<p>You are right, it is the same thing. Actually, later on the book the density over the history <span class="math-container">$s^t = [s_t, s_{t-1}, \cdots, s_0]$</span> is written as </p>

<p><span class="math-container">$$
\pi(s^t) = \pi(s_t|s_{t-1}) \cdots \pi (s_1|s_0)\pi(s_0) \tag{2.3.1}
$$</span></p>

<p>where <span class="math-container">$\pi(s_0)$</span> denotes the probability of the initial state (or <span class="math-container">$\pi_0(s_0)$</span> if you wish), and <span class="math-container">$\pi(s|s')$</span> is a transition probability.</p>

<p>Note that Eq. (2.3.1) encapsulates the the probability of obtaining a given history in a Markov process, since <span class="math-container">${\rm Prob}(s_t|s_{t-1}, s_{t-2}\cdots,s_0) = {\rm Prob}(s_t|s_{t-1}) = \pi(s_t|s_{t-1})$</span>. This just to show that <span class="math-container">$\pi_t(s^t)\color{blue}{\equiv}\pi(s^t)$</span></p>

CC BY-SA 4.0
2019-01-07T00:14:59.067Z
2019-01-07T00:14:59.067Z
caverac
<p>I am reading LS book "Recursive Macroeconomics Theory".</p>

<p>In this book, as well as in some online lecture notes on the Internet. They often denote, for example, <span class="math-container">$\pi_t(s^t)$</span> for probability that history <span class="math-container">$s^t$</span> happens at time <span class="math-container">$t$</span>.</p>

<p>But I wonder, if we are mentioning <span class="math-container">$s^t$</span>, it implies time is <span class="math-container">$t$</span>. So, why don't we just make it simple by writing <span class="math-container">$\pi(s^t)$</span> instead?</p>

<p>Thank in advance for any help!</p>

<macroeconomics><complete-markets>
Probability of history occurring at time $t$
bluesky23
# Probability of history occurring at time $t$

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

I am reading LS book "Recursive Macroeconomics Theory".

In this book, as well as in some online lecture notes on the Internet. They often denote, for example, $\pi\_t(s^t)$ for probability that history $s^t$ happens at time $t$.

But I wonder, if we are mentioning $s^t$, it implies time is $t$. So, why don't we just make it simple by writing $\pi(s^t)$ instead?

Thank in advance for any help!

**Tags:** macroeconomics, complete-markets

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## 1 answers

### Answer by username_1 on 2019-01-07. Score: 2 | Accepted answer

You are right, it is the same thing. Actually, later on the book the density over the history $s^t = [s\_t, s\_{t-1}, \cdots, s\_0]$ is written as

$$
\pi(s^t) = \pi(s\_t|s\_{t-1}) \cdots \pi (s\_1|s\_0)\pi(s\_0) \tag{2.3.1}
$$

where $\pi(s\_0)$ denotes the probability of the initial state (or $\pi\_0(s\_0)$ if you wish), and $\pi(s|s')$ is a transition probability.

Note that Eq. (2.3.1) encapsulates the the probability of obtaining a given history in a Markov process, since ${\rm Prob}(s\_t|s\_{t-1}, s\_{t-2}\cdots,s\_0) = {\rm Prob}(s\_t|s\_{t-1}) = \pi(s\_t|s\_{t-1})$. This just to show that $\pi\_t(s^t)\color{blue}{\equiv}\pi(s^t)$


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