Finding ##P(X_2 = 2)## of a Markov Chain

[user366312]

Homework Statement

If $(X_n)_{n≥0}$ is a Markov chain on $S = \{1, 2, 3\}$ with initial distribution $α = (1/2, 1/2, 0)$ and transition matrix

$\begin{bmatrix} 1/2&0&1/2\\ 0&1/2&1/2\\ 1/2&1/2&0 \end{bmatrix},$

then $P(X_2 = 2) = ?$ and $E(X_2)=?$.

Relevant Equations

Markov Chain

My solution:

$X_1 = \begin{bmatrix} 1/2&1/2&0 \end{bmatrix} \begin{bmatrix} 1/2&0&1/2\\ 0&1/2&1/2\\ 1/2&1/2&0 \end{bmatrix} = \begin{bmatrix} 1/4&1/4&1/2 \end{bmatrix}$

$X_2 = \begin{bmatrix} 1/4&1/4&1/2 \end{bmatrix} \begin{bmatrix} 1/2&0&1/2\\ 0&1/2&1/2\\ 1/2&1/2&0 \end{bmatrix} = \begin{bmatrix} 3/8&3/8&1/4 \end{bmatrix}$

So, $P(X_2=2) = 3/8$

$E(X_2=2) = 1 * 3/8 + 2 * 3/8 + 3 * 1/4 = 15/8$
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Is this solution correct?

Why or why not?

 

[tnich]

You have done all the work correctly, so your solutions are correct.

 

[RPinPA]

Is this solution correct?

Not quite for pedantic reasons.

Why or why not?

Because you wrote $E[X_2 = 2]$ and what you're calculating is the expectation of the variable $X_2$, so the proper notation is $E[X_2]$.

Pedantic, as I admit. But if your teacher is pedantic as well, you don't want to lose points for silly notational stuff like that.