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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](http://math.stackexchange.com/questions/180985) to avoid needless duplication of efforts.2012-08-10

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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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    @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.2012-08-12