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How can I show that $g(r) = (a - be^{-rt})^{+}$ is not necessarily convex nor concave in $r$ when $b > 0$?

By the way,

$$x^{+} = \begin{cases} 0 & \text{ if } x < 0 \\[.7em] x & \text{ if } x \geq 0 \end{cases} $$

I tried plotting in Desmos to get an understanding of the function. It looks really strange at some points. You can also get a plot by entering

$$f\left(r\right)=\left\{r\ge-\frac{1}{t}\ln\left(\frac{b}{a}\right):0,\left(a-be^{-rt}\right)\right\}$$

on https://www.desmos.com/calculator.

Now, I don't know a great way to explain it. A result that the book gives is $g(r) = (ae^{-rt} - b)^{+}$ is, for $a > 0$, decreasing and convex in $r$. So I can use this result, I guess, but I don't see much relation between the two functions.

Any help would be appreciated.

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  • $\begingroup$ You are being asked to disprove something. That usually calls for finding a specific counterexample -- in this case, find a set of parameters $a$, $b$, and $t$ so that the function is not convex, and another set so that it is not concave. Unless you were specifically asked to do so, most likely there is no need to provide a theoretical explanation. $\endgroup$ – LarrySnyder610 May 6 at 0:31
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First assume that $t>0$. Observe the following: $g=0$ for $r$ less then some number $r_0$. [ $\lim_{r \to -\infty} (a-be^{-rt})=a-b\lim_{r \to -\infty} e^{-rt}=a-b\infty=-\infty$]. (Hence $g'(r)=0$ for $r<r_0$. For $r$ sufficiently large $g'(r)=bte^{-rt} >0$. For a function which is convex or concave the derivative is monotonic. It follows that $g'$ is increasing. Hence $g'' \geq 0$. But $g''(r)=-bt^{2}e^{-rt} <0$ for $r$ sufficiently large. Hence $g$ is neither concave nor convex.

The case $t<0$ is similar.

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  • $\begingroup$ Sorry for the late reply, but I don't quite see why there necessarily exists an index $r_{0}$ for which $r < r_{0}$ implies $g = 0$. Can you please clarify? $\endgroup$ – user666729 May 10 at 5:10
  • $\begingroup$ @gallileo22 I am assuming that $t >0$. $e^{-rt} \to \infty$ as $r \to -\infty$ so $a-be^{-rt} <0$ for $r$ less than some $r_0$. Hence $g(r)=0$ for $r$ less than $r_0$. $\endgroup$ – Kavi Rama Murthy May 10 at 5:29
  • $\begingroup$ I thought $g'(r) = bte^{-rt}$, not $g'(r) = be^{-rt}$? Also, $g''(r) = -t^{2}be^{-rt}$? $\endgroup$ – user666729 May 10 at 5:49
  • $\begingroup$ @gallileo22 I have edited my answer. $\endgroup$ – Kavi Rama Murthy May 10 at 5:51
  • $\begingroup$ I'm trying to understand the $t < 0$ case. Can you help me? Here's what I've got $\endgroup$ – user666729 May 10 at 11:55

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