<p>The first order condition results in a differential equation in $\beta(x_1)$, </p>

<p>$$\left[ x_{1} + \alpha x_{1} + \alpha \frac{N-2}{2} x_{1} - \beta(x_{1})\right] (N-1)  - (1 + \gamma) \beta'(x_{1})x_{1} = 0$$</p>

<p>$$\implies (1 + \gamma) x_{1}\beta'(x_{1}) = -(N-1)\beta(x_{1}) + (N-1)\left(1+a \frac{N}{2}\right)x_1$$</p>

<p>or generically $$g(x_1) y' = f_1(x_1)y + f_0(x_1)$$</p>

<p>Where $y \equiv \beta(x_1)$ and the other obvious mappings.</p>

<p>This has general solution </p>

<p>$$y= \beta(x_1) = C\cdot  \exp \{F\} + \exp\{F\}\int \exp\{-F\} \frac{f_0(x_1)}{g(x_1)} dx_1, \;\;\; F\equiv \int \frac{f_1(x_1)}{g(x_1)}dx_1 $$</p>

<p>and doing the work results in </p>

<p>$$ \beta(x_1) = \beta(z^*) = Cx_1^{-(N-1)/(1+\gamma)} + \frac{(N \alpha + 2)(N-1)}{2(N + \gamma)}x_1$$</p>

<p>I guess there exists a condition in the model that leads to $C=0$ so that we get the linear expression for the $\beta$ function</p>

<p>$$ \beta(z^*) = \frac{(N \alpha + 2)(N-1)}{2(N + \gamma)}x_1$$</p>

<p>Regarding whether $\Pi(z)$ has a global maximum at the stationary point $z^* = x_1$, working the 2nd derivative with respect to $z$, and evaluating it at $z^*=x_1$ is what one can do.  </p>

<p>Then, if I haven't done any trivial mistake, it appears that a necessary condiiton for the second derivative to be negative at $z^* = x_1$ is that $\alpha(1+\gamma) &lt;1$ and, given this, the condition for a global maximum at $z^*=x_1$is </p>

<p>$$N  &gt; \frac{2+\gamma}{1-\alpha(1+\gamma)}$$</p>

<p>which appears to put a lower bound in the number of participants in the auction, which I guess is what $N$ measures. </p>

<p>But this is the 2nd time I had to guess (and a third guess is that parameters are positive), validating @denesp comment that even though the OP issue is mathematical, providing context and information on what the various symbols stand for, is needed if one is to have a good-quality question to work with. After all, this context is what would make this an economics question and not a math question.</p>

CC BY-SA 3.0
2018-03-15T15:41:27.740Z
2018-03-15T15:41:27.740Z
Alecos Papadopoulos
<p>I have the following function:</p>

<p>$$
\Pi =\int_{0}^{z}[x_{1} + \alpha y + \alpha \frac{N-2}{2}y - \beta(z) - \gamma ( \beta(z) - \beta(y))](N-1)y^{N-2} dy $$</p>

<p>The first derivative with respect to $z$ is:</p>

<p>$$
\frac{\partial \Pi}{\partial z}=[ x_{1} + \alpha z + \alpha \frac{N-2}{2}z - \beta(z)] (N-1) z^{N-2} - (1 + \gamma) \beta'(z)z^{N-1}
$$</p>

<p>at $z = x_{1}$ it must be zero so that:</p>

<p>$$
[ x_{1} + \alpha x_{1} + \alpha \frac{N-2}{2} x_{1} - \beta(x_{1})] (N-1)  - (1 + \gamma) \beta'(x_{1})x_{1} = 0
$$</p>

<p>The solution is:</p>

<p>$$
\beta(z) = \frac{(N \alpha + 2)(N-1)}{2(N + \gamma)}
$$</p>

<p>What I want to prove is that the function $\Pi$ have a global maximum in $x_{1}$. Until now I think I've only obtained a necessary condition. Can you suggest me a way to prove it?</p>

<p>EDIT: $\beta(\cdot)$ is an increasing function.</p>

CC BY-SA 3.0
2018-03-14T10:42:44.570Z
How is this an economics question?
Giskard
CC BY-SA 3.0
2018-03-14T11:02:26.433Z
This is an equilibrium strategy in a common value auction. People in economics use a lot maximization and similar tools. I'm an economic student. This is why I've chosen to post it here.
zar
CC BY-SA 3.0
2018-03-14T11:28:17.140Z
You start by assuming that the function has an extreme at $z_1$. What you want to prove is that it is is maximum, right?
caverac
CC BY-SA 3.0
2018-03-14T11:31:03.630Z
Check the [second order sufficient condition](https://catalog.flatworldknowledge.com/bookhub/13?e=mcafee-ch09_s03)
Herr K.
CC BY-SA 3.0
2018-03-14T12:07:11.367Z
@caverac Yes, I assume that the function has an extreme at $x_{1}$ and I want to prove that it is in fact a global maximum.
zar
CC BY-SA 3.0
2018-03-14T13:24:54.923Z
@caverac Why is it wrong? Anyway, $\beta()$ is a strictly increasing function.
zar
CC BY-SA 3.0
2018-03-14T14:05:39.570Z
"Anyway, $\beta()$ is a strictly increasing function." See, this is the kind of thing that should be incorporated in the body of the question. Providing context gives important information about your assumption and would also make the question more useful for future visitors.
Giskard
CC BY-SA 3.0
2018-03-15T14:33:09.873Z
@caverac The differentiation-under-the integral result is correct since we have $$\frac d{dz}\int_0^z f(t,z)dt = f(z,z) + \int_0^z \frac {\partial}{\partial z}f(t,z)dt$$
Alecos Papadopoulos
<microeconomics><optimization><auctions><profit-maximization>
How to prove that a point is a maximum point
zar
# How to prove that a point is a maximum point

