What I mean by the title is often, if we have a value function like $$V(K,I) = \max_{K',I'} F(K') +\beta V(K',I')$$ the First order conditions will give us something that depends on the derivative of the value function, say $V_1(K',I')$. We deal with this by using the B-S condition to find $V_1(K,I)$, then advance it one period?

What about when $V_1(K,I)$ still depends on $V_1(K',I')$, though? (see $(*)$ below for an example, if you wish to skip the setup)

Here is an example. Note that subscripts denote derivatives, the number in the subscript being the input with respect to which we are taking the derivative. The exception to this is negative subscripts, such as $I_{-1}$, where the subscript denotes being one period in the past (in the case of $-1$). A prime denotes being one period in the future (t+1), and two primes denotes being two periods in the future.

Back to the problem: We have value function $$V(K,I_{-1}) = \max_{I,K'',C}u(C) + \beta V(K',I) \\ \text{ s.t. } C+I \leq f(K) \\ \text{ and } I =K'' - (1-\delta)K' $$ So basically we have a simple model with investment that takes two periods to build. $f(K)$ is our production function.

The FOC's give $$ u_1(c) = \lambda_1 \text{, $\lambda_1$ is the multiplier on the first constraint }\\ \lambda_2 = 0 \text{ multiplier on second constraint is zero} \\ \beta V_2(K',I) = -\lambda_1 $$ Now to get $V_2$ we use the B-S condition, which gives $$V_2(K,I_{-1}) = \beta V_1(K',I)$$ because $K' = I_{-1} + (1-\delta)K$ from the second constraint. Use the BS condition again and get $$V_1(K,I_{-1}) =f_1(K) + (1-\delta)V_1(K',I) \tag{*}$$, again, because $K' = I_{-1} + (1-\delta)K$

The previous equation is my question To reiterate, we use B-S condition to find the derivative of the value function, but in $(*)$ the B-S condition depends on what we are trying to find. How do we handle this?

I feel like $V_1(K'I)$ may just be $1$, or $0$, but then that makes $V_2$ be $\beta$ or $0$, which seems wrong...

  • $\begingroup$ What do the primes used in the notation for capital and investment mean? $\endgroup$ Commented Dec 4, 2015 at 19:32
  • 1
    $\begingroup$ Primes indicate variables from one period ahead. They are used to eliminate the time subscripts, so $K$ is really $K_t$, and $K'$ is $K_{t+1}$. And therefore $K''$ is thus two periods ahead. That is, $K'' = K_{t+2}$ $\endgroup$
    – majmun
    Commented Dec 4, 2015 at 19:37
  • $\begingroup$ You use primes in order to indicate also a derivative, and you use a different subscript notation to denote lagged investment ( I guess because "negative prime" would be even stranger a symbol). This makes the lot confusing. Why do you wish to avoid, time subscripts? $\endgroup$ Commented Dec 4, 2015 at 19:45
  • $\begingroup$ The only place prime is used to indicate derivative is the derivative of $u$, which I will edit so that is no longer the case. There is no good reason to avoid time subscripts, it is simply how these problems are usually written, I believe. I guess it indicates the stationarity of the problem, which is important for knowing a solution exists. We are a bit off topic though; I'm just curious how the handle the B-S condition that arises in this problem. $\endgroup$
    – majmun
    Commented Dec 4, 2015 at 19:47
  • $\begingroup$ For my point of view, in order to be on-topic I have to understand what the topic is in the first place. Unfamiliar notation gets in the way (this is not how I am used to see these problems written), so I try first to clarify that so that I can see whether I have anything useful to offer. $\endgroup$ Commented Dec 4, 2015 at 19:53

1 Answer 1


I guess I will provide a basic answer for now, since I somewhat figured this out, and hopefully either someone posts a more complete answer, or I add more to this later.

Basically, we can use forward or backward iteration. For example, for forward iteration (given that we know the value of $V_1$ at $t=0$, call it $V_{initial}$), then we can use this to find the value of $V$ at time $t$ simply by using the equation over and over again. Or we can work backwards, if for some reason forward iteration doesn't work (at this moment I'm not sure why/how this is different then forward). I don't think the solution is too neat, but it works.

Or, if we are at a steady state, then $V_1(K,I_{-1}) = V_1(K'_I)$, so we can simply rearrange the equation and solve.

Again, this isn't a great answer, but it is correct.


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