### Revisit MCMC

However MCMC have vast usage in the field of Statistics, Mathematics and Computer Science, here we will discuss simple problem in Bayesian Computation , and asses why convergence of Markov Chain is Important

Let us assume that we want to estimate certain parameter $t(\theta)$ and the model is given such that $g(\theta)$ is prior density for $\theta$ and $f(y | \theta)$ is likelihood of $y = {y_1,y_2 ……y_n}$ give the value of $\theta$ then the posterior can be written as

\[g(\theta | y ) \propto g(\theta)f(y|\theta)\]Which have to be normalized , then the posterior density will be given by

\[g(\theta | y ) = \frac{g(\theta)f(y|\theta)}{\int g(\theta)f(y|\theta)d\theta}\]For the sake of simplicity let us assume $t(\theta) = \theta$ and let us assume $\hat{\theta}$ is an estimate, then take Square Error Loss Function

\[L(\theta , \hat{\theta}) = (\theta - \hat\theta)^2\]Then the Classical Risk will be given by $R_{\hat{\theta}}(\theta) = E_{\theta}(L(\theta,\hat{\theta}))$ and Bayes Risk is given by

\[r(\hat{\theta}) = \int R_{\hat{\theta}}(\theta)g(\theta)d\theta\]Now our target is to minimize bayes risk to get the bayes estimate

\[\begin{align*} r(\theta) &= \int R_{\hat{\theta}}(\theta)g(\theta)d\theta \\ &= \int E_{\theta}(L(\theta,\hat{\theta})) g(\theta)d\theta \\ &= \int \left( \int (\theta - \hat\theta)^2f(y|\theta)dy\right)g(\theta)d\theta \\ &= \int \left( \int (\theta - \hat\theta)^2f(y|\theta)g(\theta)dyd\theta\right) \\ &= \int \left( \int (\theta - \hat\theta)^2g(\theta|y)d\theta\right)f(y)dy \tag{1} \end{align*}\]The equation $({1})$ can be minimized if the inner integral is minimized, when

\[\hat{\theta} = E(\theta |y)\]Now we may not always able to calculate mean of posterior density, that means

\[\hat\theta = \int\theta g(\theta|y)d\theta\]That is when we do not know the kernel density , and integral will be complex , then we use CLT to estimate $\theta$ that is we take random samples from the kernel $g(\theta | y)$ i.e posterior kernel , and calculate the means of the samples , that can be mathematically seen as

\[X^1,X^2......X^n \ are \ samples \ from\ g(\theta|y) \ now \\ \frac{\sum X_i}{n} \to \hat\theta \ as \ n \to \infty\]## When does Markov Chain Converge ?

Now let us take $[Math Processing Error]g(\theta | y) = \pi(\theta)$ it can be assumed because y is realized and $[Math Processing Error]g(\theta | y)$ is only function of $[Math Processing Error]\theta$ , Now comes the MCMC , if we can create a chain whose stationary distribution is $[Math Processing Error]\pi(\theta)$, then we can assume that chain as a random samples which converges to $[Math Processing Error]\pi(\theta)$ and that is the reason we need Markov Chain to converge, before we move forward let us describe some definitions

Let us denote $\pi$ as a probability measure on $(\mathcal{X},\mathcal{B})$ and $\Phi = {X^0,X^1 …}$ are discrete time Markov Chain on $(\mathcal{X},\mathcal{B})$ , let us assume transition kernel P and k as transition density and can be illustrated as

\[P(x,A) = Pr(X^{i+1} \in A | X^i = x ) = \int_A k(x,y)dy\]that is $P(x,A)$ gives us the probability of one step transition probability from state x to any state in A, now the transition kernel assumes two linear operators

- $\lambda P $ where $\lambda$ is probability distribution on $(\mathcal{X},\mathcal{B})$
- Pf where f is non-negative measurable function on on $(\mathcal{X},\mathcal{B})$

where

\[\lambda P(A) = \int_{\mathcal{X}}\lambda(x)P(x,A)dx\]so if $X^i \sim \lambda$ then $\lambda P(A)$ is the marginal distribution of $X^{i+1}\in A$ and

\[Pf(x) = \int_{\mathcal{X}}P(x,dy)f(y) = E_p[f(X_{i+1})|X_i = x]\]and m-step transition probability is given by

\[P^m(x,A) = \int_A k^m(x,y)dy\]Invariant Density - $\pi$

\[\pi = \pi P \\ \Rightarrow \pi(x) = \int_{\mathcal{X}}\pi(y)k(y,x)dy\]

