I am a student of remote sensing and recently came across a problem that is essentially a mathematical statistic and I don't know how to prove it.

Here is the background. Finally, I will put this specialized problem into mathematical words, and I hope you can understand it.

Satellites take images periodically and split the images into many pixels, $x_i^j$ represents the index $A$ value of the ith image of the $j$th pixel. For example, $x_2^3=-5$ means that the index $A$ of the 2nd image of the 3rd pixel is $-5$. The index $A$ here can be understood as a physical parameter, just like temperature.

My task is to distinguish rice in $x^j$ based on the time series values of $x_i$, one growth cycle of rice is about half a year, so $i ≤ 20$. enter image description here

Physically, rice has two major characteristics compared with other ground objects (all other ground objects are classified as non-rice). As shown in Fig 1, it is not difficult to find that:

  1. $x_ i$ of rice changes a lot, and $x_ i$ of non-rice hardly changes, which means $x_ i$ of rice have greater variance and range.

  2. The minimum value of $x_ i$ of rice must be in the first three images, e.g. 13—May in Figure 1, while $x_ i$ of non-rice basically does not change with time

The first three images here correspond to the actual situation of the transplanting stage. Although rice planting in the same area is certainly not on the same day, the time difference is not great.

In actual operation, there are many pixels, $j$ is one hundred thousand or even one million.

Imagine that a curve of each type in the following figure is translated left and right, up and down respectively (the translation range cannot be too large). That is, all the scatter plots of $x^j$, rice and non-rice still fit the above two characteristics.

So someone proposed a method to define a new parameter. For each pixel, $x^j$ defines $z^j$ as


Where, $m>n$, $n$ takes $1$, $2$, $3$, and $m$ takes all $i$

The result shows that $z^j$ presents a Gaussian mixture distribution, and the intersection of two Gaussian distributions (corresponding to rice and non-rice respectively) is the threshold value, which can be used for classification

See Figure 2 for details

enter image description here

I have carried out a lot of experiments and found that each experimental result conforms to the Gaussian mixture distribution.

To turn it into a mathematical problem:

$a_ i^j$ represents the $i$th number of the $j$th number sequence, where $i≤20$. The sequence $a^j$ includes $T$ and $S$, which meet the following 2 characteristics:

  1. $a^j$ of $T$ varies greatly, while $a^j$ of $S$ hardly changes

  2. The min of $a^j$ of $T$ must be in the first three numbers, while the min of $a^j$ of $S$ random distribution

To distinguish between $T$ and $S$,let $$b^j=\max(\frac{a_m-a_n}{a_m+a_n})$$ Where, $m > n $, $n$ takes $1$, $2$, $3$, and $m$ takes all $i$

Then $b^j$ presents a Gaussian mixture distribution, and the intersection of the two Gaussian distributions (corresponding to $S$ and $T$ respectively) is the threshold value, which can be used for classification.

PS:In the experiment, the value of $b^j$ is concentrated on the interval $(-1,1)$, and a few points (points with large $a_i$ offset due to noise error and other reasons) are normalized, that is, take $1$ if more than $1$, and take $-1$ if less than $-1$.

In Figure 1, for rice, $b^j$ is in the interval $(-1,1)$, but the probability of approaching $0$ is obviously low, and the expected value of $b^j$ is greater than $0$, so the actual value range corresponds to the red curve in Figure 2. For non-rice, because $a_n$ may be not the minimum, $b^j$ is expected to be slightly greater than $0$, so it corresponds to the blue curve in Figure 2.

At least the expected value fits the distribution Figure 2, but I don't understand how to prove the distribution (it may not be a normal distribution, but the approximate distribution curve should fit the curve in Figure 2.)

  • $\begingroup$ Your "max"-formula for $z^j$ and $b^j$ doesn't have $j$ on the right side (although of course we can guess where it should be). $\endgroup$ Jan 10, 2023 at 11:11

1 Answer 1


The given information does not allow to show that you get a Gaussian mixture, or even a Gaussian separately for rice and non-rice. Rather the opposite. Values that are constrained to be in the interval $[-1,1]$ can not be distributed Gaussian (and neither Gaussian mixture), as this has nonzero probability over the whole real line. At best, distributions can be approximated well by a Gaussian/Gaussian mixture. Such a thing can be in some situations proved based on the Central Limit Theorem, but this would require far stronger assumptions than you have made until now, and I suspect they could hardly be defended here.

In particular, you haven't said anything about dependence structure over time and between the pixels. This would need to be taken into account in any conceivable argument, which very well may be complex, and I'm pretty sure dependence cannot be neglected (assumed independent) in this situation.

If you are willing to assume two Gaussian populations anyway (which from your graph doesn't look too bad as an approximation, even though I don't see any possibility to prove it mathematically), for classification in fact if variances are different, there are two intersections of the densities. The classification problem is a special case of Quadratic Discriminant Analysis. You may classify as rice what is on the right or extremely on the left side. The thresholds may be chosen as the intersections of the densities, but you may chose the likelihood ratio threshold higher or lower, depending on what loss is associated with what kind of misclassification.

Note also that for this it is not only important whether the data taken together look like a Gaussian mixture, but rather whether distributions look like single Gaussians within each class, and in particular if rice occurs not only on the right but in fact also on the left side of non-rice.

In any case it may be possible to do better using a more than one-dimensional representation that somehow takes dependence into account (e.g., principal components).

Edit: Thinking a bit more, I'm not actually sure whether your problem is a supervised or an unsupervised classification problem. My answer treats it as supervised, which would assume that you have training samples where you know what is rice and what isn't. This is not clear to me from your question. If you don't, assuming that you have a two-component mixture and that rice corresponds to the larger variance component seems bold, but anyway, then what to do regarding classification is similar, and discussed here.


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