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Suppose $x = (x_1, x_2, \dots, x_n)^t \in \mathbb{R}^n$. The Probability distribution function of $x$ is $f(x)$. My goal is to minimize the following function, \begin{equation} \underset{a \in \mathbb{R}^n}{\arg\min} \;E\frac{\|x-a\|^2}{\|x-a_p\|}, \end{equation} where $E$ denotes the expected value and $a_p$ is any given value of $a$. My approach is,

\begin{equation} \nabla_{a} E\frac{\|x-a\|^2}{\|x-a_p\|}=0 \end{equation} \begin{equation} -E \frac{x-a}{\|x-a_p\|}=0 \end{equation} I want to derive an iterative approach to calculate an optimum value of a. But after the last expression, I am stuck.

Robin
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1 Answers1

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You don't need an iterative approach. \begin{align} E\left(\frac{\|x-a\|^2}{\|x-a_p\|}\right)&= E\left(\frac{\|x-a_p+a_p-a\|^2}{\|x-a_p\|}\right)\\ &=E\left(\|x-a_p\|\right)+2\left\langle a_p-a, E\left(\frac{x-a_p}{\|x-a_p\|}\right)\right\rangle\\ &\hspace{2em}+\|a-a_p\|^2E\big(\|x-a_p\|^{-1}\big)\ . \end{align} This is a positive definite quadratic function of $\ a\ $ whose minimum is attained at $$ a=a_p+\frac{E\left(\frac{x-a_p}{\|x-a_p\|}\right)}{E\big(\|x-a_p\|^{-1}\big)}\ . $$

lonza leggiera
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