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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 $q_{i,j}(\infty,\sigma)$?
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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
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1 Answer

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.

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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
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