Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Here is an unknown luminosity function $f(x,y)$ and its integration results: $$\begin{align*} p_{i,j} &= \frac{1}{\Delta_{i,j}}\iint\limits_{D_{i,j}} \! f(x,y) \, dx \, dy,\\ \Delta_{i,j} &= \iint\limits_{D_{i,j}} \!dx\,dy\;. \end{align*}$$ Let's consider the following transformations: $$\begin{align*} F(x,y,r,\sigma)&=\frac{1}{2 \pi \sigma^2}\int\limits_{x-r}^{x+r} \int\limits_{y-r}^{x+r} \! f(u,v) \,e^{-\frac{(u-x)^2+(v-y)^2}{2\sigma^2}} \, du \, dv\;,\\ q_{i,j}(r,\sigma)&=\frac{1}{\Delta_{i,j}}\iint\limits_{D_{i,j}} \! F(x,y,r,\sigma) \, dx \, dy\;. \end{align*}$$ Is there a functional relationship between:

  1. $p_{i,j}$ and $q_{i,j}(r,\sigma)$, where $r \in [0,+\infty)$;
  2. $p_{i,j}$ and $q_{i,j}(\infty,\sigma)$?
share|cite|improve this question
When you ask a question that's closely related to an earlier question, please link to the earlier question to avoid needless duplication of efforts. – joriki Aug 10 '12 at 12:49
up vote 1 down vote accepted

No, there can't be such a relationship, since $q_{i,j}$ depends on values of $f$ outside of $D_{i,j}$ and $p_{i,j}$ doesn't.

share|cite|improve this answer
What if $r$ is chosen suchwise $[x-r,x+r]\times[y-r,y+r] \subseteq D_{i,j}$?! – Max Aug 10 '12 at 18:43
@Max: That doesn't make any sense. $x$ and $y$ are bound variables in the integral defining $q_{i,j}$; they range over all of $D_{i,j}$, so there's no non-zero $r$ that would lead to that inclusion. The whole idea of blurring is to mix in values from the neighbourhood, and if you do that at the boundary of $D_{i,j}$, it necessarily mixes in values from outside $D_{i,j}$. – joriki Aug 10 '12 at 20:14
What if $\;\bigcup\limits_{i,j} D_{i,j} = \mathbb R \times \mathbb R,\; \bigcap\limits_{i,j} D_{i,j} = \varnothing$ and values of $\Delta_{i,j}$ is small enough that we can assume that $p_{i,j} \approx f(x_i, y_j)$ and $q_{i,j}(r, \sigma) \approx F(x_i, y_j, r, \sigma)$? – Max Aug 11 '12 at 12:21
More precisely: $f(x,y) \approx p_{i,j}$ for any $(x,y) \in D_{i,j}$ and $q_{i,j}(r, \sigma) \approx F(x_i, y_j, r, \sigma)$ where $(x_i, y_j)$ is a centroid of $D_{i,j}$ – Max Aug 11 '12 at 12:41
@Max: Well, then either $\sigma$ is small compared to the extent of the $D_{i,j}$, in which case in this approximation $q_{i,j}=p_{i,j}$, or it isn't, and then you have the same problem that $q_{i,j}$ is affected by function values outside $D_{i,j}$. I don't see how you could possibly get around that without defeating the whole purpose of blurring and rendering the problem trivial. – joriki Aug 12 '12 at 10:13

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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