<p>Your first question, if it's literally correct, is easy:</p>

<p>The only way for $u'$ to be positive for c=0 is for p=1. if p =1 then sign($\phi$)=sign($\theta$) so that the product is positive. But, since $exp()&gt;0$ and $p=1$, the first term of $u''$ cancels out and the only way for u'' to be negative is for $\phi$ to be negative. Therefore $\theta&lt;0$ too. So the conditions are $\phi&lt;0$, $\theta&lt;0$ and $p=1$.</p>

<p>Your second question seems like it should have a simple answer too: </p>

<p>It seems that if $\beta R=1$, then consumption should be constant forever, which suggests that the agent consumes $(R-1)W_0$ each period.</p>

<p>The point of writing the Euler equation is to connect one period to the next, not so much to write out the bellman equation non-recursively. So, if you do that then  you have that an increase in consumption today increases utility by $U'(RW-\tilde W)$ and decreases next period's utility by $R\beta U'(R\tilde W-\tilde {\tilde W})$. So again this lets us conclude that $R\tilde W-\tilde{\tilde W}=R W-\tilde W$, i..e that consumption is constant.</p>

<p>So, now that we know that consumption is constant, can we say whether it is $(R-1)W_0$ or not? If consumption in period 1 were higher than that then consumption would have to be lower than that at some point because $W_1&lt;W_0$ and so the only way to keep consumption high is to deplete the capital stock W. If consumption were constant but lower than $(R-1)W_0$ the agent would accumulate assets forever, but a constant consumption of $(R-1)W_0$ would provide higher utility. Therefore , you can be sure that  constant $(R-1)W_0$ is the optimal consumption value.</p>

CC BY-SA 3.0
2016-05-23T20:34:19.197Z
This cannot be true if you look at the assumptions at the beginning of the problem. $R>1$, so you cannot consume more than your initial wealth every period or else you would go negative wealth in the first period.
DornerA
CC BY-SA 3.0
2016-05-23T21:45:33.567Z
Sorry, you are correct. My mistake. I meant $(R-1)W$.
Fix.B.
CC BY-SA 3.0
2016-05-23T18:03:51.890Z
2016-05-23T21:45:02.393Z
2016-05-23T21:45:02.393Z
Fix.B.
<p>Ok, first try for your first question.<br>
From $u'&gt;0\Rightarrow \phi\theta p&gt;0 \Rightarrow \phi,\theta, p\ne 0$.<br>
From $u''&lt;0\Leftrightarrow \phi \theta p c_t^{p-1}\exp(\theta c_t^p)(p-1)c_t^{-1}+\phi \theta p c_t^{p-1}c_t^{p-1}\exp(\theta c_t^p)c_t^{-1}p\theta&lt;0\Leftrightarrow \phi \theta p c_t^{p-1}\exp(\theta c_t^p)(p-1)c_t^{-1}+\phi \theta p c_t^{p-1}c_t^{p-1}\exp(\theta c_t^p)c_t^{-1}p\theta&lt;0
\Leftrightarrow \phi \theta p (p-1)+\phi \theta p c_t^{p-1}p\theta&lt;0$ 
<br>
From here, we set cases.
<br>1. $(p&lt;1 \wedge \phi &lt; 0) \wedge (p=-\gamma\theta)$ with $\gamma\in \mathbb{R}_{++}$
<br>2. $(p&lt;1 \wedge \phi &gt;0) \Rightarrow \left(\frac{1-p}{\theta p}\right)^{\frac{1}{p-1}}&gt;c_t \Rightarrow \theta p &gt;0 $
<br>3. $(p=1 \wedge \phi &lt; 0 \wedge \theta&lt;0)$
<br>4. $(p=1 \wedge \phi &gt; 0 $ (no good)
<br>5. $(p&gt;1 \wedge \phi &lt;0) \Rightarrow \theta&gt;0$
<br>6. $(p&gt;1 \wedge \phi &gt;0) \Rightarrow u''&gt;0$ (no good)
<br>
<br>
So, you get to keep conditions 1-3+5.
<br>
[Will get back when I have time]
<br>
<br>
Euler
$$\beta R u'_{t+1}=u'_t$$
(multiperiod Lagrangian, FOC wrt $c_t,c_{t+1},w_t$, get $\lambda_t=R\lambda_{t+1}$, divide the FOCs wrt $c$s and substitute the $\lambda$s $\Rightarrow$ Euler. At least that would be how I would do it.
<br>
<br>
[Bellman I would have to get my notes and I'm just here to procrastinate and not finishing an essay. I'll be back]</p>

