Consider Example 4.4.2. Suppose now that the probability...

Consider Example 4.4.2. Suppose now that the probability that a 0 is received as a 1 is $\alpha$ and the probability a 1 is received as a 0 is $\beta$, so that the transition probability matrix $P$ is given by $$ P=\left[\begin{array}{cc} 1-\alpha & \alpha \\ \beta & 1-\beta \end{array}\right] \text {. } $$ (a) Show that $\left(\pi_0, \pi_1\right)=(\beta /(\alpha+\beta, \alpha /(\alpha+\beta)$ is a stationary distribution. (b) Show that $f_0^{(1)}=(1-\alpha)$ and $f_0^{(n)}=\alpha \beta(1-\beta)^{n-2}$ for $n=$ $2,3, \ldots$. (c) Calculate the mean recurrence time $m_0=\sum_{n=1}^{\infty} n f_0^{(n)}$ and verify that $\pi_0=1 / m_0$.

Consider Example 4.4.2. Suppose now that the probability that a 0 is received as a 1 is $\alpha$ and the probability a 1 is received as a 0 is $\beta$, so that the transition probability matrix $P$ is given by
$$
P=\left[\begin{array}{cc}
1-\alpha & \alpha \\
\beta & 1-\beta
\end{array}\right] \text {. }
$$
(a) Show that $\left(\pi_0, \pi_1\right)=(\beta /(\alpha+\beta, \alpha /(\alpha+\beta)$ is a stationary distribution.
(b) Show that $f_0^{(1)}=(1-\alpha)$ and $f_0^{(n)}=\alpha \beta(1-\beta)^{n-2}$ for $n=$ $2,3, \ldots$.
(c) Calculate the mean recurrence time $m_0=\sum_{n=1}^{\infty} n f_0^{(n)}$ and verify that $\pi_0=1 / m_0$.

Key Concepts

Mean Recurrence Time

The mean recurrence time for a state is the expected number of steps it takes for the process to return to that state for the first time, based on the first return probabilities. Mathematically, it is expressed as m_i = ? n f_i^(n). An important relationship in Markov chain theory is that for positive recurrent states, the stationary probability of a state is the reciprocal of its mean recurrence time, i.e., ?_i = 1 / m_i.

Markov Chain

A Markov chain is a stochastic process in which the probability of transitioning to any particular state depends solely on the current state and not on the sequence of events that preceded it. This property, known as the Markov property, simplifies analysis of systems that evolve randomly over time.

Transition Probability Matrix

The transition probability matrix is a square matrix used to describe the probabilities of moving from one state to another in a Markov chain. Each entry in the matrix represents the probability of transitioning from one state to a certain state in one step, and each row of the matrix typically sums to one, reflecting the total probability distribution for the next step from any given state.

Stationary Distribution

A stationary distribution is a probability distribution over the states of a Markov chain that remains unchanged after a transition. Formally, if ? is the stationary distribution and P is the transition matrix, then ? must satisfy the equation ?P = ?. It characterizes the long-run or equilibrium behavior of the Markov chain, provided that the chain is irreducible and aperiodic.

First Return Probability

The first return probability, f_i^(n), is the probability that, starting from state i, the process returns to state i for the first time exactly after n steps. This concept is crucial in understanding the recurrence properties of states within a Markov chain and forms the basis for further analysis like determining the mean recurrence time.

Transcript

00:01 Here we are given a markov chain with two states, 0 and 1, and the probability of going from state 0 to 1 is alpha, and the probability of going from state 1 to 0 is beta.

00:15 For part a we are asked to determine the steady state probabilities of states 0 and 1 in terms of alpha and beta.

00:25 We can first construct our one -step transition matrix for this markov chain.

00:30 So the one -step probability of transitioning from 0 to 1 is alpha, and from 1 to 0 is beta.

00:48 And the roles must sum to 1, so we know that the 1 -step transition probability going from 0 to 0 must be 1 minus alpha, and from 1 to 1 must be 1 minus beta.

01:04 For the steady -state distribution, pi times the 1 -step transition matrix is equal to pi.

01:13 So we can say that pi 1, pi sub 2, the product of these two matrices is equal to pi sub 1, pi sub 2.

01:46 And so this gives us some linear equations.

02:15 And the third equation that we have is that we know that the sum of pi 1 and pi 2 is equal to 1.

02:32 So from equation 3, we can isolate pi sub 2.

02:37 It's equal to 1 minus pi sub 1.

02:42 And then if we substitute this value for pi sub 2 into equation 1, we get the following.

03:35 And then collecting like terms and rearranging, we get.

04:01 And now if we substitute this expression of pi sub 1 into this equation, we get value for pi sub 2 in terms of alpha and beta.

04:13 So these give us our steady -state distribution for states 0 and 1 in terms of alpha and beta.

04:58 And now for part b, we're asked what happens if alpha and or beta equals 0 or 1.

05:10 So if alpha and beta are both equal to 1, then we have a transition matrix that looks like this.

05:24 So if you're in state 0, you must go to state 1...

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