<p>Since this is a Markov chain</p>

<p><span class="math-container">\begin{eqnarray}
\pi_t(0,0,\cdots,1,1) &amp;=&amp;  \color{blue}{\pi(s_t = 0 | s_{t - 1} = 0)} \cdots \color{magenta}{\pi(s_2 = 0 | s_{1} = 1)} \color{red}{\pi(s_1 = 1 | s_{0} = 1)}\color{orange}{\pi(s_0 = 1)} \\
&amp;=&amp; \color{blue}{1} \times \cdots\times \color{magenta}{0.5}\times\color{red}{1}\times \color{orange}{1} = 0.5
\end{eqnarray}</span></p>

<p>But you are right if <span class="math-container">$t &gt; 2$</span> then <span class="math-container">$\pi(s_{t + 1}  = 0 | s_{t} = 0) = 1$</span>, so if <span class="math-container">$s_t = 0$</span> then <span class="math-container">$s_{t + 1} = 0$</span>, but at <span class="math-container">$t = 2$</span> there's a 50/50 chance that the state changes from <span class="math-container">$1$</span> to <span class="math-container">$0$</span>, so the state may remain <span class="math-container">$1$</span> as in the first history, or switch to <span class="math-container">$0$</span> as the second one</p>

CC BY-SA 4.0
2019-01-07T00:47:49.977Z
2019-01-07T00:47:49.977Z
caverac
<p>I am reading LS book, on Chapter 8 - Complete Markets (3rd version). There is an example on page 264 that I am quite confused. </p>

<p>I attach the picture of the example here under.</p>

<p>I don't understand why we have a history (0,0,0,...,0,1,1)? since if <span class="math-container">$s_t=0$</span>, then <span class="math-container">$s_{t+1} = 0$</span> for sure, right?</p>

<p>Or the authors means the history here is in order <span class="math-container">$(s_t, s_{t-1},....,s_1,s_0)$</span>,
so it makes sense since we know that <span class="math-container">$s_0=1, s_1=1$</span>, but isn't it quite weird, since we usually write a history <span class="math-container">$h_t=(s_0,s_1,...s_t)$</span> right?</p>

<p>Anyone has an idea? Really appreciate your help!</p>

<p><a href="https://i.sstatic.net/Ni5qb.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/Ni5qb.png" alt="enter image description here"></a></p>

<complete-markets>
LS book (Recursive Macro Theory). Chapter8: Complete Market
bluesky23
# LS book (Recursive Macro Theory). Chapter8: Complete Market

(Asked by: username_0 on 2019-01-06)

I am reading LS book, on Chapter 8 - Complete Markets (3rd version). There is an example on page 264 that I am quite confused.

I attach the picture of the example here under.

I don't understand why we have a history (0,0,0,...,0,1,1)? since if $s\_t=0$, then $s\_{t+1} = 0$ for sure, right?

Or the authors means the history here is in order $(s\_t, s\_{t-1},....,s\_1,s\_0)$,
so it makes sense since we know that $s\_0=1, s\_1=1$, but isn't it quite weird, since we usually write a history $h\_t=(s\_0,s\_1,...s\_t)$ right?

Anyone has an idea? Really appreciate your help!

[![enter image description here](https://i.sstatic.net/Ni5qb.png)](https://i.sstatic.net/Ni5qb.png)

**Tags:** complete-markets

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## 1 answers

### Answer by username_1 on 2019-01-07. Score: 1

Since this is a Markov chain

\begin{eqnarray}
\pi\_t(0,0,\cdots,1,1) &=& \color{blue}{\pi(s\_t = 0 | s\_{t - 1} = 0)} \cdots \color{magenta}{\pi(s\_2 = 0 | s\_{1} = 1)} \color{red}{\pi(s\_1 = 1 | s\_{0} = 1)}\color{orange}{\pi(s\_0 = 1)} \\
&=& \color{blue}{1} \times \cdots\times \color{magenta}{0.5}\times\color{red}{1}\times \color{orange}{1} = 0.5
\end{eqnarray}

But you are right if $t > 2$ then $\pi(s\_{t + 1} = 0 | s\_{t} = 0) = 1$, so if $s\_t = 0$ then $s\_{t + 1} = 0$, but at $t = 2$ there's a 50/50 chance that the state changes from $1$ to $0$, so the state may remain $1$ as in the first history, or switch to $0$ as the second one


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