<p>We are given that $u$ is increasing and concave, and $u(0) = 0$. This implies that $\dfrac{u(t)}{t}$ is decreasing in $t$, and also, $\dfrac{u(t)}{t} &gt;  u'(t)$ for all $t$.</p>

<p>Nicole's maximization problem :
\begin{eqnarray*} \max_{x} \  q(x)(w-t(x))\end{eqnarray*}</p>

<p>FOC :
$q'(x)(w-t(x)) = q(x)t'(x)$</p>

<p>Suppose $x_N$ solves Nicole's problem. Therefore, it satisfy</p>

<p>$q'(x_N)(w-t(x_N)) = q(x_N)t'(x_N)$
or equivalently,</p>

<p>$\dfrac{q'(x_N)}{q(x_N)t'(x_N)} = \dfrac{1}{w-t(x_N)} \tag{1}$</p>

<p>Andrew's maximization problem :
\begin{eqnarray*} \max_{x} \  q(x)u(w-t(x))\end{eqnarray*}</p>

<p>FOC :
$q'(x)u(w-t(x)) = q(x)t'(x)u'(w-t(x))$</p>

<p>or equivalently,</p>

<p>$\dfrac{q'(x)u(w-t(x))}{q(x)t'(x)} = u'(w-t(x)) \tag{2}$</p>

<p>Let us compare the two sides of the above FOC (2) at $x = x_N$,</p>

<p>LHS  = $\dfrac{q'(x_N)u(w-t(x_N))}{q(x_N)t'(x_N)} = \dfrac{u(w-t(x_N))}{w-t(x_N)} &gt; u'(w-t(x_N)) =$ RHS</p>

<p>Since at $x = x_N$ LHS of (2) is greater than RHS of (2), we need to reduce the value of $x$ to make them equal. So, there is $x_A &lt; x_N$ such that </p>

<p>$\dfrac{q'(x_A)u(w-t(x_A))}{q(x_A)t'(x_A)} = u'(w-t(x_A)) \tag{3}$</p>

<p>Therefore, Nicole buys larger quantity of technology. </p>

<p>Observing the FOCs, we can also conclude that increase in $w$ lead to increase in optimal quantity chosen by both.</p>

