Let $N = \{1,\ldots,n\}$. For a given $(r_1, \ldots, r_n) \in \mathbb R_{++}^n$. I need to solve \begin{align} \max_{(k_1, \ldots, k_n)\in \mathbb R_{++}^n} \prod_{i \in N}{\left[\sum_{j \in N}{(r_j - k_j)} + r_i\ln\left(\frac{k_i}{r_i}\right)\right]}. \end{align} I could not come up with a closed form solution of the maximizers by using first order conditions, because of the log. Is it possible to tell, that there does not exist a closed form solution and we thus have to maximize the product numerically.
Asked
Active
Viewed 57 times
1 Answers
0
$\ln$ is a strictly increasing monotonic function then with $x_k > 0$
$$ \max \prod_k x_x \equiv \max\sum_k \ln(x_k) $$
$$ \begin{align} \max_{(k_1, \ldots, k_n)\in \mathbb R_{++}^n} f(k)=\prod_{i \in N}{\left[\sum_{j \in N}{(r_j - k_j)} + r_i\ln\left(\frac{k_i}{r_i}\right)\right]}. \end{align} $$
so under the hypothesis that
$$ \sum_{j \in N}{(r_j - k_j)} + r_i\ln\left(\frac{k_i}{r_i}\right) > 0 $$
the problem presents a more amenable formulation as
$$ \max F(k) = \ln(f(k)) = \sum_{i\in N}\ln\left(\sum_{j \in N}{(r_j - k_j)} + r_i\ln\left(\frac{k_i}{r_i}\right)\right) $$
with
$$ \frac{\partial F}{\partial k_{\nu}} = \frac{r_{\nu}k_{\nu}-1}{\sum_{j \in N}{(r_j - k_j)} + r_i\ln\left(\frac{k_i}{r_i}\right) }=0\Rightarrow r_{\nu} = k_{\nu} $$
and the stationary global point is attained at
$$ r_{\nu} = k_{\nu}, \forall\nu\in N $$
Cesareo
- 33,252
-
This does not seem to be correct. Your solution gives $f(k) = 0$ for all $r,k,n$. Consider, e.g., $k_i = 1/n^2$ and $r_i = 1/n$ for all $i \in N$. Then $f(k) > 0$ if $1-(1+\ln(n))/n>0$ which is true for all $n>1$. – clueless May 18 '18 at 09:27
-
What I did say was that a stationary global point is attained at. I dind't say that this global (maximum?, minimum? saddle?) was the solution to the problem. Any case, you can use the gradient information to proceed with a steepest ascent/descent method. – Cesareo May 18 '18 at 10:14