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 $t(x)<w$ 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

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


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

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

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

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

Risk natural



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

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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. $(t_N(x_N)$


$$t_N(x_N)={q’(x_N)kx_N-q(x_N)[t’_N(x_N)-k]\over q’(x_N)}$$

Similar for Andrew

$$t_A(x_A)={q’(x_A)kx_A-q(x_A)[t’_A(x_A)-k]\over q’(x_A)}$$

  • $\begingroup$ Question is incomplete. You haven't specified clearly what $t(x)$ is. $\endgroup$
    – Amit
    Commented May 24, 2018 at 10:56
  • $\begingroup$ 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. $\endgroup$
    – b11bb
    Commented May 24, 2018 at 11:13
  • $\begingroup$ 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 $\endgroup$
    – AndrewC
    Commented May 25, 2018 at 14:17
  • $\begingroup$ @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 $\endgroup$
    – b11bb
    Commented May 25, 2018 at 22:13
  • $\begingroup$ @Amit I updated my question. Whenever you find time, please help me. Thanks. $\endgroup$
    – b11bb
    Commented May 26, 2018 at 23:30

1 Answer 1


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.

  • $\begingroup$ 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! :) $\endgroup$
    – b11bb
    Commented May 28, 2018 at 11:10

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