<p>Now try to solve the remaining two parts.</p>

CC BY-SA 4.0
2018-05-28T11:10:40.220Z
Amit your solution really helps me thank you. We have tried to solve. And I added this solution below of the question box. Please can you look at? What do think? True? And please help me to complete missing ones of this solution. Many thanks! :)
b11bb
CC BY-SA 4.0
2018-05-27T12:09:27.960Z
2018-05-27T12:09:27.960Z
Amit
<blockquote>
<p>Please imagine that Nicole is uncertain of her future wealth. Her wealth in the bad state of the world is zero. Her wealth in the good state is <span class="math-container">$w&gt;0$</span>.  Each state is initially equally likely. However nikole can purchase technology (such as safety equipment) that will make the good state more likely. If she buys <span class="math-container">$x&gt;0$</span> units, the probability of the good state becomes <span class="math-container">$q(x)&gt;1/2$</span> where <span class="math-container">$q(x)$</span>is strictly increasing and strictly concave in x, with <span class="math-container">$q(x)&lt;1$</span> for all x. <em>To purchase x units, Nichole commits to pay <span class="math-container">$t(x)&lt;w$</span> if the good state materialized. She doesn’t pay anything in the bad state</em> Nicole is risk natural. Andrew faces the exact same problem however Andrew is risk averse. Normalize andrew’s utility in the bad state to <span class="math-container">$u(0)=0$</span>. Assume that <span class="math-container">$q(x) $</span> and <span class="math-container">$t(x)$</span> are such that both Nichole and Andrew would like to buy a strictly positive quantity f the technology.</p>
<p>So thinking of x as a choice variable formulate Nicole ‘s and Andrew ‘s maximization problem. How can I prove who buys the larger quantity of the technology? And what does this result tells to me intuitively? And secondly how can I prove how the optimal quantity for ether agents depends on <span class="math-container">$w$</span>?</p>
<p>In the second case, I assume that the technology is provided by a risk national firm whose costs of supplying x units of the technology is <span class="math-container">$kx$</span> where <span class="math-container">$k&gt;0$</span>. The firm knows Nichole ‘s and Andrew ‘s preferences and is allowed to make different offers to them. If we imagine that the firm decides to offer Nichole a fixed quantity of <span class="math-container">$x_N&gt;0$</span> and Andrew a fixed quantity of <span class="math-container">$x_A&gt;0$</span>. Let  <span class="math-container">$t_N(x_N)$</span>  and  <span class="math-container">$t_A(x_A)$</span> denote the optimal price to charge Nichole and Andrew respectively (where I should remember this is allocated only in the good state).</p>
<p>Given a fixed <span class="math-container">$x_N$</span> , I need to derive  <span class="math-container">$t_N(x_N)$</span>and the implicit expression of  <span class="math-container">$t_A(x_A)$</span> for a fixed <span class="math-container">$x_A$</span>. If <span class="math-container">$x_N=x_A$</span>, can I say that  <span class="math-container">$t_N(x_N)$</span>  is greater than  <span class="math-container">$t_A(x_A)$</span>? Or not?  I want to prove it and what can I say about this intuitively? And I want to derive  <span class="math-container">$t_N’(x_N)$</span> and  <span class="math-container">$t_A’(x_A)$</span>. Finally how can I derive an implicit characterization of the optimal quantity to offer to Nichole <span class="math-container">$x_N$</span>. And how can prove whether the optimal value of <span class="math-container">$x_A$</span> is greater than the optimal value of <span class="math-container">$x_N$</span> or not?</p>
