<p>(facepalm)
Multiply both sides with $\beta^{1-t}$:
$$ \beta^{1-t+t} \lambda_{t+1} \frac{\partial f(k_{t+1})}{\partial k_{t+1}} +  \beta^{1-t+t} \lambda_{t+1}  (1 - \delta) - \beta^{1-t+t-1}\lambda_t = 0$$</p>

CC BY-SA 3.0
2015-03-02T15:29:29.710Z
2015-03-02T15:29:29.710Z
Konstantinos
<p>In an intertemporal maximization problem, we seek to find the optimal <em>sequence</em>  of the control and the state variables. It is the recursive nature of the problem that permits us to consider a "typical" point in time and just one condition per variable.  </p>

<p>For each such problem, we need to find out (carefully) <em>in how many distinct periods a specific realization of a variable appears</em>. To do this properly we should distinguish between the "absolute" index, and a "running" index. In the formulation of the Lagrangean as appears in the question, this is not done (and it is usual practice not to, but it may become confusing).  </p>

<p>So I would use the $t$ symbol as the absolute index (to arrive at same-looking first-order conditions), and some other symbol for the running index, say</p>

<p>$$\mathcal{L_t} = \sum_{j=0}^\infty \beta^{j}\left\{u(c_{t+j}) + \lambda_{t+j} \left[ f(k_{t+j}) + (1 - \delta)k_{t+j} - c_{t+j} - k_{t+j+1}\right]\right\} $$</p>

<p>Note that $t$ no longer affects  the discount factor $\beta$, and this is because the discount factor has to do with looking at the future, which is represented by the index $j$. Also, note that $j$ starts at zero, indicating that the first period is the $t$ period.  </p>

<p>Written this way, the Lagrangean says "we are at some point in time indicated by $t$ (that can take the value zero or whatever positive value), and we are looking forward period by period counted by index $j$".</p>

<p>For <em>any given</em> $j$ we have</p>

<p>$$\mathcal{L}_t =...+ \beta^{j}\Big\{u(c_{t+j}) + \lambda_{t+j} \left[ f(k_{t+j}) + (1 - \delta)k_{t+j} - c_{t+j} - k_{t+j+1}\right]\Big\} + \beta^{j+1}\Big\{u(c_{t+j+1}) + \lambda_{t+j+1} \left[ f(k_{t+j+1}) + (1 - \delta)k_{t+j+1} - c_{t+j+1} - k_{t+j+2}\right]\Big\} + ...$$</p>

<p>Pondering this, we realize that the variable $k_{t+j+1}$ will appear in only two consecutive periods, and so the first order condition for a "typical" element of the sequence $\{k_{t+j}\}_{j=0}^{\infty}$ can be exrpessed by differentiating only these two periods with respect to $k_{t+j+1}$. Doing so we get</p>

<p>$$\frac {\partial \mathcal{L}_t}{\partial k_{t+j+1}} =  -\beta^{j} \lambda_{t+j}  + \beta^{j+1}\Big\{ \lambda_{t+j+1} \left[ f'(k_{t+j+1}) + (1 - \delta)\right]\Big\} $$</p>

<p>Take common factors (which will simplify the discount factor) and set equal to zero</p>

<p>$$\frac {\partial \mathcal{L}_t}{\partial k_{t+j+1}} = \beta^{j} \Big[-\lambda_{t+j}  + \beta\Big\{ \lambda_{t+j+1} \left[ f'(k_{t+j+1}) + (1 - \delta)\right]\Big\}\Big] = 0$$</p>

<p>To lighten the indexing burden, we can express this for $j=0$, to obtain</p>

<p>$$\frac {\partial \mathcal{L}_t}{\partial k_{t+1}} = 0 \implies -\lambda_{t}  + \beta\Big\{ \lambda_{t+1} \left[ f'(k_{t+1}) + (1 - \delta)\right]\Big\} = 0$$</p>