CC BY-SA 3.0
2016-05-21T04:39:11.920Z
2016-05-21T05:10:56.420Z
2016-05-21T05:10:56.420Z
user_newbie10
<p>Your first question (regarding constraints on the parameters) can be answered through first and second derivative analysis. In order to satisfy strictly increasing, we need $u'&gt;0$ and to satisfy strictly concave, we need $u''&lt;0$. What does this actually mean?</p>

<p>$$u'(\frac{}{})=\phi p\theta c_t^{p-1}e^{\theta c_t^p}&gt;0$$
Since we know that $c_t^{p-1}e^{\theta c_t^p}&gt;0$ automatically, this reduces to:
$$\phi p\theta&gt;0$$
Next:
$$u''(\frac{}{})=\big[\phi p\theta (p-1) c_t^{p-2}+\phi p\theta c_t^{p-1}\big]e^{\theta c_t^p}&lt;0$$
Because we know that $e^{\theta c_t^p}&gt;0$ and $\phi p\theta c_t^{p-1}&gt;0$, this must mean: 
$$\phi p\theta (p-1)c_t^{p-2}&lt;0$$</p>

<p>By virtue of $(1)$ and the fact that $c_t^{p-2}&gt;0$, we know that:
$$p-1&lt;0$$
or
$$p&lt;1$$
Now,  what we can do is think about it critically. If $p=0$ then utility does not change when consumption changes. This is clearly untrue. If $p&lt;0$ then utility decreases when consumption increases which also seems incorrect. Therefore, our condition for $p$ is:
$$0&lt;p&lt;1$$
Now, what does this mean for $\phi \;\text{and}\; \theta$? It means: 
$$either\quad\phi , \theta&gt;0\quad or\quad \phi ,\theta&lt;0$$
Now, this looks like a typical exponential utility function, so we know that in these cases, both the coefficient for the exponent and the coefficient in the power will be negative (I will not bother proving this as I am sure it is somewhere in literature). Therefore we have the following three conditions for the parameters:
$$0&lt;p&lt;1$$
$$\theta &lt;0$$
$$\phi &lt;0$$</p>

<p>Now, the next step is to show that we do in fact have constant consumption. Taking our Euler Equation approach, we substitute in the constraint and take two terms in the sum,
$$...+\beta^t\phi e^{\theta (Rw_{t-1}-w_t)^p}+\beta^{t+1}\phi e^{\theta(Rw_t-w_{t+1})^p}+...$$
Differentiating w.r.t $w_t$ and doing some simplifying, we get:
$$(Rw_{t-1}-w_t)^{p-1}e^{\theta(Rw_{t-1}-w_t)^p}=\beta R(Rw_t-w_{t+1})^{p-1}e^{\theta(Rw_t-w_{t+1})^p}$$
Substituting consumption back in, we get:
$$c_t^{p-1}e^{\theta c_t^p}=\beta Rc_{t+1}^{p-1}e^{\theta c_{t+1}^p}$$
Now, since it is specified that $\beta R=1$ in the problem, we can see that the only way this equality holds is if $c_t=c_{t+1}=c^\star$</p>

<p>What does this mean for wealth? Will wealth also be constant? Well, let's conduct a thought experiment and suppose not. Suppose wealth increases each period ($w_{t+1}&gt;w_t\;\forall t$). This implies that as $t\rightarrow \infty$, we will see wealth grow to be infinite as well, which is wasteful and would not satisfy the transversality condition. Now suppose wealth is decreasing ($w_{t+1}&lt;w_t\;\forall t$). We can clearly see that eventually we will not have enough wealth to sustain constant consumption. Therefore we can see that $w_t=w_{t+1}=w^\star$. Now, what are the relationship between these two? Well, going back to our condition for wealth, we can see
$$w^\star=Rw^\star-c^\star$$
Rearranging, we get:
$$c^\star=(R-1)w^\star$$ </p>

<p>Because we have steady state wealth, we know that $w^\star=w_0$</p>

<p>Now, knowing that we have steady state consumption, we know our utility function will be the same value for every period (discounted by $\beta$). So our value function simply becomes:</p>

<p>$$V(w)=\frac{\phi e^{((R-1)w_0)^p}}{1-\beta}$$ </p>

<p>Therefore, we do not need to use the method of undetermined coefficients to find the value function.</p>