</blockquote>
<hr />
<p>What is only done as a solution is</p>
<p>For Nichole</p>
<p><span class="math-container">$$\max_{x} \ \ q(x) u(w-xt(x))+(1-q(x))u(0)$$</span></p>
<p>FOCs</p>
<p><span class="math-container">$$q’(x) u(w-xt(x))+q(x) u’(w-xt(x))-(w-xt(x))’+(1-q’(x))’u(0)=0$$</span></p>
<p><span class="math-container">$$q’(x) u(w-xt(x))+q(x) (-t(x)-x)u’(w-xt(x))=u(0)q’(x)$$</span></p>
<p><span class="math-container">$$q’(x) u(w-xt(x))-q’(x) u(0)=q( x )u’(w-xt(x))(t(x)+x)$$</span>
<span class="math-container">$$q’(x)[ u(w-xt(x))- u(0)]=q( x )u’(w-xt(x))(t(x)+x)$$</span>
<span class="math-container">$$q’(x)[(1/q(x))[u(q(x)(w-xt(x)))- u(0)]]=q( x )u’(w-xt(x))(t(x)+x)$$</span></p>
<p><span class="math-container">$$(q’(x)/q(x))[u(q(x)(w-xt(x)))- u(0)]=q( x )u’(w-xt(x))(t(x)+x)$$</span></p>
<p>Risk natural</p>
<p><span class="math-container">$$q(x)(w-xt(x)+(1-q(x))u(0)=u(q(x)(w-xt(x))$$</span></p>
<p><span class="math-container">$$[u(w-xt(x))-u(0)]=(1/q(x))[u(q(x)(w-xt(x))-u(0)]$$</span></p>
<p>Risk averse</p>
<p><span class="math-container">$$q(x)u(w-xt(x))\le u(q(x)(w-xt(x))$$</span></p>
<p>Any help is appreciated. Thank you.</p>
<hr />
<p>The original version of question</p>
<p><a href="https://i.sstatic.net/sDRro.jpg" rel="nofollow noreferrer"><img src="https://i.sstatic.net/sDRro.jpg" alt="enter image description here" /></a></p>
<hr />
<p><strong>Edit: for adding my solution for remaining parts</strong></p>
<p>Take that Nichole is risk neutral. I end up.</p>
<p><span class="math-container">$${\partial u \over \partial x}[q’(x)(w-t(x))-q(x)t’(x)]=0$$</span>
<span class="math-container">$$q’(x)(w-5(x))-q(x)t’(x )=0$$</span> and differentiate this w.r.t. w.
<span class="math-container">$${\partial q’(x) \over \partial x}{\partial x \over \partial w}(w-t(x))+q’(x)[{\partial w \over \partial w}-{\partial t \over \partial x}{\partial x \over \partial w}-\frac{\partial q(x)}{\partial x}{\partial x \over \partial w}t’(x)-q(x){\partial t’(x) \over \partial x}{\partial x \over \partial w}=0$$</span></p>
<p>After calculations</p>
<p><span class="math-container">$${\partial x \over \partial w}=+{q’(x)\over -q”( x)(w-t(x)+2q’(x)t’(x)+q(x)t”(x)}&gt;0$$</span></p>
<hr />
<p><em>Another part’s solution:</em></p>
<p><span class="math-container">$$\text{profit}=t_N(x_N)k-Nx_N$$</span>
Subject to <span class="math-container">$q(x)u(w-t_N(x_N))\ge 0$</span></p>
<p>But again for this part I couldn’t set up the equation.</p>
<p>What I thought is</p>
<p><span class="math-container">$$max_{(t_N(x_N), t_A(x_A))} \ \ q(x_A)[t_A(x_A)-kx_A]+q(x_N)[t_N(x_N)-kx_N]$$</span></p>
<p>First order conditions</p>
<p>W.r.t. <span class="math-container">$(t_N(x_N)$</span></p>
<p><span class="math-container">$$q’(x_N)(t_N(x_N)-kx_N)+q(x_N)[t’_N(x_N)-k]=0$$</span></p>
<p><span class="math-container">$$t_N(x_N)={q’(x_N)kx_N-q(x_N)[t’_N(x_N)-k]\over q’(x_N)}$$</span></p>
<p>Similar for Andrew</p>
<p><span class="math-container">$$t_A(x_A)={q’(x_A)kx_A-q(x_A)[t’_A(x_A)-k]\over q’(x_A)}$$</span></p>