(Asked by: username_0 on 2018-03-14)

I have the following function:

$$
\Pi =\int\_{0}^{z}[x\_{1} + \alpha y + \alpha \frac{N-2}{2}y - \beta(z) - \gamma ( \beta(z) - \beta(y))](N-1)y^{N-2} dy $$

The first derivative with respect to $z$ is:

$$
\frac{\partial \Pi}{\partial z}=[ x\_{1} + \alpha z + \alpha \frac{N-2}{2}z - \beta(z)] (N-1) z^{N-2} - (1 + \gamma) \beta'(z)z^{N-1}
$$

at $z = x\_{1}$ it must be zero so that:

$$
[ x\_{1} + \alpha x\_{1} + \alpha \frac{N-2}{2} x\_{1} - \beta(x\_{1})] (N-1) - (1 + \gamma) \beta'(x\_{1})x\_{1} = 0
$$

The solution is:

$$
\beta(z) = \frac{(N \alpha + 2)(N-1)}{2(N + \gamma)}
$$

What I want to prove is that the function $\Pi$ have a global maximum in $x\_{1}$. Until now I think I've only obtained a necessary condition. Can you suggest me a way to prove it?

EDIT: $\beta(\cdot)$ is an increasing function.

**Tags:** microeconomics, optimization, auctions, profit-maximization

> **Comments:**
> **username_1**: How is this an economics question?  
> **username_0**: This is an equilibrium strategy in a common value auction. People in economics use a lot maximization and similar tools. I'm an economic student. This is why I've chosen to post it here.  
> **username_2**: You start by assuming that the function has an extreme at $z_1$. What you want to prove is that it is is maximum, right?  
> **username_3**: Check the [second order sufficient condition](https://catalog.flatworldknowledge.com/bookhub/13?e=mcafee-ch09_s03)  
> **username_0**: @username_2 Yes, I assume that the function has an extreme at $x_{1}$ and I want to prove that it is in fact a global maximum.  
> **username_0**: @username_2 Why is it wrong? Anyway, $\beta()$ is a strictly increasing function.  
> **username_1**: "Anyway, $\beta()$ is a strictly increasing function." See, this is the kind of thing that should be incorporated in the body of the question. Providing context gives important information about your assumption and would also make the question more useful for future visitors.  
> **username_4**: @username_2 The differentiation-under-the integral result is correct since we have $$\frac d{dz}\int_0^z f(t,z)dt = f(z,z) + \int_0^z \frac {\partial}{\partial z}f(t,z)dt$$  

---

## 1 answers

### Answer by username_4 on 2018-03-15. Score: 2

The first order condition results in a differential equation in $\beta(x\_1)$,

$$\left[ x\_{1} + \alpha x\_{1} + \alpha \frac{N-2}{2} x\_{1} - \beta(x\_{1})\right] (N-1) - (1 + \gamma) \beta'(x\_{1})x\_{1} = 0$$

$$\implies (1 + \gamma) x\_{1}\beta'(x\_{1}) = -(N-1)\beta(x\_{1}) + (N-1)\left(1+a \frac{N}{2}\right)x\_1$$

or generically $$g(x\_1) y' = f\_1(x\_1)y + f\_0(x\_1)$$

Where $y \equiv \beta(x\_1)$ and the other obvious mappings.

This has general solution

$$y= \beta(x\_1) = C\cdot \exp \{F\} + \exp\{F\}\int \exp\{-F\} \frac{f\_0(x\_1)}{g(x\_1)} dx\_1, \;\;\; F\equiv \int \frac{f\_1(x\_1)}{g(x\_1)}dx\_1 $$

and doing the work results in

$$ \beta(x\_1) = \beta(z^\*) = Cx\_1^{-(N-1)/(1+\gamma)} + \frac{(N \alpha + 2)(N-1)}{2(N + \gamma)}x\_1$$

I guess there exists a condition in the model that leads to $C=0$ so that we get the linear expression for the $\beta$ function

$$ \beta(z^\*) = \frac{(N \alpha + 2)(N-1)}{2(N + \gamma)}x\_1$$

Regarding whether $\Pi(z)$ has a global maximum at the stationary point $z^\* = x\_1$, working the 2nd derivative with respect to $z$, and evaluating it at $z^\*=x\_1$ is what one can do.

Then, if I haven't done any trivial mistake, it appears that a necessary condiiton for the second derivative to be negative at $z^\* = x\_1$ is that $\alpha(1+\gamma) <1$ and, given this, the condition for a global maximum at $z^\*=x\_1$is

$$N > \frac{2+\gamma}{1-\alpha(1+\gamma)}$$

which appears to put a lower bound in the number of participants in the auction, which I guess is what $N$ measures.

But this is the 2nd time I had to guess (and a third guess is that parameters are positive), validating @denesp comment that even though the OP issue is mathematical, providing context and information on what the various symbols stand for, is needed if one is to have a good-quality question to work with. After all, this context is what would make this an economics question and not a math question.


---

