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I'm having difficulty imagining (or explaining to myself) how the partial sums a two dimensional Gaussian noise can produce a surface. According to equation (20) of the paper, "On two-dimensional fractional Brownian motion and fractional Brownian random field" by Qian et.al. "the two-dimensional fractional Brownian random field is defined as a partial sum of the two dimensional fGn

$B_{h,k} = \sum_{i = 1}^h \sum_{j = 1}^k X_{i,j}$

In light of this relation, is it I am correct to say that

$B_{3,2} = X_{1,1} + X_{1,2} + X_{2,1} + X_{2,2} + X_{3,1} + X_{3,2}$

Would $B_{3,2}$ be the height of the field at $(3,2)$? ... And if the $X_{(i,.)}$s are (random) steps taken in the x-direction and the $X_{(.,j)}$s are steps in the y-direction, is it correct to add them?

I understand the concept of a random walk but the extension beyond one dimension baffles me.

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The notion of ordinary Brownian motion beyond one dimension is related to the notion of the Gaussian Free Field. See e.g. this survey arxiv.org/abs/math/0312099 by Scott Sheffield. –  j.c. Nov 4 '10 at 14:18

2 Answers 2

The interpretation concerning the double sum seems right. That is, $B_{h,k}$ would be the height of the field at $(h,k)$. As for the second question, moving along the $x$-direction gives the process $B_h = \sum\nolimits_{i = 1}^h {\sum\nolimits_{j = 1}^k {X_{i,j} } }$, $k$ fixed, and moving along the $y$-direction gives the process $B_k = \sum\nolimits_{j = 1}^k {\sum\nolimits_{i = 1}^h {X_{i,j} } }$, $h$ fixed. Then, you may be interested to check whether the processes $B_h = B_{h;k}$ and $B_k = B_{k;h}$ are ordinary fBm's (with variance parameter $\sigma^2$ depending on $k$ and $h$, respectively). This is what one would expect from a good definition of fractional Brownian random field (but needs verifying). Hope this will help you a little.

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I don't understand how $\sum_{i = 1}^h \sum{j = 1}^k X_{i,j}$ amounts to fixing $k$. Also, if $B_{i,j}$ is the height of the field at the point $(i,j)$ given as partial sum above, it seems superfluous to sum over two terms instead of one. I'm still struggling to see why the double summation is necessary. –  Olumide Nov 4 '10 at 17:32

Since, for some reason, Add Comment did not work, I post my reply to Olumide's comment as an Answer.

As for the first point, you first fix $k$; this results in $\sum\nolimits_{i = 1}^h {\sum\nolimits_{j = 1}^k {X_{i,j} } }$ being a one-parameter process ($B_h = B_{h;k}$). Even more specifically, if $k$ is fixed, then we may put $\tilde X_i = \sum\nolimits_{j = 1}^k {X_{i,j} } $, leading to $\sum\nolimits_{i = 1}^h {\sum\nolimits_{j = 1}^k {X_{i,j} } } = \sum\nolimits_{i = 1}^h {\tilde X_i }$, which we can denote by $B_h\,(=B_{h;k})$. As for the second point, the double summation is necessary in order to define the two-parameter random field (discrete random surface, defined on $\mathbb{Z}_+ \times \mathbb{Z}_+$)---how would you do it otherwise? (For example, consider the covariance Cov$(B_{h_1,k_1},B_{h_2,k_2})$.)

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