Now there are several way to ensure $\pi$ is invariant (or stationary ) distribution one of the way is , to satisfy the balance condition i.e

\[\pi(x)k(x,y) = \pi(y)k(y,x) \ \ \ \ \ \ \ \ \ \ \ for \ all \ x,y \in \mathcal{X}\]**Proof**

Suppose $\pi$ satisfy the balance condition then

\[\begin{align*} \pi(x)k(x,y) = \pi(y)k(y,x) \ \ \ \ \ \ \ \ \ \ \ for \ all \ x,y \in \mathcal{X} \\ \\ \int_{\mathcal{X}}\pi(y)k(y,x)dy = \int_{\mathcal{X}}\pi(x)k(x,y)dy = \pi(x) \ \ \ \ \ \ \ \ \ \ \ \ \end{align*}\]However Balance condition is not necessary condition it is only sufficient that means Reversibility is not required for $\pi$ to be invariant, suppose $X^i \sim \pi$ and it preserve it distribution over any number of transition , then we say that the Markov chain is stationary and hence it converges to $\pi$ that is required for MCMC

Let us Define

**$\phi$-irreducible** A Markov Chain is for some measure $\phi$ on $\mathcal{X},\mathcal{B}$ if for all $x \in X$ and $A \in \mathcal{B}$ for which $\phi(A) > 0$ , there exist n for which $P^n(x,A)>0$

*A Chain is Aperiodic if Period is 1*

**Harris Recurrent** A $\phi$- irreducible Markov Chain is Harris Recurrent if a $\phi$ positive set A, the chain reaches set A with probability 1

**Harris Ergodic** A Markov Chain is said to be Harris ergodic if it is $\phi$ irreducible , aperiodic , Harris Recurrent and posses invariant distribution $\pi$ for some measure $\phi$ and $\pi$

**Total Variation Distance** The Total Variation distance between two measures $\mu(.) \ and \ v(.)$ is defined by

#### What does Harris Ergodicity Guarantees ?

- Guaranteed to explore entire space without getting stuck
- Strong Consistency of Markov Chain Average
- Convergence of Markov Chain to stationary in total Variation Distance

The following two theorems are very important for MCMC

**Ergodic Theorem** A Markov chain $\Phi$ is Harris ergodic with Invariant Distribution $\pi$ and $E_{\pi} | g(X) | < \infty$ for some function $g : \mathcal{X} \to \Bbb{R}$ Then for any starting value $x \in \mathcal{X}$ , then

and that is the main requirement that we use generally in MCMC

Birkhoff, George D. “Proof of the Ergodic Theorem.”

Proceedings of the National Academy of Sciences of the United States of America, vol. 17, no. 12, 1931, pp. 656–660.JSTOR, www.jstor.org/stable/86016. Accessed 9 Apr. 2021.

The other Theorem is as follows

*Suppose Markov chain $\Phi$ is Harris ergodic with invariant distribution of $\pi$ Then for any starting value $x \in \mathcal{X}$ . $\Phi$ will converge to $\pi$ in total variation distance , i.e

\[||P^n(x,.) - \pi(.)|| \to 0 \ as \ n \to \infty\]*further $ | | P^n(x,.) - \pi(.)| | $ is monotonically non-increasing in n*

## Rate of Convergence

The Ergodic Theorem tells us about convergence of Markov chain however it does not declare anything about the rate of convergence, we define a Markov Chain converging at geometric rate as **geometrically ergodic**, i.e there exist $M:\mathcal{X} \to \Bbb{R}$ and some constant $t \in (0,1)$ that satisfy

If M is bounded , the Markov chain is uniformally ergodic

- As long as the starting value of x , such that M(x) is not large, geometric ergodicity guarantees quick convergence of Markov Chain
- Geometric Ergodicity holds for every irreducible and aperiodic Markov chain on finite space

### What is Needed for Geometric Ergodicity

#### Drift and Minorization Condition

A Type 1 drift condition holds if there exist some non-negative function $V:\mathcal{X} \to \Bbb{R}_{\geq 0}$ and constant $0 < \gamma <1$ and $L < \infty$

\[PV(x) \leq \gamma V(x) + L \ \ \ \ \ \ \ \ \ \ \ \ for \ any \ x \in \mathcal{X}\]Further we call V a drift function and a $\gamma$ a drift rate

