<p>Consider a two-player model with P and A. A can engage in criminal activities and P can catch that by putting effort into it. The more effort P puts into it, the more likely he's going to catch the criminal, so the chance of catching the criminal is a function of P's effort. </p>

<p>I model the probability of catching the criminal as follows: </p>

<p><span class="math-container">$$p(x) = \dfrac{e^x-1}{e^x}, x \in [0, \infty),$$</span>
where <span class="math-container">$x$</span> is the principal's effort choice. In this way, we have <span class="math-container">$p(0) = 0$</span>, and <span class="math-container">$\lim_{x\to\infty} p(x) = 1$</span>.</p>

<p>Then I need to model the cost function of effort. In order to have an analytical solution, I need an exponential cost function as follows: <span class="math-container">$c(x) = e^x - 1$</span>. </p>

<p>I have two questions: </p>

<ol>
<li>How legitimate to have an exponential cost function of efforts like this? </li>
<li>Are there any better alternatives for these two specifications? For example, are there other choices to model the catching chance function <span class="math-container">$p(x)$</span> such that <span class="math-container">$p(0) = 0$</span>, and <span class="math-container">$\lim_{x\to\infty} p(x) = 1$</span>?</li>
</ol>

<p>Thanks a lot for any comments in advance. </p>

CC BY-SA 4.0
2020-06-05T14:02:13.850Z
It certainly depends on the specific literature your model is placed in and the contribution you are adding to that literature. The larger the contribution of your insights, the more likely it is that your specifications are accepted. That being said, you could also make the specification even simpler with the catching function just being $p$ and the cost for the principal being $c(p)=\frac{k}{2} p^2$ with $k\geq 0$ being a cost function parameter.
Matthias 
CC BY-SA 4.0
2020-06-06T00:28:18.733Z
Thanks for the comments. This quadratic cost function is widely used because it is convex, and has a nice form after being differentiated.  However, as I said in the question, with this quadratic cost function, I can't get an analytic solution with the current catching function $p$. Then I can only use simulation to get the solution given a certain set of parameters, which is not good for a general analysis and discussion.
Lin Jing
<cost-functions>
Exponential cost function (of effort) and something else
Lin Jing
# Exponential cost function (of effort) and something else

(Asked by: username_0 on 2020-06-05)

Consider a two-player model with P and A. A can engage in criminal activities and P can catch that by putting effort into it. The more effort P puts into it, the more likely he's going to catch the criminal, so the chance of catching the criminal is a function of P's effort.

I model the probability of catching the criminal as follows:

$$p(x) = \dfrac{e^x-1}{e^x}, x \in [0, \infty),$$
where $x$ is the principal's effort choice. In this way, we have $p(0) = 0$, and $\lim\_{x\to\infty} p(x) = 1$.

Then I need to model the cost function of effort. In order to have an analytical solution, I need an exponential cost function as follows: $c(x) = e^x - 1$.

I have two questions:

1. How legitimate to have an exponential cost function of efforts like this?
2. Are there any better alternatives for these two specifications? For example, are there other choices to model the catching chance function $p(x)$ such that $p(0) = 0$, and $\lim\_{x\to\infty} p(x) = 1$?

Thanks a lot for any comments in advance.

**Tags:** cost-functions
> **Comments:**
> **username_1**: It certainly depends on the specific literature your model is placed in and the contribution you are adding to that literature. The larger the contribution of your insights, the more likely it is that your specifications are accepted. That being said, you could also make the specification even simpler with the catching function just being $p$ and the cost for the principal being $c(p)=\frac{k}{2} p^2$ with $k\geq 0$ being a cost function parameter.  
> **username_0**: Thanks for the comments. This quadratic cost function is widely used because it is convex, and has a nice form after being differentiated.  However, as I said in the question, with this quadratic cost function, I can't get an analytic solution with the current catching function $p$. Then I can only use simulation to get the solution given a certain set of parameters, which is not good for a general analysis and discussion.  

---

## 0 answers

