# Use Lag Operator to find Lifetime Budget Constraint

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!

• 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. Jul 14, 2018 at 8:00
• 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
Jul 16, 2018 at 14:18

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

$$s_{t+1} = \sum_{i=0}^\infty \lambda_{t-i}^{i}(w_{t-i}(1 - l_{t-i}) - c_{t-i} - \tau_{t-i})$$

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