CC BY-SA 4.0
2018-05-24T10:56:52.273Z
Question is incomplete. You haven't specified clearly what $t(x)$ is.
Amit
CC BY-SA 4.0
2018-05-24T11:13:11.323Z
Sorry @Amit I missed this part. But I corrected it and also I posted the original version of the question. Really thank you for your reply.
b11bb
CC BY-SA 4.0
2018-05-25T14:17:40.900Z
Just to clarify- what exactly are you interested in for the first question? Are you looking for help interpreting the FOCs/RN/RA formulas you gave (and how to use them to see who buys more of the technology)? 

Also- on that note: I think you're missing some close-parentheses and close-brackets in those RN/RA formulas
AndrewC
CC BY-SA 4.0
2018-05-25T22:13:47.197Z
@Andrew I tried to edit some wrong parenthesis. We try to the first part but could not proceed it. Can you want to help? Thanks
b11bb
CC BY-SA 4.0
2018-05-26T23:30:08.933Z
@Amit I updated my question. Whenever you find time, please help me. Thanks.
b11bb
<microeconomics><uncertainty><mechanism-design><principal-agent><adverse-selection>
Question on uncertainity
b11bb
# Question on uncertainity

(Asked by: username_0 on 2018-05-24)

> Please imagine that Nicole is uncertain of her future wealth. Her wealth in the bad state of the world is zero. Her wealth in the good state is $w>0$. Each state is initially equally likely. However nikole can purchase technology (such as safety equipment) that will make the good state more likely. If she buys $x>0$ units, the probability of the good state becomes $q(x)>1/2$ where $q(x)$is strictly increasing and strictly concave in x, with $q(x)<1$ for all x. *To purchase x units, Nichole commits to pay $t(x) if the good state materialized. She doesn’t pay anything in the bad state* Nicole is risk natural. Andrew faces the exact same problem however Andrew is risk averse. Normalize andrew’s utility in the bad state to $u(0)=0$. Assume that $q(x) $ and $t(x)$ are such that both Nichole and Andrew would like to buy a strictly positive quantity f the technology.
>
> So thinking of x as a choice variable formulate Nicole ‘s and Andrew ‘s maximization problem. How can I prove who buys the larger quantity of the technology? And what does this result tells to me intuitively? And secondly how can I prove how the optimal quantity for ether agents depends on $w$?
>
> In the second case, I assume that the technology is provided by a risk national firm whose costs of supplying x units of the technology is $kx$ where $k>0$. The firm knows Nichole ‘s and Andrew ‘s preferences and is allowed to make different offers to them. If we imagine that the firm decides to offer Nichole a fixed quantity of $x\_N>0$ and Andrew a fixed quantity of $x\_A>0$. Let $t\_N(x\_N)$ and $t\_A(x\_A)$ denote the optimal price to charge Nichole and Andrew respectively (where I should remember this is allocated only in the good state).
>
> Given a fixed $x\_N$ , I need to derive $t\_N(x\_N)$and the implicit expression of $t\_A(x\_A)$ for a fixed $x\_A$. If $x\_N=x\_A$, can I say that $t\_N(x\_N)$ is greater than $t\_A(x\_A)$? Or not? I want to prove it and what can I say about this intuitively? And I want to derive $t\_N’(x\_N)$ and $t\_A’(x\_A)$. Finally how can I derive an implicit characterization of the optimal quantity to offer to Nichole $x\_N$. And how can prove whether the optimal value of $x\_A$ is greater than the optimal value of $x\_N$ or not?

---

What is only done as a solution is

For Nichole

$$\max\_{x} \ \ q(x) u(w-xt(x))+(1-q(x))u(0)$$

FOCs

$$q’(x) u(w-xt(x))+q(x) u’(w-xt(x))-(w-xt(x))’+(1-q’(x))’u(0)=0$$

$$q’(x) u(w-xt(x))+q(x) (-t(x)-x)u’(w-xt(x))=u(0)q’(x)$$

$$q’(x) u(w-xt(x))-q’(x) u(0)=q( x )u’(w-xt(x))(t(x)+x)$$
$$q’(x)[ u(w-xt(x))- u(0)]=q( x )u’(w-xt(x))(t(x)+x)$$
$$q’(x)[(1/q(x))[u(q(x)(w-xt(x)))- u(0)]]=q( x )u’(w-xt(x))(t(x)+x)$$

$$(q’(x)/q(x))[u(q(x)(w-xt(x)))- u(0)]=q( x )u’(w-xt(x))(t(x)+x)$$

Risk natural

$$q(x)(w-xt(x)+(1-q(x))u(0)=u(q(x)(w-xt(x))$$

$$[u(w-xt(x))-u(0)]=(1/q(x))[u(q(x)(w-xt(x))-u(0)]$$

Risk averse

$$q(x)u(w-xt(x))\le u(q(x)(w-xt(x))$$

Any help is appreciated. Thank you.

---

The original version of question

[![enter image description here](https://i.sstatic.net/sDRro.jpg)](https://i.sstatic.net/sDRro.jpg)

---

**Edit: for adding my solution for remaining parts**

Take that Nichole is risk neutral. I end up.

$${\partial u \over \partial x}[q’(x)(w-t(x))-q(x)t’(x)]=0$$
$$q’(x)(w-5(x))-q(x)t’(x )=0$$ and differentiate this w.r.t. w.
$${\partial q’(x) \over \partial x}{\partial x \over \partial w}(w-t(x))+q’(x)[{\partial w \over \partial w}-{\partial t \over \partial x}{\partial x \over \partial w}-\frac{\partial q(x)}{\partial x}{\partial x \over \partial w}t’(x)-q(x){\partial t’(x) \over \partial x}{\partial x \over \partial w}=0$$

After calculations

$${\partial x \over \partial w}=+{q’(x)\over -q”( x)(w-t(x)+2q’(x)t’(x)+q(x)t”(x)}>0$$

---

*Another part’s solution:*

$$\text{profit}=t\_N(x\_N)k-Nx\_N$$
Subject to $q(x)u(w-t\_N(x\_N))\ge 0$

But again for this part I couldn’t set up the equation.