CC BY-SA 3.0
2016-05-27T18:29:39.233Z
2016-07-20T17:41:46.950Z
2016-07-20T17:41:46.950Z
DornerA
<p><em>Full disclosure</em>: this problem was part of a final exam that none of our class could really solve definitively. Below the general form is a specific utility function we worked with that I'll try to replicate my work for. Any help on the solution method would be good; specifically, it seems Euler's equations/steady states would be better for this particular form of question rather than using a Bellman, but any guidance on either method would be appreciated.</p>

<hr>

<p>Suppose a representative agent is solving</p>

<p>$$
\begin{align}
&amp; \max_{\left\{ c_t, w_t \right\}^\infty_{t=0}} \sum_{t=0}^\infty \beta^t u(c_t) \\
&amp; \text{s.t.} \\
&amp; w_t = Rw_{t-1} - c_t \\
&amp; c_t \geq 0 \\
&amp; w_t \geq 0 \\
\end{align}
$$</p>

<p>For all $t$, and $w_{-1}$. Additionally, $0 &lt; \beta &lt; 1$, $R &gt; 1$. </p>

<p>$c_t$ is consumption at time $t$, and $w_t$ is wealth at the end of time $t$. $\beta$ is a discount factor.</p>

<p>Let 
$$u(c_t) = \phi \mathrm{e}^ {\theta c_t^p}$$
where $\phi, \theta, p$ are constant parameters.</p>

<blockquote>
  <p>My first question is what values of the constants $\phi, \theta$, and $p$ are needed to ensure that $u(c_t)$ are strictly increasing and strictly concave in $c_t$, for all values of $c_t$. <strong>Later on the problem asks us to assume those above conditions.</strong></p>
</blockquote>

<p>So taking the first and second derivative of $u(c_t)$ gives us:</p>

<p>$$
\begin{align}
u'(c_t) &amp; = \phi \theta p c_t^{p-1}\mathrm{e}^ {\theta c_t^p} \\
u''(c_t) &amp; = \phi \theta p(p-1) c_t^{p-2}\mathrm{e}^ {\theta c_t^p} + \phi \theta ^2 p^2 c_t^{(p-1)^2}\mathrm{e}^ {\theta c_t^p} \\
&amp; = \phi \theta p c_t^{p-1} \mathrm{e}^{\theta c_t^p}\left[ \frac{p-1}{c_t} + \theta p c_t^{p-1} \right] \\
\end{align}$$</p>

<p>So $u'(c_t) &gt; 0$ in order to satisfy the strictly increasing constraint. $u''(c_t) &lt; 0$ must be true for strict concavity. I suggested that $\phi, \theta &gt; 0$ and $p = 1$ for this to be true. I'm not sure how this ugly second derivative can be simplified to find the appropriate conditions.</p>

<hr>

<p>After doing this, we had to express Bellman's equation and Euler's equation for the problem.</p>

<p><strong>Bellman:</strong></p>

<p>$$\boxed{V(w) = \max_\tilde{w} \left[ \phi \mathrm{e}^{\theta (Rw - \tilde{w})^p} + \beta V(\tilde{w})\right]}$$</p>

<p><strong>Euler:</strong></p>

<p>$$\boxed{\sum_{t=0}^\infty \beta^t (Rw_{t-1} - w_t)^p = \cdots \beta^t (Rw_{t-1} - w_t)^p + \beta^{t+1}(Rw_t - w_{t+1})^p + \cdots }$$</p>

<hr>

<p>Finally, we got to say that $\beta \cdot R = 1$. We were then asked to find the value function and optimal decision rules for our choice variable(s). I used the Bellman to try to solve the problem, where I guessed the function form of the value function:</p>

<p>$$V(\tilde{w}) = F \mathrm{e}^{M\tilde{w}} + H$$</p>

<p>where $F, M, H$ are undetermined coefficients. After substituting this into the Bellman and putting in appropriate arguments, taking the derivative to get an optimal choice, substituting the optimal choices back into the Bellman, I got stuck trying to solve for $F, M, H$ in a way that would give me a sensible consumption choice at the end.</p>

<p>The little trick with $\beta R = 1$ is supposed to make consumption constant across all periods, so I assume that the point of getting the Euler's equation is so we can solve it using steady states and make life easier, but my algebra there hasn't borne any fruit either it seems.</p>

