1
$\begingroup$

The fixed-point of a vector-valued function $f: \mathbb{R}^p \rightarrow \mathbb{R}^p$ is a value $\textbf{b}$ such that $f(\textbf{b}) = \textbf{b}$ (this value is not necessarily unique). Any ideas on how to compute the multivariate fixed point for the following function \begin{align*} f(\textbf{b}) = \Gamma^{-1}\textbf{z}\left(y - \frac{e^{\textbf{z}^\intercal \textbf{b}}}{c + e^{\textbf{z}^\intercal \textbf{b}}}\right) \end{align*} where $\textbf{z} \in \mathbb{R}^p$, $y \in \{0, 1\}$, $c \in \mathbb{R}$? You can see from one of my previous posts of a similar question, which is similar in nature and express something in terms of the Lambert $W$ function.

$\endgroup$
5
  • 1
    $\begingroup$ So, uh, what is "generalized" about your definition of a fixed point? And are you interested in a closed form expression for the fixed point? Because otherwise, Newton's method for $g(b) := f(b)-b$? $\endgroup$
    – Hannes
    Apr 5, 2018 at 18:01
  • 1
    $\begingroup$ Apologies, I would like a closed-form solution for the fixed point, if possible. Solving for this fixed-point is part of several layers of other optimizations going on, so I would like this to be as fast as possible. $\endgroup$
    – Tom Chen
    Apr 5, 2018 at 18:26
  • $\begingroup$ What does $\Gamma$ mean here? $\endgroup$ Apr 5, 2018 at 18:49
  • $\begingroup$ $\Gamma \in \mathbb{R}^{p \times p}$ is positive-definite symmetric $\endgroup$
    – Tom Chen
    Apr 5, 2018 at 19:02
  • $\begingroup$ @Hannes Also, I corrected the generalized to just a fixed point. I realize what you mean, since a fixed point is just fixed point, but I meant "generalized" here in the context of multivariate. $\endgroup$
    – Tom Chen
    Apr 5, 2018 at 19:06

1 Answer 1

1
$\begingroup$

The right side is a scalar multiple of $\Gamma^{-1} {\bf z}$, and its dependence on $\bf b$ is only through ${\bf z} ^\intercal \bf b$, so this is really just a one-dimensional problem. Let ${\bf a} = \Gamma^{-1} {\bf z}$ and $k = {\bf z}^\intercal \bf a$. Then with ${\bf b} = x \bf a$, the fixed-point equation $f({\bf b}) = \bf b$ becomes $$x = y - \frac{e^{k x}}{c + e^{kx}}$$ AFAIK this does not have closed-form solutions (even using LambertW).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.