<p>If the variables are constant, everything in the equation is time-invariant, while <span class="math-container">$a^t$</span> will still grow or fall over time (unless <span class="math-container">$a=1$</span>). This is a contradiction and no steady state exists, unless <span class="math-container">$a=1$</span>.</p>

CC BY-SA 4.0
2021-02-21T16:15:17.060Z
Thank you for your comment. This case is a bit confusing for me. I am explaining why: If I defined a new variable as $A_{h} \equiv a^{h}$, $h=t,t+1,...$ then the steady-state would be $1 = A \beta (1+r)$? Is this correct or Is my understanding wrong here?
Giordano
CC BY-SA 4.0
2021-02-21T16:43:02.643Z
No. Simply redefining the object that grows as $A$ does not allow you to assume that it is constant over time. If $A$ has a trend, then $r$ must have one as well.
jpfeifer
CC BY-SA 4.0
2021-02-21T09:49:45.677Z
2021-02-21T16:37:59.423Z
2021-02-21T16:37:59.423Z
jpfeifer
<p>I am wondering how to get the steady-state for the following Euler equation. I know that we can get rid of time in subscripts. However, here I have a constant (a) to the power of <span class="math-container">$t$</span>. Does anyone know if there is a way to get rid of <span class="math-container">$t$</span> in power? Or can I just consider <span class="math-container">$a^t$</span> a new constant, say <span class="math-container">$\bar{a}$</span>?</p>
<p><span class="math-container">$$ \frac{1}{c_{t}}=a^t \beta E_t \Big[(1+r_{t+1})\dfrac{1}{c_{t+1}}\Big] $$</span></p>

<macroeconomics><steady-state>
Constant to the power of t in steady state
Giordano
# Constant to the power of t in steady state

(Asked by: username_0 on 2021-02-19)

I am wondering how to get the steady-state for the following Euler equation. I know that we can get rid of time in subscripts. However, here I have a constant (a) to the power of $t$. Does anyone know if there is a way to get rid of $t$ in power? Or can I just consider $a^t$ a new constant, say $\bar{a}$?

$$ \frac{1}{c\_{t}}=a^t \beta E\_t \Big[(1+r\_{t+1})\dfrac{1}{c\_{t+1}}\Big] $$

**Tags:** macroeconomics, steady-state

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## 1 answers

### Answer by username_1 on 2021-02-21. Score: 3 | Accepted answer

If the variables are constant, everything in the equation is time-invariant, while $a^t$ will still grow or fall over time (unless $a=1$). This is a contradiction and no steady state exists, unless $a=1$.

> **Comments:**
> **username_0**: Thank you for your comment. This case is a bit confusing for me. I am explaining why: If I defined a new variable as $A_{h} \equiv a^{h}$, $h=t,t+1,...$ then the steady-state would be $1 = A \beta (1+r)$? Is this correct or Is my understanding wrong here?  
> **username_1**: No. Simply redefining the object that grows as $A$ does not allow you to assume that it is constant over time. If $A$ has a trend, then $r$ must have one as well.  

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