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I am trying to find the expected connectivity of two line segments, $L_1$,$L_2$, based on 'functional' distance, $E(f(d))$. For context, consider a fish travelling between two segments on a river where $f(d)$ represents the probability of the fish reaching $L_2$ from $L_1$ based on an inverse relationship with distance.

Assume that $L_1$ and $L_2$ do not overlap and are segments along the line, $L$. The distances between $L_1$ and $L_2$ are assumed to be from pairs of points, $(P_1,P_2)$, randomly drawn and uniformly distributed on each segment. The 'functional' distance, $f(d)$ ($[0,1]$) may be one of several non-linear distance functions that approach $f(d)=0$ as $d$ increases.

Without a distance function, I believe $E(d)$ is simply the distance from the midpoints of $L_1$ and $L_2$; it should be the centroid of a rectangle, $(P_1,P_2)$, formed from the set of all possible point-pairs. I am concluding that from this answer, which seems applicable because the random points on $L_1$ and $L_2$ are uniformly distributed and independent, therefore $(P1,P2)$ is uniformly distributed on a rectangle.

However, how do I evaluate $E(f(d))$? The probability distribution of all $d$'s intuitively seems symmetrical and centred on $E(d)$ but in contrast it seems that $(f(d))$ is probably asymmetrical. intuitively it seems it will lead to a problem where integrating (f(d)) is required.

I am assuming this is more tractable than the related question asked and answered here. If I understand it correctly, my situation is simplified because $L_1$ and $L_2$ are assumed to be aligned along $L$ and do not overlap. This is well outside my area of expertise, so any help is much appreciated.

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It's a bit misleading to call it a distance, isn't it, if $f(0) \not= 0$? Just say it's a function.

Anyway, if I understand you correctly, suppose $L_1 = [a, b]$ and $L_2 = [c, d]$ with $L1 \cap L2 = \varnothing$. Suppose $X_1$ and $X_2$ are points drawn uniformly (and independently) from $L_1$ and $L_2$ respectively. Then

$$ E(f(|X_2-X_1|)) = \frac{1}{(b-a)(d-c)} \int_{x_1=a}^b \int_{x_2=c}^d f(|x_2-x_1|) \,dx_2 \,dx_1 $$

with the usual assumptions of integrability. Without knowing more about what $f$ does, it's hard to say much more than that, I suspect.

ETA: Things don't change much if we remove the assumption that the function is invariant under translation, or that it's symmetric with respect to its two implicit arguments. Just replace $f(|X_2-X_1|)$ and $f(|x_2-x_1|)$ with $f(X_1, X_2)$ and $f(x_1, x_2)$, respectively.

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  • $\begingroup$ Thanks - I follow! It'd be better to call it an inverse distance function, right? But just calling it a function simplifies it. $\endgroup$
    – r4gt4g
    Commented Feb 12, 2022 at 20:59

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