A Minorization condition holds on set $C \in \mathcal{B}$ if there exist some positive integer $m ,\epsilon > 0$ and probability measure Q in $(\mathcal{X},\mathcal{B})$ for which

\[P^m(x,A) \geq \epsilon Q(A)\]we can also call this m step minorization condition, here C is called small, It imply the following condition

\[k^m(x,y) \geq \epsilon q(A)\]**Proposition**

Suppose Markov chain $\Phi$ is irreducible and periodic with invariant distribution $\pi$ , Then $\Phi$ is geometrically ergodic if the following two conditions are met:

- Type I drift condition hold
- There exists some constants $d > 2L(1-\gamma)$ for which one step minorization condition holds on set $C= {x:V(x)\leq d}$

This Proposition is a Corollary of Rosenthal(1995a)

*Let $\Phi$ be a a periodic and irreducible Markov chain with invariant distribution $\pi$*

Let us suppose the Condition 1&2 of Proposition holds and $X^0 = x_0$ be the starting value and define

\[\alpha = \frac{1+d}{1+2L+\gamma d} \ \ \ \ \ \ and \ \ \ \ \ \ \ U = 1+2(\gamma d+L)\]Then for any $r \in (0,1)$

\[||P^n(x_0 ,.) - \pi(.)|| \leq (1-\epsilon)^{rn} +\left(\frac{U^r}{\alpha^{1-r}} \right)^n\left(1 + \frac{L}{1-\gamma} + V(x_0)\right)\]We can rearrange this to see that is satisfy geometric ergodicity condition

- V(x) + 1 is proportion to M(x) hence starting point should minimize V(x)

**Type II Drift Condition** : If there exist some function W : $\mathcal{X} \to [1,\infty)$ finite at some x $\in \mathcal{X}$, some set $D \in \mathcal{B}$ , and constants $0 < \rho < 1$ and $b < \infty$ for which

It is easy to show that *Type I Drift Condition $\Leftarrow\Rightarrow$ Type II Drift Condition*

Finally we can say that

**Suppose Markov Chain $\Phi$ is aperiodic and $\phi-$irreducible with invariant distribution $\pi$. Then $\Phi$ is geometrically ergodic if there exist some small set D, the drift function $W: \mathcal{X} \to [1,\infty)$ and some constants $0 < \rho < 1$ and $b < \infty$ for which a type II drift conditions hold**

Now Let me reinstate the earlier theorem

*Suppose Markov chain $\Phi$ is Harris ergodic with invariant distribution of $\pi$ Then for any starting value $x \in \mathcal{X}$ . $\Phi$ will converge to $\pi$ in total variation distance , i.e*

*further $ | | P^n(x,.) - \pi(.)| | $ is monotonically non-increasing in n*

**Jain and Jamison (1967) have shown that for every $\phi-irreducible$ Markov chain on $(\mathcal{X},\mathcal{B})$ . Then there exists some small set $C \in \mathcal{B}$ for which $\phi(C) > 0$.Furthermore , the corresponding minorization measure Q(.) can be defined so that Q(C) > 0**

the Jain and Jamison allow us to assume $C \in \mathcal{B}$ such that

\[P(x , A) \geq \epsilon Q(A) \ \ \ \ \ for \ all \ x \in C\]That is one step minorization condition , Now we can write

\[P(x,A) = \epsilon Q(A) + (1-\epsilon)R(x,A) \ \ \ \ \ \ \ for \ all \ x \in C \ and \ A \in \mathcal{B}\]Here $R(x,.)$ is probability measure for $(\mathcal{X},\mathcal{B})$ , then this allow us to construct two separate chain which couple with probability 1

\[\Phi(X) = \{X^0,X^1 ...........\} \\ \Phi(Y) = \{Y^0,Y^1 ............\}\]Now $(X^{n},Y^n) \to (X^{n+1},Y^{n+1})$ with the following algorithm

- While $X^n \neq Y^n$
- If $(X^n,Y^n) \not\in C \times C$
- Draw $X^{n+1} \sim P(X^n,.)$ and$Y^{n+1} \sim P(Y^n,.)$ independently

- If $(X^n,Y^n) \in C \times C$
- Draw $\delta_n \sim Bern(\epsilon)$
- If $\delta_{n} = 0$ , Draw $X^{n+1} \sim R(X^n,.)$ and$Y^{n+1} \sim R(Y^n,.)$ independently
- otherwise , draw $X^{n+1} = Y^{n+1} \sim P(x,.)$

- If $(X^n,Y^n) \not\in C \times C$
- Once $X^n = x = Y^n $,draw $X^{n+1} = Y^{n+1} \sim P(x,.)$

Now define coupling time T such that T denotes n for which first time $(X^{n-1},Y^{n-1}) \in C \times C$ and $\delta_{n-1}=1$ , once the chain couples it will remain equal

