I have a function as $$E=\int_\Omega -\log\big( p_i(x)\big) dx$$ where $p_i(x)$ is density distribution which estimated by Parzen window method. $$p_i(x)=\frac{1}{\Omega_i} \int_{\Omega_i}K_\sigma\big(I(x)-I(y)\big) dy$$ where $\Omega_i$ is the image domain, with $$K_\sigma=\frac{1}{\sqrt{2\pi}\sigma}\exp\left(\frac{-z^2}{2\sigma^2}\right)$$ and $\sigma$ is scalar parameter. $I$ is an image $I(x):\Omega \to \mathbb{R}$, $i$ is given.

So, which ones is correct for left side: $E(\sigma),E(x),E(x,\sigma),E(p_i)$ or other form?

Update: I upload the original equation for better visualization enter image description here

  • $\begingroup$ You should correctly define E(.) as a function of all the variables which effect its value. Hence it is $E(\sigma, z, p_i)$ Or may be you want to treat some of them as "parameters" instead of arguments and call it as $E_{\sigma, p_i}(z)$ it really depends on your taste of it ;) $\endgroup$
    – Fat32
    Apr 26, 2016 at 0:01
  • $\begingroup$ Thank you. But $\sigma$ is scalar and $z$ is just pixel location. $\endgroup$
    – John
    Apr 26, 2016 at 0:42
  • 2
    $\begingroup$ Please tighten up your notation. It seems (to me) to be all over the place. See my answer below. $\endgroup$
    – Peter K.
    Apr 26, 2016 at 12:56

1 Answer 1


Based on your new explanations, I think the answer is just $$ E_{u_i} $$ given that what you're really trying to find a shorthand for is the expression inside the summation sign in (5) of the screenshot. That is:

$$ \min_{U} \left\{ \lambda \sum_{i=1}^N \int_\Omega r_i(\chi, p_i(\chi)) u_i(\chi) d\chi \right\} = \min_{U} \left\{ \lambda \sum_{i=1}^N E_{u_i} \right\} $$

so that $$ E_{u_i} = \int_\Omega r_i(\chi, p_i(\chi)) u_i(\chi) d\chi $$

Original answer below for reference

As per @Fat32's comment, any of:

  • $E(\sigma, z, p_i)$
  • $E_{\sigma,p_i}(z)$
  • $E(z; \sigma,p_i)$

is acceptable.

But you need to tighten up your notation. What is $x$ vs $z$? How does $\Omega$ relate to them? What is $y$?

  • $\begingroup$ Thank you Peter K. for your answer. Let $I$ is an image, $\Omega$ is the image domain. The $x$ is pixel location. For each pixel $x$, we can approximate the density distribution of the pixel in each region $i$ by using Parzen window method. I called the density distribution as $p_i(x)$. The Parzen window method used $\sigma$ as scalar parameter to estimate the density. Thus, as my knowledge, I think $\sigma$ is tuning parameter, we don't need to put it as variable. $y$ is other pixel in the image, except the pixel $x$. My goal is that formulates the energy $E$ and optimize it. $\endgroup$
    – John
    Apr 26, 2016 at 13:27
  • $\begingroup$ Please refer the Eq. (5) in the paper sciencedirect.com/science/article/pii/S003132031100197X $\endgroup$
    – John
    Apr 26, 2016 at 13:42
  • $\begingroup$ Could you check my comment? Is it clear? $\endgroup$
    – John
    Apr 26, 2016 at 22:46
  • $\begingroup$ @user8264 Yes, that is clear. Can you add it to the question? $\endgroup$
    – Peter K.
    Apr 26, 2016 at 22:51
  • $\begingroup$ So, how can I represent the energy $E$ in term of which variable? I think $\sigma$ is given. $z$ is just denotes for Gaussian function $K(z)$ $\endgroup$
    – John
    Apr 27, 2016 at 11:42

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