# Edge correction of Ripley's K-function for two 1D point processes?

I am just beginng an investigation involving characterizing the dependence between two 1D stochastic point processes $x$, $y$. The natural approach seems to involve Ripley's 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 $y$ across the interval $T$. Deviation from $K(t)=t$ is an indication of correlation between the two point processes.

However, it's not clear to me how to estimate the edge correction $w(x_i,y_j)$ the 1D case. Some papers refer to Hani Doss' 1989 papers, but in the JSTOR paper he explicity states that ' this edge correction will not concern us.' FWIW-I am currently using a correction weight of 2, but intuitively that seems excessive in my case.

• I am confused by the apparent contradiction between "1D" (one-dimensional?) and "bivariate" (two-dimensional, of course). Would you mind explaining what you mean by these? – whuber Aug 22 '12 at 17:05
• @whuber: there seems to be various ways to refer this situation (bivariate, cross, multivariate, etc.) and, with apologies, I tried to hit them all. Hopefully the edit will clear up the confusion. I believe you have referred to it as 'cross-K' in a slightly related posting. – Aengus Aug 22 '12 at 19:35
• You may be interested in some more recent work that looks at Ripley's K for point processes on 1-D road networks. See Ang et al. 2011 for one example. They have implemented their corrections in the spatstat package for R. – Andy W Aug 23 '12 at 12:34
• @Andy: Thanks much for the reference. I had not seen this; there's some good ideas in here and, while not directly applicable to the immediate problem, will apply to some related work. – Aengus Aug 23 '12 at 13:39
• The best edge correction for 1D Ripley's functions seems to be to weight by dividing the counted points by the proportion of the interval that was actually sampled. For a 2t interval around i that gets out of the transect start (zero) it will be w=(i+t)/2t; on the other side, w=(t-i+[top limit])/2t. So this "inflates" the actual counted points within 2t to compensate for the "uncounted" (though expected under stationarity) that were out of bounds [Sorry, I can't find the proper reference now, but will add it if it eventually finds me again] – FairMiles Sep 2 '15 at 16:06

In his vignette, synchronicity is $$K(t) > 2t$$ and asynchrony between $$2t$$ and $$0$$. I believe the $$2t$$ arises from the evaluation of the pair in both ways which is necessary as Gavin explains that sometimes the edge correction $$w(t_i,t_j)\neq w(t_j,t_i)$$.