<p>I'm currently learning the RBC model, but as someone with a Math's background am struggling to understand the lack of rigour around the expectation. It's clearly explained that you can view the RBC as all decisions being taken at time 0, so expectations begin from time 0.</p>
<p>To take a simple example, the objective for the firm is</p>
<p><span class="math-container">$$
\max_{\{k_t\}} E_0 \sum_{t=1}^\infty \pi(k_t)
$$</span></p>
<p>The argument given is that we can ignore the expectation here, and take a simple first order condition for each <span class="math-container">$k_t$</span>, that is</p>
<p><span class="math-container">$$
\frac{\partial \pi}{\partial k_t} ( k_t) = 0
$$</span></p>
<p>On the household side of the model, we show some equivalence between <span class="math-container">$E_0$</span> and <span class="math-container">$E_t$</span> which is similarly confusing to me.</p>
<p>I think what is difficult is determining what is a random variable in this context and what is not. Focusing just on <span class="math-container">$k_t$</span>, the argument I've found argues that you can treat</p>
<p><span class="math-container">$$ k_t = k_t(s_1, s_2, s_3 , \dots , s_t)$$</span></p>
<p>For some set of states up to time <span class="math-container">$t$</span>.</p>
<p>However, trying to treat this rigorously, I would then write the expectation <span class="math-container">$E_0$</span> as</p>
<p><span class="math-container">$$ E_0(x) = \int_{\{s_i\}} x p(s_1, s_2, \dots, s_t, \dots) ds_1 ds_2 \dots$$</span></p>
<p>over the infinite p.d.f of states.</p>
<p>For the first order condition on <span class="math-container">$k_t$</span>, (and writing the derivative wrt to <span class="math-container">$k$</span> as <span class="math-container">$\pi_K(k)$</span> we then have</p>
<p><span class="math-container">$$ \int_{\{s_i\}} \pi_K(k_t(s_1, \dots, s_t)) p(s_1, \dots, s_t, \dots) ds_1 ds_2 \dots $$</span></p>
<p>or alternatively</p>
<p><span class="math-container">$$ E_0 [ \pi_K(k_t(s_1, \dots, s_t)) ] = 0 $$</span></p>
<p>This FOC is different to the one without expectation, quite clearly.</p>
<p>I've seen some arguments around simplification when using discrete state sets (and conditional distributions) - but can anyone explain this step rigorously in probability theory terms?</p>

CC BY-SA 4.0
2023-11-06T00:00:02.127Z
Can you give any reference where the argument you find strange occurs? Your question isn't very clear right now.
Michael Greinecker
CC BY-SA 4.0
2023-11-06T07:24:07.367Z
Sure - for example
- Pages 1091-1093 here, this author does a similar derivation https://web.econ.ku.dk/okocg/VM/VM-general/Kapitler%20til%20bog/Ch30-2016-1.pdf
- Slide 9 here https://www.karlwhelan.com/MAMacro/part7.pdf - the author makes a handwaving argument for ignoring the expectation which is what I'm looking to make more rigorous.
mwaddoups
CC BY-SA 4.0
2023-11-06T15:49:07.940Z
@mwaddoups : Hi. I understand the argument in your question but I don't see the same thing happening in the examples you sent. Could you be more specific in where that occurs in those examples. Thanks.
mark leeds
CC BY-SA 4.0
2023-11-06T18:54:14.337Z
Thanks for getting back. I think the second article I linked has the clearest example of the argument.

Slide 11, final point, the author states "Then, the time t + n variables appear exactly as the time t variables do, except that they are in expectation form and they are multiplied by the discount rate βn. But this means the FOCs for the time t + n variables will be identical to those for the time t variables. So differentiating this equation gives us the equations for the optimal dynamics at all times"

My reading is he argues you can ignore the expectation in the FOC.
mwaddoups
CC BY-SA 4.0
2023-11-06T23:14:08.917Z
@mwaddoups I don't think this is what is meant here. The idea is that from period $t$ onward, you will just behave as if $t$ would be the initial period, and your optimal continuation will satisfy a similar FOC conditional on the capital stock then.
Michael Greinecker
CC BY-SA 4.0
2023-11-07T06:09:46.787Z
@mwaddoups : Michael is following what's going on with this better than I am so I'll let you and Michael talk about it.
mark leeds
CC BY-SA 4.0
2023-11-08T06:23:40.717Z
mwaddoups: Assuming that I  understand Michael's comment,  then I think he's right. It's along the same lines as Bellman's equation but there are no choices to make. So, once you are there, the value collapses to a sum.  But, assuming that Michael is correct,  then, when they took the expectation out,  the intended notation would have been clearer if they had used something like $\forall ~ t = 1, ... \infty$. Thanks to Michael for clearing that up.
mark leeds
<rbc>
Ignoring expectation in first order conditions for RBC model
mwaddoups
# Ignoring expectation in first order conditions for RBC model

