<p>ANSWER CHANGED TO EXPRESSION IMMEDIATELY BELOW on 7/29/2021 BASED ON BrsG's QUESTION REGARDING WHAT HAPPENED TO THE LAG OPERATOR.</p>
<p><span class="math-container">$s_{t+1} = \sum_{t=0}^{\infty} \lambda_{t-i}^{i} (w_{t-i}(1-l_{t-i}) -c_{t-i} - \tau_{t-i})$</span></p>
<p>where <span class="math-container">$\lambda_{t} = (1 + r_t)$</span>.</p>
<p>DETAILS OF ANSWER FOLLOW. THANKS to BrsG FOR ASKING QUESTION WHICH LED TO CORRECTION AND DELETION OF ORIGINAL ANSWER.</p>
<p>The first step  was to make the substitution that, by definition,</p>
<p><span class="math-container">$s_{t+1} = (1 + r_t) s_{t}$</span></p>
<p><span class="math-container">$ = \lambda_t s_{t}$</span> where <span class="math-container">$\lambda_t = (1 + r_t)$</span>. Hopefully this definition is correct because the answer relies on it.</p>
<p>Then, given this relation and the lag operator,  the initial expression given in the question can be written as</p>
<p><span class="math-container">$ s_{t+1} (1 - \lambda_{t} L) = (w_t(1-l_{t}) - c_t - \tau_t)$</span>.</p>
<p>Next, I divide both sides by <span class="math-container">$(1 - \lambda_t L)$</span> which results in</p>
<p><span class="math-container">$ s_{t+1} = \frac{w_t(1-l_{t}) - c_t - \tau_t}{\left(1 - \lambda_t L\right)}$</span>.</p>
<p>But now the RHS can be re-written as an infinite sum if one makes the assumption that the absolute value of <span class="math-container">$\lambda_t L$</span> is less than 1.0.</p>
<p>So, writing out the RHS as an infinite sum results in</p>
<p><span class="math-container">$ s_{t+1} = \sum_{i=0}^\infty \lambda_{t-i}^{i}(w_{t-i}(1 - l_{t-i}) - c_{t-i} - \tau_{t-i})$</span></p>

CC BY-SA 4.0
2021-07-29T13:12:38.733Z
May I suggest to put your comments (those in capital) as a simple comment below.
BrsG
CC BY-SA 4.0
2021-07-29T14:25:03.880Z
I modified my original answer on 7/29/2021 after BrsG asked me a question regarding how the lag operator was eliminated. Thanks to him for the question because it led to the change in the answer which is written immediately above.
mark leeds
CC BY-SA 4.0
2021-07-29T13:08:27.640Z
2021-07-29T13:08:27.640Z
mark leeds
<p>The budget constraint is </p>

<p>$c_t + \tau_t + s_{t+1} =w_t(1-l_t) +(1+r_t)s_t$</p>

<p>And assume</p>

<p>$\underset{t \longrightarrow \infty}{lim} \ \displaystyle{\frac{s_t}{\Pi_{i=1}^{t-1} (1+r_i)}} = 0$</p>

<p>Lag Operator $L$ is defined as $L \cdot x_{t+1} = x_t$</p>

<p>How can I get lifetime budget constraint using the Lag Operator?</p>

<p>Many Thanks!</p>

CC BY-SA 4.0
2018-07-14T08:00:53.170Z
T.G. I'm going to print this out and work on it on scrap paper but if you get $s_{t}$ alone on the left hand side, then you can divide by (1-something L) and obtain an infinite sum of the other stuff. Obviously, this is very vague but I will play around and see what I come up with.
mark leeds
CC BY-SA 4.0
2018-07-16T14:18:50.763Z
Please confirm the variable definitions. Is $c_t$ is consumption, $\tau_t$ is lump sum taxes, $s_t$ is savings / wealth, $r_t$ is the return on wealth, $w_t$ is the wage rate, and $l_t$ is the quantity of leisure?
BKay
<macroeconomics><representative-agent>
Use Lag Operator to find Lifetime Budget Constraint
T. G.
# Use Lag Operator to find Lifetime Budget Constraint

(Asked by: username_0 on 2018-05-29)

The budget constraint is

$c\_t + \tau\_t + s\_{t+1} =w\_t(1-l\_t) +(1+r\_t)s\_t$

And assume

$\underset{t \longrightarrow \infty}{lim} \ \displaystyle{\frac{s\_t}{\Pi\_{i=1}^{t-1} (1+r\_i)}} = 0$

Lag Operator $L$ is defined as $L \cdot x\_{t+1} = x\_t$

How can I get lifetime budget constraint using the Lag Operator?

Many Thanks!

**Tags:** macroeconomics, representative-agent
> **Comments:**
> **username_1**: T.G. I'm going to print this out and work on it on scrap paper but if you get $s_{t}$ alone on the left hand side, then you can divide by (1-something L) and obtain an infinite sum of the other stuff. Obviously, this is very vague but I will play around and see what I come up with.  
> **username_2**: Please confirm the variable definitions. Is $c_t$ is consumption, $\tau_t$ is lump sum taxes, $s_t$ is savings / wealth, $r_t$ is the return on wealth, $w_t$ is the wage rate, and $l_t$ is the quantity of leisure?  

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## 1 answers

### Answer by username_1 on 2021-07-29. Score: 1

ANSWER CHANGED TO EXPRESSION IMMEDIATELY BELOW on 7/29/2021 BASED ON BrsG's QUESTION REGARDING WHAT HAPPENED TO THE LAG OPERATOR.

$s\_{t+1} = \sum\_{t=0}^{\infty} \lambda\_{t-i}^{i} (w\_{t-i}(1-l\_{t-i}) -c\_{t-i} - \tau\_{t-i})$

where $\lambda\_{t} = (1 + r\_t)$.

DETAILS OF ANSWER FOLLOW. THANKS to BrsG FOR ASKING QUESTION WHICH LED TO CORRECTION AND DELETION OF ORIGINAL ANSWER.

The first step was to make the substitution that, by definition,

$s\_{t+1} = (1 + r\_t) s\_{t}$

$ = \lambda\_t s\_{t}$ where $\lambda\_t = (1 + r\_t)$. Hopefully this definition is correct because the answer relies on it.

Then, given this relation and the lag operator, the initial expression given in the question can be written as

$ s\_{t+1} (1 - \lambda\_{t} L) = (w\_t(1-l\_{t}) - c\_t - \tau\_t)$.

Next, I divide both sides by $(1 - \lambda\_t L)$ which results in

$ s\_{t+1} = \frac{w\_t(1-l\_{t}) - c\_t - \tau\_t}{\left(1 - \lambda\_t L\right)}$.

But now the RHS can be re-written as an infinite sum if one makes the assumption that the absolute value of $\lambda\_t L$ is less than 1.0.

So, writing out the RHS as an infinite sum results in

$ s\_{t+1} = \sum\_{i=0}^\infty \lambda\_{t-i}^{i}(w\_{t-i}(1 - l\_{t-i}) - c\_{t-i} - \tau\_{t-i})$

> **Comments:**
> **username_3**: May I suggest to put your comments (those in capital) as a simple comment below.  
> **username_1**: I modified my original answer on 7/29/2021 after BrsG asked me a question regarding how the lag operator was eliminated. Thanks to him for the question because it led to the change in the answer which is written immediately above.  

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