Now let us assume

\[X^0 = x \ and \ Y^0 \sim \pi\]And $Pr_x$ denotes the probability with respect to starting point x, then $\Phi(y)$ is stationary

\[\begin{align*} |P^n(x,A) - \pi(A)| &= |Pr_x(X^n \in A) - Pr_x(Y^n \in A)| \\ &= |Pr_x(X^n \in A,X^n = Y^n) +Pr_x(X^n \in A,X^n \neq Y^n)- Pr_x(Y^n \in A,X^n \neq Y^n)- Pr_x(Y^n \in A,X^n = Y^n)| \\ &= |Pr_x(X^n \in A,X^n \neq Y^n)- Pr_x(Y^n \in A,X^n \neq Y^n)| \\ &\leq max\{Pr_x(X^n \in A,X^n \neq Y^n)- Pr_x(Y^n \in A,X^n \neq Y^n)\} \\ &\leq Pr_x(X^n \neq Y^n) \\ &= Pr_x(T > n) \end{align*}\]Thus

\[||P^n(x,.) - \pi(.)|| \leq Pr_x(T>n)\]Now Let us Suppose Minorization condition hold over entire space i.e $C = \mathcal{X}$ in this case every couple generated belongs to $C \times C$ for all n then

\[T \sim Geo(\epsilon) \\ P(T>n) = (1-\epsilon)^n\]so

\[||P^n(x,.) - \pi(.)|| \leq (1-\epsilon)^n\]so when C = $\mathcal{X}$ , $ | |P^n(x,.) - \pi(.) | | \to 0 \ as \ n \to \infty$

and When $C \neq \mathcal{X}$ , the distribution of $P(X>t)$ is complicated and beyond the scope of this presentation

#### Deterministic Update Gibbs Sampler (DUGS)

Let us assume our Target Distribution is $\pi(\theta)$ such that $\theta = (\theta_1,\theta_2….\theta_d)$

**Notation : $\theta_{-i}$ is vector of parameter except $\theta_i$**

**Initialization :** $\theta^0 = (\theta_1^0,\theta_2^0……\theta_d^0)$

**Iteration:** For $i \geq 1$

- Sample $\theta_1^i \sim \pi(\theta_1^i | \theta^2_{-1})$
- Sample $\theta_2^i \sim \pi(\theta_2 | \theta_1^i , \theta^{i-1}_{-(1,2)})$
- … .. ……………………………
- … … ……………………………
- Sample $\theta_1^i \sim \pi(\theta_d | \theta^{i}_{-d})$

The Transition Kernel for two parameter will be given by

\[k((\theta_1,\theta_2),(\tilde{\theta}_1,\tilde{\theta}_2)) = \pi(\tilde\theta_1|\theta_2)\cdot \pi(\tilde\theta_2|\tilde\theta_1)\]Let us check the stationarity for two parameter

\[\begin{align*} \int\int \pi(\theta_1,\theta_2)k((\theta_1,\theta_2),(\tilde{\theta}_1,\tilde{\theta}_2))d\theta_1d\theta_2 &= \int\int \pi(\theta_1,\theta_2)\pi(\tilde\theta_1|\theta_2)\cdot \pi(\tilde\theta_2|\tilde\theta_1)d\theta_1d\theta_2 \\ &= \int \pi(\theta_2)\pi(\tilde\theta_1|\theta_2)\cdot \pi(\tilde\theta_2|\tilde\theta_1)d\theta_2 \\ &= \int \pi(\tilde\theta_1,\theta_2)\cdot \pi(\tilde\theta_2|\tilde\theta_1)d\theta_2 \\ &= \pi(\tilde\theta_1)\cdot \pi(\tilde\theta_2|\tilde\theta_1) \\ &= \pi(\tilde\theta_2,\tilde\theta_1) \\ \end{align*}\]However this does not suffices for for the convergence, **Aperiodicity needed for surety that the samples are not repeating hence leads to exploring whole space and Irreducibility confirms that it will not stuck** If we are to prove the balance condition the we are assured that it will converge, Let $\Phi_i={\theta_i^0,\theta_i^1……..}$ and let $k_1(\tilde\theta_1,\theta_1)$ be the transition density in $\Phi_i$ , then

