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I want to measure the correlation between two 1D point processes $x$ and $y$. Ordinarily I could use the bivariate K-function

$K(t) = \frac{T}{n_xn_y} \sum_{i=1}^{n_x} \sum_{j=1}^{n_y} w(x_i,y_j) I[d(x_i,y_j)<t]$

where $n_x$ is the number of observations in $x$ and $n_y$ is the number of observations in $y$. Deviation from $K(t)=t$ is an indication of correlation between the two point processes.

However, my analysis is complicated by the fact that the unconditional distributions of the point processes are non-uniform. They each have a characteristic distribution of intervals between observations, which could be different for $x$ and $y$.

To make things more complicated, the intensities vary over the time period (for example, the intensity might be low around midnight and higher during daylight hours).

Assuming I have enough data to get good estimates of the unconditional distributions of $x$ and $y$, and of the way that intensities vary across time, is there a way to take this into account?

whuber
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Chris Taylor
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2 Answers2

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Yes. The key is that how you measure correlation does not change, but the expected value does depend on the hypothesized underlying process.

$K$ (and cross-$K$, which is what you are looking at) is defined generally for inhomogeneous point processes. For a good discussion with examples see Philip Dixon's analysis. Because all his examples are 2D, the specific formulas for expectations do not apply directly to your 1D data, but all the general ideas do apply. Note that with non-uniform distributions the definition (and computation) of $K$ do not change, but the expectation does change. With non-stationary distributions you also have to take care to distinguish the two forms of cross-K between your two series of data.

whuber
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  • Thanks. So let's say that the $K$ functions for processes $x$ and $y$ are given by $K_x(t)$ and $K_y(t)$. How do I test the hypothesis that $x$ is independent of $y$? Under the assumption of stationarity I can simply look at $L_{xy}(t) = \sqrt{K_{xy}(t)/\pi}$ and look for departures from $L_{xy}(t)-t=0$. For non-stationary distributions I see that $K_{xy}(t)$ will in general be different from $K_{yx}(t)$, but I don't see what the impact of this on my hypothesis testing should be. – Chris Taylor Feb 01 '11 at 10:07
  • You have to redo the calculations for 1D and your particular assumptions, Chris: the formulas you quote are for 2D and a homogeneous process. Dixon's references provide the theory. – whuber Feb 01 '11 at 14:37
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It's a little weird to answer my own question, but...

To address the non-stationarity issue, it seems as though it's not particularly tricky to compute the expected $K$-function under the hypothesis of independent event times. In particular, let's say that we're looking at point processes on the interval $I=[0,T]$, and our processes $X$ and $Y$ have unconditional distributions $f_X(x)$ and $f_Y(y)$. We'll also denote the unit ball in one dimension by $B(x,t) = \{y\in I:d(x,y)<t\}$. Then

$K_{XY}(t) = \mathbb{E}[\textrm{number of Y points in a neighbourhood t of a randomly chosen X point}]$

$K_{XY}(t) = \mathbb{E}_X[\mathbb{E}_Y[\textrm{number of Y points within t of x}]]$

$K_{XY}(t) = \int_I dx \int_{B(x,t)} dy \, f_X(x) f_Y(y)$

So deviations from this function in the estimated $\hat{K}_{XY}(t)$ will be an indicator of correlation between the point processes.

Does this sound reasonable?

Edit: In fact, I think this all holds if you take $I\subseteq\mathbb{R}^n$ and interpret $x$, $y$ as vectors.

Chris Taylor
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