<blockquote>
  <p>For the dynamic programming method, have I set the functional form of the value function correctly? If not, what should be the guess? And finally, how does all the nasty algebra work out?</p>
  
  <p>For the steady states method, what does the path of wealth look like? Does it end up constant as well? (I suppose that question applies for the Bellman approach as well.)</p>
</blockquote>

<p>(If it's pedagogically interesting or useful--or if you don't trust that I'm not begging for homework help--I can show my work for both methods.)</p>

<p>Please let me know if anything is unclear.</p>

CC BY-SA 3.0
2016-05-23T07:04:30.213Z
Have you tried the approach that I gave in my answer?
MathUser
<macroeconomics><dynamic-programming><steady-state>
Solution Method for Infinite-Horizon Maximization Problem
Kitsune Cavalry
# Solution Method for Infinite-Horizon Maximization Problem

(Asked by: username_0 on 2016-05-16)

*Full disclosure*: this problem was part of a final exam that none of our class could really solve definitively. Below the general form is a specific utility function we worked with that I'll try to replicate my work for. Any help on the solution method would be good; specifically, it seems Euler's equations/steady states would be better for this particular form of question rather than using a Bellman, but any guidance on either method would be appreciated.

---

Suppose a representative agent is solving

$$
\begin{align}
& \max\_{\left\{ c\_t, w\_t \right\}^\infty\_{t=0}} \sum\_{t=0}^\infty \beta^t u(c\_t) \\
& \text{s.t.} \\
& w\_t = Rw\_{t-1} - c\_t \\
& c\_t \geq 0 \\
& w\_t \geq 0 \\
\end{align}
$$

For all $t$, and $w\_{-1}$. Additionally, $0 < \beta < 1$, $R > 1$.

$c\_t$ is consumption at time $t$, and $w\_t$ is wealth at the end of time $t$. $\beta$ is a discount factor.

Let
$$u(c\_t) = \phi \mathrm{e}^ {\theta c\_t^p}$$
where $\phi, \theta, p$ are constant parameters.

> My first question is what values of the constants $\phi, \theta$, and $p$ are needed to ensure that $u(c\_t)$ are strictly increasing and strictly concave in $c\_t$, for all values of $c\_t$. **Later on the problem asks us to assume those above conditions.**

So taking the first and second derivative of $u(c\_t)$ gives us:

$$
\begin{align}
u'(c\_t) & = \phi \theta p c\_t^{p-1}\mathrm{e}^ {\theta c\_t^p} \\
u''(c\_t) & = \phi \theta p(p-1) c\_t^{p-2}\mathrm{e}^ {\theta c\_t^p} + \phi \theta ^2 p^2 c\_t^{(p-1)^2}\mathrm{e}^ {\theta c\_t^p} \\
& = \phi \theta p c\_t^{p-1} \mathrm{e}^{\theta c\_t^p}\left[ \frac{p-1}{c\_t} + \theta p c\_t^{p-1} \right] \\
\end{align}$$

So $u'(c\_t) > 0$ in order to satisfy the strictly increasing constraint. $u''(c\_t) < 0$ must be true for strict concavity. I suggested that $\phi, \theta > 0$ and $p = 1$ for this to be true. I'm not sure how this ugly second derivative can be simplified to find the appropriate conditions.

---

After doing this, we had to express Bellman's equation and Euler's equation for the problem.

**Bellman:**

$$\boxed{V(w) = \max\_\tilde{w} \left[ \phi \mathrm{e}^{\theta (Rw - \tilde{w})^p} + \beta V(\tilde{w})\right]}$$

**Euler:**

$$\boxed{\sum\_{t=0}^\infty \beta^t (Rw\_{t-1} - w\_t)^p = \cdots \beta^t (Rw\_{t-1} - w\_t)^p + \beta^{t+1}(Rw\_t - w\_{t+1})^p + \cdots }$$

---

Finally, we got to say that $\beta \cdot R = 1$. We were then asked to find the value function and optimal decision rules for our choice variable(s). I used the Bellman to try to solve the problem, where I guessed the function form of the value function:

$$V(\tilde{w}) = F \mathrm{e}^{M\tilde{w}} + H$$

where $F, M, H$ are undetermined coefficients. After substituting this into the Bellman and putting in appropriate arguments, taking the derivative to get an optimal choice, substituting the optimal choices back into the Bellman, I got stuck trying to solve for $F, M, H$ in a way that would give me a sensible consumption choice at the end.

