2
$\begingroup$

I would like to minimize the function $$ (\alpha,\theta) \mapsto F(\alpha,\theta) := -\theta x^\alpha + \sum_{k=1}^N \ln(1+p_k(e^{\theta \ell_k^\alpha}-1)) $$ where $\theta \in (0,+\infty)$, $\alpha \in (0,1)$ and the parameters are such that $p_k,x,\ell_k \in (0,1)$, with $\sum_{k=1}^N \ell_k = 1$.

I took the partial derivatives with respect to $\alpha$ and $\theta$ and I obtained $$ \partial F/\partial \theta = -x^\alpha + \sum_{k=1}^N \ell_k^\alpha \frac{p_ke^{\theta \ell_k^\alpha}}{1+p_k(e^{\theta \ell_k^\alpha}-1)} $$ $$ \partial F/\partial \alpha = -\theta \ln(x) x^\alpha + \theta \sum_{k=1}^N \ell_k^\alpha \ln(\ell_k) \frac{p_k e^{\theta \ell_k^\alpha}}{1+p_k(e^{\theta \ell_k^\alpha}-1)} $$ but I'm stuck in trying to find a pair $(\alpha^*,\theta^*)$ that makes them null.

Can I say something about the minimum of $F$ (it if exists)?

$\endgroup$

1 Answer 1

-1
$\begingroup$

You try to minimize a function on an open set, so there is no guarantee that such a minimum actually exist.

Looking for the zero of the gradient is a necessary condition, but not necessarily sufficient since your function F is not convex. Indeed, this points can also be a local maximum or a saddle point.

Now, looking for points that satisfy this necessary condition, you can simplify your expressions considering that the first equation gives $$\partial F/\partial \theta=0 \rightarrow x^{\alpha}=\sum_{k=1}^N l_k^{\alpha} \frac{p_k e^{\theta l_k^{\alpha}}}{1+p_k(e^{\theta l_k^{\alpha}}-1)}$$ using this in the second equation leads to $$ \partial F/\partial \theta=0 \rightarrow 0=-\theta \ln(x) \sum_{k=1}^N l_k^{\alpha} \frac{p_k e^{\theta l_k^{\alpha}}}{1+p_k(e^{\theta l_k^{\alpha}}-1)}+\theta \sum_{k=1}^N l_k^{\alpha} \ln(l_k) \frac{p_k e^{\theta l_k^{\alpha}}}{1+p_k(e^{\theta l_k^{\alpha}}-1)} $$ simplifying and taking into account that $\theta>0$ $$ 0= \sum_{k=1}^N l_k^{\alpha} (\ln(l_k)-\ln(x)) \frac{p_k e^{\theta l_k^{\alpha}}}{1+p_k(e^{\theta l_k^{\alpha}}-1)} $$ and now the results highly depend on the $l_k$.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .