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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!

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  • $\begingroup$ 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. $\endgroup$
    – mark leeds
    Jul 14, 2018 at 8:00
  • $\begingroup$ 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? $\endgroup$
    – BKay
    Jul 16, 2018 at 14:18

1 Answer 1

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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})$

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  • $\begingroup$ May I suggest to put your comments (those in capital) as a simple comment below. $\endgroup$
    – BrsG
    Jul 29, 2021 at 13:12
  • $\begingroup$ 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. $\endgroup$
    – mark leeds
    Jul 29, 2021 at 14:25

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