The little trick with $\beta R = 1$ is supposed to make consumption constant across all periods, so I assume that the point of getting the Euler's equation is so we can solve it using steady states and make life easier, but my algebra there hasn't borne any fruit either it seems.

> For the dynamic programming method, have I set the functional form of the value function correctly? If not, what should be the guess? And finally, how does all the nasty algebra work out?
>
> For the steady states method, what does the path of wealth look like? Does it end up constant as well? (I suppose that question applies for the Bellman approach as well.)

(If it's pedagogically interesting or useful--or if you don't trust that I'm not begging for homework help--I can show my work for both methods.)

Please let me know if anything is unclear.

**Tags:** macroeconomics, dynamic-programming, steady-state

> **Comments:**
> **username_1**: Have you tried the approach that I gave in my answer?  

---

## 3 answers

### Answer by username_3 on 2016-05-27. Score: 4 | Accepted answer

Your first question (regarding constraints on the parameters) can be answered through first and second derivative analysis. In order to satisfy strictly increasing, we need $u'>0$ and to satisfy strictly concave, we need $u''<0$. What does this actually mean?

$$u'(\frac{}{})=\phi p\theta c\_t^{p-1}e^{\theta c\_t^p}>0$$
Since we know that $c\_t^{p-1}e^{\theta c\_t^p}>0$ automatically, this reduces to:
$$\phi p\theta>0$$
Next:
$$u''(\frac{}{})=\big[\phi p\theta (p-1) c\_t^{p-2}+\phi p\theta c\_t^{p-1}\big]e^{\theta c\_t^p}<0$$
Because we know that $e^{\theta c\_t^p}>0$ and $\phi p\theta c\_t^{p-1}>0$, this must mean:
$$\phi p\theta (p-1)c\_t^{p-2}<0$$

By virtue of $(1)$ and the fact that $c\_t^{p-2}>0$, we know that:
$$p-1<0$$
or
$$p<1$$
Now, what we can do is think about it critically. If $p=0$ then utility does not change when consumption changes. This is clearly untrue. If $p<0$ then utility decreases when consumption increases which also seems incorrect. Therefore, our condition for $p$ is:
$$0<p<1$$
Now, what does this mean for $\phi \;\text{and}\; \theta$? It means:
$$either\quad\phi , \theta>0\quad or\quad \phi ,\theta<0$$
Now, this looks like a typical exponential utility function, so we know that in these cases, both the coefficient for the exponent and the coefficient in the power will be negative (I will not bother proving this as I am sure it is somewhere in literature). Therefore we have the following three conditions for the parameters:
$$0<p<1$$
$$\theta <0$$
$$\phi <0$$

Now, the next step is to show that we do in fact have constant consumption. Taking our Euler Equation approach, we substitute in the constraint and take two terms in the sum,
$$...+\beta^t\phi e^{\theta (Rw\_{t-1}-w\_t)^p}+\beta^{t+1}\phi e^{\theta(Rw\_t-w\_{t+1})^p}+...$$
Differentiating w.r.t $w\_t$ and doing some simplifying, we get:
$$(Rw\_{t-1}-w\_t)^{p-1}e^{\theta(Rw\_{t-1}-w\_t)^p}=\beta R(Rw\_t-w\_{t+1})^{p-1}e^{\theta(Rw\_t-w\_{t+1})^p}$$
Substituting consumption back in, we get:
$$c\_t^{p-1}e^{\theta c\_t^p}=\beta Rc\_{t+1}^{p-1}e^{\theta c\_{t+1}^p}$$
Now, since it is specified that $\beta R=1$ in the problem, we can see that the only way this equality holds is if $c\_t=c\_{t+1}=c^\star$

What does this mean for wealth? Will wealth also be constant? Well, let's conduct a thought experiment and suppose not. Suppose wealth increases each period ($w\_{t+1}>w\_t\;\forall t$). This implies that as $t\rightarrow \infty$, we will see wealth grow to be infinite as well, which is wasteful and would not satisfy the transversality condition. Now suppose wealth is decreasing ($w\_{t+1}<w\_t\;\forall t$). We can clearly see that eventually we will not have enough wealth to sustain constant consumption. Therefore we can see that $w\_t=w\_{t+1}=w^\star$. Now, what are the relationship between these two? Well, going back to our condition for wealth, we can see
$$w^\star=Rw^\star-c^\star$$
Rearranging, we get:
$$c^\star=(R-1)w^\star$$

Because we have steady state wealth, we know that $w^\star=w\_0$

Now, knowing that we have steady state consumption, we know our utility function will be the same value for every period (discounted by $\beta$). So our value function simply becomes:

$$V(w)=\frac{\phi e^{((R-1)w\_0)^p}}{1-\beta}$$

Therefore, we do not need to use the method of undetermined coefficients to find the value function.


