I am studying a certain maximisation problem (coming from some sort of likelyhood estimations); after a number of generalisations I need to examine

$$f(x_1, x_2,y) = a_1\ln x_1 +a_2 \ln x_2 + \ln y - \left(x_1 y+ x_2 y + cx_1+\frac {x_2}{c} \right). $$

We have $c>0$, $c\ne 1$, $x_i>0$, $y>0$, $a_i\ge 1$.

It is quite easy to show that the equation $\nabla f=0$ has a unique solution and that in that point $f$ is strictly concave, hence this point is a local maximum. I can give exact analytical expressions; they are a bit cumbersome, however.

One can also show that this function is bounded, because $$f(x_1,x_2,y)= \frac{a_1}{a_1+a_2} \ln (x_1y)-x_1 y\\ +\frac{a_2}{a_1+a_2} \ln (x_2y)-x_2 y\\+ a_1\left(1-\frac{1}{a_1+a_2}\right)\ln x_1 - cx_1\\ +a_2\left(1-\frac{1}{a_1+a_2}\right)\ln x_2 - x_2/c.$$

Each of the four terms in this sum admits a unique global maximum, hence $f$ is bounded.

I'd appreciate any hints and/or links helping to show that $f$ admits a global maximum.

  • 2
    $\begingroup$ From the decomposition, you see that for every $C$ there are $0 < m_i < M_i$ such that $x_i < m_i \lor x_i > M_i \implies f(x_1,x_2,y) < C$. Then you find $0 < n < N$ so that for $m_i \leqslant x_i \leqslant M_i$ you have $y < n \lor y > N \implies f(x_1,x_2,y) < C$. $\endgroup$ Jun 26 '15 at 13:17
  • $\begingroup$ @DanielFischer can we say that $L_i = \sup_{y} \left(\frac{a_i}{a_1+a_2} \ln (x_iy)-x_i y \right)= \frac{a_i}{a_1+a_2}\ln \frac{a_i}{a_1+a_2}-\frac{a_i}{a_1+a_2}$ and then write $f(x_1,x_2,y)\le L_1+L_2+a_1\left(1-\frac{1}{a_1+a_2}\right)\ln x_1 - cx_1 +a_2\left(1-\frac{1}{a_1+a_2}\right)\ln x_2 - x_2/c$. After that we say that for an arbitrary $C<0$ there exist closed intervals $I_i\subset (0,+\infty)$ such that $f\le C$ whenever $x_i\notin I_i$ and conclude without finding the boundaries for $y$? $\endgroup$ Jun 26 '15 at 14:29
  • 1
    $\begingroup$ If you don't constrain $y$, how would you know that $f$ never exceeds your local maximum for $x_i \in I_i$ and $y$ sufficiently large/sufficiently close to $0$? I'd think it's simpler to get the boundaries for $y$, then you have $f \leqslant C$ for $(x_1,x_2,y) \notin I_1\times I_2 \times I_y$. Choose $C$ small enough, and you can conclude that your local maximum is global, for $f$ attains a maximum in the interior of the cuboid, and that is the global maximum and a local maximum. $\endgroup$ Jun 26 '15 at 14:38
  • $\begingroup$ @DanielFischer but of course! If you post your comment as an answer, I will be able to upvote and accept it. Thank you! $\endgroup$ Jun 26 '15 at 14:49

If you show the existence of a compact set $K$ such that outside $K$ the values of $f$ are strictly smaller than the value at the critical point, it follows that the (unique) global maximum is attained at the (unique) critical point. For $f$ attains a maximum on $K$ by continuity, and that maximum is then the global (and hence also local) maximum, thus it can only be attained at the critical point.

The shape of $f$ allows to deduce the existence of such a compact set. For $A,B > 0$, the function $g_{A,B}(x) = A\ln x - Bx$ tends to $-\infty$ as $x\to 0$ or $x\to +\infty$, hence is bounded above. You have decomposed $f$ as the sum of four functions of that type.

From the term

$$a_1\biggl(1-\frac{1}{a_1+a_2}\biggr)\ln x_1 - cx_1,$$

by the boundedness from above of the other terms, it follows that for any $C\in \mathbb{R}$ there is a compact interval $I_1 \subset (0,+\infty)$ such that $f(x_1,x_2,y) < C$ for $x_1 \notin I_1$. Ditto we find a compact $I_2 \subset (0,+\infty)$ such that $f(x_x,x_2,y) < C$ for $x_2 \notin I_2$.

Then we can use

$$\frac{a_1}{a_1+a_2}\ln (x_1y) - x_1y \leqslant \frac{a_1}{a_1+a_2}\ln (M_1y) - m_1y$$

where $I_1 = [m_1,M_1]$ to deduce the existence of a compact $I_y \subset (0,+\infty)$ such that $f(x_1,x_2,y) < C$ whenever $x_1\in I_1$ and $y\notin I_y$. The product $I_1\times I_2 \times I_y$ satisfies our requirements if we choose $C$ less than (or equal to) the value at the critical point.


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