During the course of physics research on nonequilibirum statistical mechanics involving the theory of majorization, I have come across a linear transformation on a d-dimensional vector space that I suspect has been explored by mathematicians, but which is unfamiliar to me.
The transformation is the matrix whose columns are: (1,0,…), (1/2,1/2,0,…), (1/3,1/3,1/3,0,…),…, (1/d,1/d,…,1/d). In other words:
$$
\left[\begin{array}{cccc}1 & 1/2 & 1/3 & 1/4 & \ldots \\ 0 & 1/2 & 1/3 & 1/4 & \ldots \\ 0 & 0 & 1/3 & 1/4 & \ldots \\ 0 & 0 & 0 & 1/4 & \ldots \\ \ldots & \ldots & \ldots & \ldots & \ldots \end{array}\right]
$$
The inverse transformation is the matrix that has the sequence {1,2,3,…,d} along the main diagonal, and {-1,-2,-3,…,-(d-1)} along the diagonal above the main diagonal. In other words
$$
\left[ \begin{array}{ccccc} 1 & -1 & 0 & 0 & \ldots \\ 0 & 2 & -2 & 0 & \ldots \\ 0 & 0 & 3 & -3 & \ldots \\ 0 & 0 & 0 & 4 & \ldots \\ \ldots & \ldots & \ldots & \ldots & \ldots \end{array}\right]
$$ I would be keen to hear from anyone who has encountered this transformation before and can point me to any relevant literature. Thanks!
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1$\begingroup$ What sort of things do you want to know about this transformation? $\endgroup$– Alex BeckerCommented Aug 8, 2013 at 18:27
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$\begingroup$ Aside from the fact that this object has all the properties of an upper-triangular column-stochastic matrix, do you have something specific in mind? $\endgroup$– Vidit NandaCommented Aug 8, 2013 at 18:46
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2$\begingroup$ What's interesting about this transformation is that if it is fed an arbitrary probability distribution, then it outputs a probability distribution whose components are organized in descending order of magnitude. $\endgroup$– user38383Commented Aug 8, 2013 at 19:27
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$\begingroup$ In majorization theory, the important object is not the distribution x, but the permutation of that distribution, call it x^\downarrow, that puts the components in descending order. However, if z is a tensor product of two distributions, x and y, then one cannot easily infer the z^{\downarrow} from x^{\downarrow} and y^{\downarrow}. The inverse transformation described above provides a bijective map between the set of ordered distributions and the full set of distributions. I suspect that it might get around the reordering problems one has when working with the ordered distributions. $\endgroup$– user38383Commented Aug 8, 2013 at 19:28
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$\begingroup$ So one question I have is this: what is the set of distributions that is the image of the product distributions under the inverse transformation? I'm happy to work this out myself, I just wanted to make sure that there isn't already a body of work on this that I could just refer to. $\endgroup$– user38383Commented Aug 8, 2013 at 19:29
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1 Answer
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Maybe the following comment is not entirely useless.
If we consider the finite $n\times n$ version of your matrix, we see that your matrix is the upper triangular part of the following kernel matrix
\begin{equation*} M_{ij} = \frac{1}{\max(i,j)}. \end{equation*} This matrix is positive definite (a brief exercise shows this). This matrix is congruent to the "well-known" Brownian-bridge kernel matrix $[\min(i,j)]$ (The kernel function $\min(x,y)$ is called the Brownian-bridge).
However, the twist is that you are only considering the upper triangle, so I need to search a bit more to see if I can dig up something more relevant.
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$\begingroup$ is there a notion of non-symmetric Brownian motion (so that a particle can move only to the right)? $\endgroup$– user6976Commented Aug 8, 2013 at 20:26
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$\begingroup$ I don't know about "entirely useless", but this answer sounds like "your matrix would be better understood if we took out half its entries and put in different ones". $\endgroup$ Commented Aug 8, 2013 at 20:29
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$\begingroup$ @Vidit: not sure what you mean; but since the upper triangle under question is the upper triangle of an object that has been studied in stochastics, it might provide some leads to the author (who seems to also come from a probabilistic / stochastic process motivation...) $\endgroup$– SuvritCommented Aug 8, 2013 at 21:07
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$\begingroup$ @Mark: People have studied asymmetric random walks; but I am not aware if the exact transition matrix in the above question has been studied. Seems natural, but I am largely ignorant about the random walk literature, so I don't have a more precise thing to say. $\endgroup$– SuvritCommented Aug 8, 2013 at 21:11
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$\begingroup$ I think that although it is not an answer to the question, still it shows where to look for an answer. +1 $\endgroup$– user6976Commented Aug 8, 2013 at 21:45