---

### Answer by username_2 on 2016-05-23. Score: 3

Your first question, if it's literally correct, is easy:

The only way for $u'$ to be positive for c=0 is for p=1. if p =1 then sign($\phi$)=sign($\theta$) so that the product is positive. But, since $exp()>0$ and $p=1$, the first term of $u''$ cancels out and the only way for u'' to be negative is for $\phi$ to be negative. Therefore $\theta<0$ too. So the conditions are $\phi<0$, $\theta<0$ and $p=1$.

Your second question seems like it should have a simple answer too:

It seems that if $\beta R=1$, then consumption should be constant forever, which suggests that the agent consumes $(R-1)W\_0$ each period.

The point of writing the Euler equation is to connect one period to the next, not so much to write out the bellman equation non-recursively. So, if you do that then you have that an increase in consumption today increases utility by $U'(RW-\tilde W)$ and decreases next period's utility by $R\beta U'(R\tilde W-\tilde {\tilde W})$. So again this lets us conclude that $R\tilde W-\tilde{\tilde W}=R W-\tilde W$, i..e that consumption is constant.

So, now that we know that consumption is constant, can we say whether it is $(R-1)W\_0$ or not? If consumption in period 1 were higher than that then consumption would have to be lower than that at some point because $W\_1<W\_0$ and so the only way to keep consumption high is to deplete the capital stock W. If consumption were constant but lower than $(R-1)W\_0$ the agent would accumulate assets forever, but a constant consumption of $(R-1)W\_0$ would provide higher utility. Therefore , you can be sure that constant $(R-1)W\_0$ is the optimal consumption value.

> **Comments:**
> **username_3**: This cannot be true if you look at the assumptions at the beginning of the problem. $R>1$, so you cannot consume more than your initial wealth every period or else you would go negative wealth in the first period.  
> **username_2**: Sorry, you are correct. My mistake. I meant $(R-1)W$.  

---

### Answer by username_4 on 2016-05-21. Score: 1

Ok, first try for your first question.  
From $u'>0\Rightarrow \phi\theta p>0 \Rightarrow \phi,\theta, p\ne 0$.  
From $u''<0\Leftrightarrow \phi \theta p c\_t^{p-1}\exp(\theta c\_t^p)(p-1)c\_t^{-1}+\phi \theta p c\_t^{p-1}c\_t^{p-1}\exp(\theta c\_t^p)c\_t^{-1}p\theta<0\Leftrightarrow \phi \theta p c\_t^{p-1}\exp(\theta c\_t^p)(p-1)c\_t^{-1}+\phi \theta p c\_t^{p-1}c\_t^{p-1}\exp(\theta c\_t^p)c\_t^{-1}p\theta<0
\Leftrightarrow \phi \theta p (p-1)+\phi \theta p c\_t^{p-1}p\theta<0$
  
From here, we set cases.
  
1. $(p<1 \wedge \phi < 0) \wedge (p=-\gamma\theta)$ with $\gamma\in \mathbb{R}\_{++}$
  
2. $(p<1 \wedge \phi >0) \Rightarrow \left(\frac{1-p}{\theta p}\right)^{\frac{1}{p-1}}>c\_t \Rightarrow \theta p >0 $
  
3. $(p=1 \wedge \phi < 0 \wedge \theta<0)$
  
4. $(p=1 \wedge \phi > 0 $ (no good)
  
5. $(p>1 \wedge \phi <0) \Rightarrow \theta>0$
  
6. $(p>1 \wedge \phi >0) \Rightarrow u''>0$ (no good)
  
  
So, you get to keep conditions 1-3+5.
  
[Will get back when I have time]
  
  
Euler
$$\beta R u'\_{t+1}=u'\_t$$
(multiperiod Lagrangian, FOC wrt $c\_t,c\_{t+1},w\_t$, get $\lambda\_t=R\lambda\_{t+1}$, divide the FOCs wrt $c$s and substitute the $\lambda$s $\Rightarrow$ Euler. At least that would be how I would do it.
  
  
[Bellman I would have to get my notes and I'm just here to procrastinate and not finishing an essay. I'll be back]


---

