Computing skewness derivative in terms of variance In the Portilla Simoncelli paper (page 18):
http://www.cns.nyu.edu/pub/lcv/portilla99-reprint.pdf
They go about calculating the derivative of the skewness $\eta(x)$ of a distribution (2D matrix in the case of an image: $x$) through the following expression:
$\eta(x) = \frac{\mu_3(x)}{\mu_2(x)^{1.5}}$
where $\mu_2(x)$ is the variance and $\mu_1(x)$ is the mean, which they set to zero for their calculations.
The problem I am having is that when they calculate:
$\frac{d\eta_3(x)}{dx(i,j)} = \frac{3(x^2(i,j)-\mu(x)^{0.5}\eta(x) x(i,j) -\mu_2(x))}{|L|\mu_2^{3/2}}$,
where $x(i,j)$ is an individual element of the matrix $x$, is that I don't get the $-\mu_2(x)$ term.
So far I have this:
$\frac{\partial \mu_3(x)}{\partial x(i,j)} = \frac{3x^2(i,j)-6x(i,j) + 3\mu_1^2}{|L|}$
$\frac{\partial \mu_2(x)}{\partial x(i,j)} = \frac{2(x(i,j)-\mu_1(x))}{|L|}$
Then:
$\frac{d\eta_3(x)}{dx(i,j)} = \frac{\frac{\partial \mu_3(x)}{\partial x(i,j)} \mu_2(x) - \mu_3(x)\frac{\partial \mu_2(x)^{1.5}}{\partial x(i,j)}}{\mu_2^3(x)}$
Replacing with the previsouly calculated expressions I get:
$\frac{d\eta_3(x)}{dx(i,j)} = \frac{3x(i,j)^2 \mu_2^{1.5} - \mu_2^{0.5}(x)\mu_3(x)3 x(i,j)}{|L|\mu_2^3(x)}$
Then I replace with the definition of $\eta(x) = \frac{\mu_3(x)}{\mu_2(x)^{1.5}}$, and with $\mu_1(x)=0$ to get back where I started from. Where's my missing $\mu_2(x)$ Did I oversimplify somewhere?
 A: I just read the same paper. Initially I made the same mistake as you, which is how I came across your question. It's been more than a year since your post but, in case you are still curious, here is how to work out the derivation. The key is to properly compute $\frac{\partial \mu_n}{\partial x_{ij}}$ for n=2,3, which requires pulling $x_{ij}$ out from the sum:
$$
\begin{align}
\frac{\partial \mu_n}{\partial x_{ij}} &= \frac{1}{L} \frac{\partial}{\partial x_{ij}}\Big(\sum_{p,q \neq i,j}(x_{pq} - \mu_1)^n + (x_{ij}-\mu_1)^n\Big)\\
&= \frac{1}{L} \Big(\sum_{p,q \neq i,j}-n(x_{pq} - \mu_1)^{n-1} \frac{\partial}{\partial x_{ij}} \mu_1 + n(x_{ij}-\mu_1)^{n-1}(1-\frac{\partial}{\partial x_{ij}} \mu_1)\Big)\\
&= \frac{1}{L} \Big(\frac{-n}{L}\sum_{p,q\neq i,j} (x_{pq} - \mu_1)^{n-1} + n(x_{ij}-\mu_1)^{n-1}\frac{(L-1)}{L}\Big)\\
&= \frac{1}{L} \Big(\frac{-n}{L} \sum_{p,q} x_{pq}^{n-1} + n x^{n-1}_{ij}\Big) \quad \textrm{[use fact that }\mu_1=0]\\
&= \frac{n}{L}(x^{n-1}_{ij} - \mu_{n-1})
\end{align}
$$
Subbing the above into:
$$\frac{\partial \eta(x)}{\partial x_{ij}} = \mu^{-3/2}_2\frac{\partial \mu_3}{\partial x_{ij}} - \frac{3}{2}\mu_3\mu^{-5/2}_2\frac{\partial \mu_2}{\partial x_{ij}}$$
Gives:
$$\frac{\partial \eta(x)}{\partial x_{ij}} = \mu^{-3/2}_2\frac{3}{L}(x^2_{ij} - \mu_2) - \frac{3}{2}\mu_3\mu^{-5/2}_2\frac{2}{L}x_{ij}$$
Factoring out $\mu^{-3/2}_2\frac{3}{L}$ gives:
$$\frac{\partial \eta(x)}{\partial x_{ij}} = \mu^{-3/2}_2\frac{3}{L}(x^2_{ij} - \eta(x)\mu^{1/2}_2 x_{ij} - \mu_2)$$
which is the what is reported in the paper.