What I thought is

$$max\_{(t\_N(x\_N), t\_A(x\_A))} \ \ q(x\_A)[t\_A(x\_A)-kx\_A]+q(x\_N)[t\_N(x\_N)-kx\_N]$$

First order conditions

W.r.t. $(t\_N(x\_N)$

$$q’(x\_N)(t\_N(x\_N)-kx\_N)+q(x\_N)[t’\_N(x\_N)-k]=0$$

$$t\_N(x\_N)={q’(x\_N)kx\_N-q(x\_N)[t’\_N(x\_N)-k]\over q’(x\_N)}$$

Similar for Andrew

$$t\_A(x\_A)={q’(x\_A)kx\_A-q(x\_A)[t’\_A(x\_A)-k]\over q’(x\_A)}$$

**Tags:** microeconomics, uncertainty, mechanism-design, principal-agent, adverse-selection
> **Comments:**
> **username_1**: Question is incomplete. You haven't specified clearly what $t(x)$ is.  
> **username_0**: Sorry @Amit I missed this part. But I corrected it and also I posted the original version of the question. Really thank you for your reply.  
> **username_2**: Just to clarify- what exactly are you interested in for the first question? Are you looking for help interpreting the FOCs/RN/RA formulas you gave (and how to use them to see who buys more of the technology)?   Also- on that note: I think you're missing some close-parentheses and close-brackets in those RN/RA formulas  
> **username_0**: @Andrew I tried to edit some wrong parenthesis. We try to the first part but could not proceed it. Can you want to help? Thanks  
> **username_0**: @Amit I updated my question. Whenever you find time, please help me. Thanks.  

---

## 1 answers

### Answer by username_1 on 2018-05-27. Score: 3 | Accepted answer

We are given that $u$ is increasing and concave, and $u(0) = 0$. This implies that $\dfrac{u(t)}{t}$ is decreasing in $t$, and also, $\dfrac{u(t)}{t} > u'(t)$ for all $t$.

Nicole's maximization problem :
\begin{eqnarray\*} \max\_{x} \ q(x)(w-t(x))\end{eqnarray\*}

FOC :
$q'(x)(w-t(x)) = q(x)t'(x)$

Suppose $x\_N$ solves Nicole's problem. Therefore, it satisfy

$q'(x\_N)(w-t(x\_N)) = q(x\_N)t'(x\_N)$
or equivalently,

$\dfrac{q'(x\_N)}{q(x\_N)t'(x\_N)} = \dfrac{1}{w-t(x\_N)} \tag{1}$

Andrew's maximization problem :
\begin{eqnarray\*} \max\_{x} \ q(x)u(w-t(x))\end{eqnarray\*}

FOC :
$q'(x)u(w-t(x)) = q(x)t'(x)u'(w-t(x))$

or equivalently,

$\dfrac{q'(x)u(w-t(x))}{q(x)t'(x)} = u'(w-t(x)) \tag{2}$

Let us compare the two sides of the above FOC (2) at $x = x\_N$,

LHS = $\dfrac{q'(x\_N)u(w-t(x\_N))}{q(x\_N)t'(x\_N)} = \dfrac{u(w-t(x\_N))}{w-t(x\_N)} > u'(w-t(x\_N)) =$ RHS

Since at $x = x\_N$ LHS of (2) is greater than RHS of (2), we need to reduce the value of $x$ to make them equal. So, there is $x\_A < x\_N$ such that

$\dfrac{q'(x\_A)u(w-t(x\_A))}{q(x\_A)t'(x\_A)} = u'(w-t(x\_A)) \tag{3}$

Therefore, Nicole buys larger quantity of technology.

Observing the FOCs, we can also conclude that increase in $w$ lead to increase in optimal quantity chosen by both.

Now try to solve the remaining two parts.

> **Comments:**
> **username_0**: Amit your solution really helps me thank you. We have tried to solve. And I added this solution below of the question box. Please can you look at? What do think? True? And please help me to complete missing ones of this solution. Many thanks! :)  

---