CC BY-SA 3.0
2015-03-02T17:02:19.423Z
A comment: when we add uncertainty ($E_t$) we obviously do not change its subscript, because the subscript refers to the variable for which we take the expectation, right? I mean we do not write $E_{t+1}$ instead of $E_{t}$.
Konstantinos
CC BY-SA 3.0
2015-03-02T18:45:21.570Z
@pidosaurus Indeed, because $E_t$ applies to the whole objective function -which is an additional reason why the notation I used is "more" correct ("we stand at time $t$" and so expectations are with respect to information set at $t$). In the alternative notation, invoke the Law of Iterated Expectations to "return back" to $E_{t}$.
Alecos Papadopoulos
CC BY-SA 3.0
2015-03-02T16:27:30.687Z
2015-03-02T16:27:30.687Z
Alecos Papadopoulos
<p>Suppose we have a simple maximization problem as described in Equation 1.1 <a href="http://www.wouterdenhaan.com/teach/ch1.pdf" rel="noreferrer">here</a> or <a href="http://www.wouterdenhaan.com/teach/ch1.pdf" rel="noreferrer">here</a>. This leads us to the Lagrangian Equation 1.3: $$\begin{align*}\mathcal{L} &amp;= \sum_{t=1}^\infty \beta^{t-1}\left\{u(c_t) + \lambda_t \left[ f(k_t) + (1 - \delta)k_t - c_t - k_{t+1}\right]\right\} \\ &amp;= \sum_{t=1}^\infty \left[\beta^{t-1} u(c_t) - \beta^{t-1}\lambda_t c_t + \beta^{t-1} \lambda_t f(\mathbf{k_t}) + \beta^{t-1}\lambda_t(1-\delta)\mathbf{k_t}  - \beta^{t-1}\lambda_t \mathbf{k_{t+1}}\right]  \end{align*}
$$</p>

<p>When we derive the first order condition with respect to $k_{t+1}$, which is:
$$\frac{\partial \mathcal{L} (\cdot)}{\partial k_{t+1}} = 0 :   \beta \lambda_{t+1} \frac{\partial f(k_{t+1})}{\partial k_{t+1}} +  \beta \lambda_{t+1} (1 - \delta) -\lambda_t=0$$</p>

<p>why do we use the subscript $\phantom{.}_{t+1}$ in  $\lambda_{t+1}$ and why does $\beta^{t-1}$ becomes $\beta$? I cannot understand how the first two terms are combined with the last one ($-\lambda_t$). </p>

<p>The relevant terms (with $k$) of the Lagrangian in period $\phantom{.}_{t+1}$ are: $$ \beta^{(t+1)-1} \lambda_{t+1} f(k_{t+1}) +  \beta^{(t+1)-1} \lambda_{t+1} k_{t+1} (1 - \delta) - \beta^{(t+1)-1} \lambda_{t+1} k_{(t+1)+1}$$ so for this part of the sum we do not "care" about the last term when we take the derivative with respect to $k_{t+1}$. So for this period this part of the sum is $$\frac{\partial \mathcal{L}_{t+1}}{\partial k_{t+1}} = \beta^{t} \lambda_{t+1} \frac{\partial f(k_{t+1})}{\partial k_{t+1}} +  \beta^{t} \lambda_{t+1}  (1 - \delta)$$ </p>

<p>The relevant terms (with $k$)  of the Lagrangian in period $\phantom{.}_{t}$ are:
$$\beta^{t-1} \lambda_t f({k_t}) + \beta^{t-1}\lambda_t(1-\delta){k_t}  - \beta^{t-1}\lambda_t k_{t+1}$$
so for this period the part of the sum is $$\frac{\partial \mathcal{L}_t}{\partial k_{t+1}} =  - \beta^{t-1}\lambda_t$$
Now the First Order Condition with respect to $k_{t+1}$ should be:  $$\frac{\partial \mathcal{L}_{t+1}}{\partial k_{t+1}} +  \frac{\partial \mathcal{L}_t}{\partial k_{t+1}} = \beta^{t} \lambda_{t+1} \frac{\partial f(k_{t+1})}{\partial k_{t+1}} +  \beta^{t} \lambda_{t+1}  (1 - \delta) - \beta^{t-1}\lambda_t = 0$$right?</p>