#### Example

Let us suppose

\[Y_1 , Y_2 ..... Ym \sim^{iid} N(\mu, \theta)\]where $m \geq 5$ , Let us assume the joint prior density as

\[g(\mu,\theta) \propto \frac{1}{\sqrt{\theta}}\]Let y = $(y_1,y_1 ……y_m)$ as a sample data with mean $\bar y$ and variance $s^2 = \sum(y_i - \bar y)^2$ the the posterior will be given by

\[g(\mu , \theta | y) \propto \theta^{-\frac{m+1}{2}}exp \bigg( -\frac{1}{2\theta} \sum_{j=1}^m (y_j - \mu)^2\bigg)\]and

\[\theta | \mu,y \sim IG\left(\frac{m-1}{2}, \frac{s^2+m(\mu -\bar{y})^2}{2}\right) \\ \mu | \theta ,y \sim N(\bar y,\frac{\theta}{m})\]We know Inverse Gamma have kernel $x^{-(a+1)}e^{-bx}$ with parameter (a,b)

Let us use DUGS Sampler in the following update scheme

\[(\theta^{'},\mu{'}) \to (\theta^{},\mu{'}) \to (\theta^{},\mu{})\]so the kernel density will be given by

\[k((\mu^{'},\theta^{'}),(\mu,\theta)) = \pi(\theta|\mu^{'},y)\pi(\mu|\theta,y)\]**Type 1 Drift Condition**

Let us define $V(\mu , \theta) = (\mu - \bar{y})^2$

\[E[V(\mu,\theta)|\mu^{'},\theta^{'}] = E[V(\mu,\theta)|\mu^{'}] =E[E[V(\mu,\theta)|\theta]|\mu^{'}]\]where

\[E[V(\mu,\theta)|\theta] = E[(\mu-\bar{y})^2|\theta] = Var[\mu|\theta] = \frac{\theta}{m}\]Then

\[E[V(\mu,\theta)|\mu^{'},\theta^{'}] = E\left[\frac\theta m | \mu^{'}\right] \\ \Rightarrow \frac{1}{m} \frac{s^2+m(\mu^{'}-\bar{y})^2}{m-3} \\ \Rightarrow \frac{(\mu^{'}-\bar{y})^2}{m-3} \frac{s^2}{m(m-3)} \\ \Rightarrow \frac{1}{m-3}V(\mu^{'},\theta{'}) + \frac{s^2}{m(m-3)}\]now $m \geq 5$ guarantees that $\frac{1}{m-3} < 1$ hence

\[PV(\mu^{'},\theta^{'}) =E[V(\mu,\theta)|\mu^{'},\theta^{'}] \leq \frac{1}{m-3}V(\mu^{'},\theta{'}) + \frac{s^2}{m(m-3)}\]So its satisfy drift condition with $\gamma \in (1/(m-3),1) $ and $L^2 =s^2/(m(m-3))$

**Minorization Condition**

Let us assume $C = {(\mu,\theta) : V(\mu,\theta) \leq d }$ for $d \geq 2L/(1-\gamma)$ if there exist density q and $\epsilon > 0$ for which

\[k((\mu^{'},\theta^{'}),(\mu, \theta)) \geq \epsilon q(\mu,\theta)\ for \ all \ (\mu^{'},\theta^{'}) \in C \ and \ (\mu, \theta) \in \Bbb{R} \times \Bbb{R}_+\] \[k((\mu^{'},\theta^{'}),(\mu, \theta)) = \pi(\mu|\theta,y)\pi(\theta | \mu^{'},y) \geq \pi(\mu|\theta,y) \inf_{(\mu{'},\theta^{'}) \in C} \pi(\theta | \mu^{'},y)\]Let us assume $IG(a,b ; x)$ denote the density at $ x>0$

\[g(\theta) =\inf_{(\mu{'},\theta^{'}) \in C} \pi(\theta | \mu^{'},y) \\ \Rightarrow IG\left(\frac{m-1}{2},\frac{s^2}{2}+\frac{m}{2}(\mu^{'}-\bar{y})^2;\theta\right) \\ \Rightarrow \left\{ \begin{array}{c} IG(\frac{m-1}{2},\frac{s^2}{2}+\frac{md}{2} ; \theta ) \ \ if \ \theta < \theta^* \\IG(\frac{m-1}{2},\frac{s^2}{2} ; \theta ) \ \ if \ \theta \geq \theta^*\\ \end{array} \right.\]where $\theta^{*} = md[(m-1)log(1+md/s^2)]^{-1}$

\[k((\mu^{'},\theta^{'}),(\mu, \theta)) \geq \pi(\mu | \theta,y)g(\theta) = \epsilon q(\mu,\theta)\]Where $q(\mu , \theta) = \epsilon^{-1}\pi(\mu | \theta,y)g(\theta)$

Hence the Minorization conditions hold

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