(Asked by: username_0 on 2023-11-05)

I'm currently learning the RBC model, but as someone with a Math's background am struggling to understand the lack of rigour around the expectation. It's clearly explained that you can view the RBC as all decisions being taken at time 0, so expectations begin from time 0.

To take a simple example, the objective for the firm is

$$
\max\_{\{k\_t\}} E\_0 \sum\_{t=1}^\infty \pi(k\_t)
$$

The argument given is that we can ignore the expectation here, and take a simple first order condition for each $k\_t$, that is

$$
\frac{\partial \pi}{\partial k\_t} ( k\_t) = 0
$$

On the household side of the model, we show some equivalence between $E\_0$ and $E\_t$ which is similarly confusing to me.

I think what is difficult is determining what is a random variable in this context and what is not. Focusing just on $k\_t$, the argument I've found argues that you can treat

$$ k\_t = k\_t(s\_1, s\_2, s\_3 , \dots , s\_t)$$

For some set of states up to time $t$.

However, trying to treat this rigorously, I would then write the expectation $E\_0$ as

$$ E\_0(x) = \int\_{\{s\_i\}} x p(s\_1, s\_2, \dots, s\_t, \dots) ds\_1 ds\_2 \dots$$

over the infinite p.d.f of states.

For the first order condition on $k\_t$, (and writing the derivative wrt to $k$ as $\pi\_K(k)$ we then have

$$ \int\_{\{s\_i\}} \pi\_K(k\_t(s\_1, \dots, s\_t)) p(s\_1, \dots, s\_t, \dots) ds\_1 ds\_2 \dots $$

or alternatively

$$ E\_0 [ \pi\_K(k\_t(s\_1, \dots, s\_t)) ] = 0 $$

This FOC is different to the one without expectation, quite clearly.

I've seen some arguments around simplification when using discrete state sets (and conditional distributions) - but can anyone explain this step rigorously in probability theory terms?

**Tags:** rbc
> **Comments:**
> **username_1**: Can you give any reference where the argument you find strange occurs? Your question isn't very clear right now.  
> **username_0**: Sure - for example - Pages 1091-1093 here, this author does a similar derivation https://web.econ.ku.dk/okocg/VM/VM-general/Kapitler%20til%20bog/Ch30-2016-1.pdf - Slide 9 here https://www.karlwhelan.com/MAMacro/part7.pdf - the author makes a handwaving argument for ignoring the expectation which is what I'm looking to make more rigorous.  
> **username_2**: @mwaddoups : Hi. I understand the argument in your question but I don't see the same thing happening in the examples you sent. Could you be more specific in where that occurs in those examples. Thanks.  
> **username_0**: Thanks for getting back. I think the second article I linked has the clearest example of the argument.  Slide 11, final point, the author states "Then, the time t + n variables appear exactly as the time t variables do, except that they are in expectation form and they are multiplied by the discount rate βn. But this means the FOCs for the time t + n variables will be identical to those for the time t variables. So differentiating this equation gives us the equations for the optimal dynamics at all times"  My reading is he argues you can ignore the expectation in the FOC.  
> **username_1**: @mwaddoups I don't think this is what is meant here. The idea is that from period $t$ onward, you will just behave as if $t$ would be the initial period, and your optimal continuation will satisfy a similar FOC conditional on the capital stock then.  
> **username_2**: @mwaddoups : Michael is following what's going on with this better than I am so I'll let you and Michael talk about it.  
> **username_2**: mwaddoups: Assuming that I  understand Michael's comment,  then I think he's right. It's along the same lines as Bellman's equation but there are no choices to make. So, once you are there, the value collapses to a sum.  But, assuming that Michael is correct,  then, when they took the expectation out,  the intended notation would have been clearer if they had used something like $\forall ~ t = 1, ... \infty$. Thanks to Michael for clearing that up.  

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## 0 answers