<macroeconomics><dynamic-programming><self-study>
Understanding subscripts in first order conditions of dynamic optimization problems
Konstantinos
# Understanding subscripts in first order conditions of dynamic optimization problems

(Asked by: username_0 on 2015-03-02)

Suppose we have a simple maximization problem as described in Equation 1.1 [here](http://www.wouterdenhaan.com/teach/ch1.pdf) or [here](http://www.wouterdenhaan.com/teach/ch1.pdf). This leads us to the Lagrangian Equation 1.3: $$\begin{align\*}\mathcal{L} &= \sum\_{t=1}^\infty \beta^{t-1}\left\{u(c\_t) + \lambda\_t \left[ f(k\_t) + (1 - \delta)k\_t - c\_t - k\_{t+1}\right]\right\} \\ &= \sum\_{t=1}^\infty \left[\beta^{t-1} u(c\_t) - \beta^{t-1}\lambda\_t c\_t + \beta^{t-1} \lambda\_t f(\mathbf{k\_t}) + \beta^{t-1}\lambda\_t(1-\delta)\mathbf{k\_t} - \beta^{t-1}\lambda\_t \mathbf{k\_{t+1}}\right] \end{align\*}
$$

When we derive the first order condition with respect to $k\_{t+1}$, which is:
$$\frac{\partial \mathcal{L} (\cdot)}{\partial k\_{t+1}} = 0 : \beta \lambda\_{t+1} \frac{\partial f(k\_{t+1})}{\partial k\_{t+1}} + \beta \lambda\_{t+1} (1 - \delta) -\lambda\_t=0$$

why do we use the subscript $\phantom{.}\_{t+1}$ in $\lambda\_{t+1}$ and why does $\beta^{t-1}$ becomes $\beta$? I cannot understand how the first two terms are combined with the last one ($-\lambda\_t$).

The relevant terms (with $k$) of the Lagrangian in period $\phantom{.}\_{t+1}$ are: $$ \beta^{(t+1)-1} \lambda\_{t+1} f(k\_{t+1}) + \beta^{(t+1)-1} \lambda\_{t+1} k\_{t+1} (1 - \delta) - \beta^{(t+1)-1} \lambda\_{t+1} k\_{(t+1)+1}$$ so for this part of the sum we do not "care" about the last term when we take the derivative with respect to $k\_{t+1}$. So for this period this part of the sum is $$\frac{\partial \mathcal{L}\_{t+1}}{\partial k\_{t+1}} = \beta^{t} \lambda\_{t+1} \frac{\partial f(k\_{t+1})}{\partial k\_{t+1}} + \beta^{t} \lambda\_{t+1} (1 - \delta)$$

The relevant terms (with $k$) of the Lagrangian in period $\phantom{.}\_{t}$ are:
$$\beta^{t-1} \lambda\_t f({k\_t}) + \beta^{t-1}\lambda\_t(1-\delta){k\_t} - \beta^{t-1}\lambda\_t k\_{t+1}$$
so for this period the part of the sum is $$\frac{\partial \mathcal{L}\_t}{\partial k\_{t+1}} = - \beta^{t-1}\lambda\_t$$
Now the First Order Condition with respect to $k\_{t+1}$ should be: $$\frac{\partial \mathcal{L}\_{t+1}}{\partial k\_{t+1}} + \frac{\partial \mathcal{L}\_t}{\partial k\_{t+1}} = \beta^{t} \lambda\_{t+1} \frac{\partial f(k\_{t+1})}{\partial k\_{t+1}} + \beta^{t} \lambda\_{t+1} (1 - \delta) - \beta^{t-1}\lambda\_t = 0$$right?

**Tags:** macroeconomics, dynamic-programming, self-study


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## 2 answers

### Answer by username_1 on 2015-03-02. Score: 7 | Accepted answer

In an intertemporal maximization problem, we seek to find the optimal *sequence* of the control and the state variables. It is the recursive nature of the problem that permits us to consider a "typical" point in time and just one condition per variable.

For each such problem, we need to find out (carefully) *in how many distinct periods a specific realization of a variable appears*. To do this properly we should distinguish between the "absolute" index, and a "running" index. In the formulation of the Lagrangean as appears in the question, this is not done (and it is usual practice not to, but it may become confusing).

So I would use the $t$ symbol as the absolute index (to arrive at same-looking first-order conditions), and some other symbol for the running index, say

$$\mathcal{L\_t} = \sum\_{j=0}^\infty \beta^{j}\left\{u(c\_{t+j}) + \lambda\_{t+j} \left[ f(k\_{t+j}) + (1 - \delta)k\_{t+j} - c\_{t+j} - k\_{t+j+1}\right]\right\} $$

Note that $t$ no longer affects the discount factor $\beta$, and this is because the discount factor has to do with looking at the future, which is represented by the index $j$. Also, note that $j$ starts at zero, indicating that the first period is the $t$ period.

Written this way, the Lagrangean says "we are at some point in time indicated by $t$ (that can take the value zero or whatever positive value), and we are looking forward period by period counted by index $j$".

For *any given* $j$ we have

$$\mathcal{L}\_t =...+ \beta^{j}\Big\{u(c\_{t+j}) + \lambda\_{t+j} \left[ f(k\_{t+j}) + (1 - \delta)k\_{t+j} - c\_{t+j} - k\_{t+j+1}\right]\Big\} + \beta^{j+1}\Big\{u(c\_{t+j+1}) + \lambda\_{t+j+1} \left[ f(k\_{t+j+1}) + (1 - \delta)k\_{t+j+1} - c\_{t+j+1} - k\_{t+j+2}\right]\Big\} + ...$$

Pondering this, we realize that the variable $k\_{t+j+1}$ will appear in only two consecutive periods, and so the first order condition for a "typical" element of the sequence $\{k\_{t+j}\}\_{j=0}^{\infty}$ can be exrpessed by differentiating only these two periods with respect to $k\_{t+j+1}$. Doing so we get

$$\frac {\partial \mathcal{L}\_t}{\partial k\_{t+j+1}} = -\beta^{j} \lambda\_{t+j} + \beta^{j+1}\Big\{ \lambda\_{t+j+1} \left[ f'(k\_{t+j+1}) + (1 - \delta)\right]\Big\} $$

Take common factors (which will simplify the discount factor) and set equal to zero

$$\frac {\partial \mathcal{L}\_t}{\partial k\_{t+j+1}} = \beta^{j} \Big[-\lambda\_{t+j} + \beta\Big\{ \lambda\_{t+j+1} \left[ f'(k\_{t+j+1}) + (1 - \delta)\right]\Big\}\Big] = 0$$

To lighten the indexing burden, we can express this for $j=0$, to obtain

$$\frac {\partial \mathcal{L}\_t}{\partial k\_{t+1}} = 0 \implies -\lambda\_{t} + \beta\Big\{ \lambda\_{t+1} \left[ f'(k\_{t+1}) + (1 - \delta)\right]\Big\} = 0$$

> **Comments:**
> **username_0**: A comment: when we add uncertainty ($E_t$) we obviously do not change its subscript, because the subscript refers to the variable for which we take the expectation, right? I mean we do not write $E_{t+1}$ instead of $E_{t}$.  
> **username_1**: @pidosaurus Indeed, because $E_t$ applies to the whole objective function -which is an additional reason why the notation I used is "more" correct ("we stand at time $t$" and so expectations are with respect to information set at $t$). In the alternative notation, invoke the Law of Iterated Expectations to "return back" to $E_{t}$.  

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### Answer by username_0 on 2015-03-02. Score: 2

(facepalm)
Multiply both sides with $\beta^{1-t}$:
$$ \beta^{1-t+t} \lambda\_{t+1} \frac{\partial f(k\_{t+1})}{\partial k\_{t+1}} + \beta^{1-t+t} \lambda\_{t+1} (1 - \delta) - \beta^{1-t+t-1}\lambda\_t = 0$$


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