The travel cost method, a method used in economics to estimate the value of parks and other recreational sites, requires (as one step) estimating the relation between $VR_i$, the visit rate to the site for the $i$'th individual (or sometimes zone of origin), and $TC_i$, the corresponding travel cost to the site. Travel cost is often roughly proportional to distance. Sometimes the visit rate is influenced by a second site, so that the true regression equation is of the form:

$$VR_i = B_0 + B_1TC_i + B_2TCS_i$$

where $TCS_i$ is the travel cost to the second site. Regression theory implies that if the variable $TCS_i$ is omitted when estimating the relation between $VR_i$ and $TC_i$, then the estimates of $B_0$ and $B_1$ are likely to be biased, and that the amount and direction of bias will depend partly on the sign and magnitude of correlation between $TC_i$ and $TCS_i$.


Abstracting from the above, suppose $A$ is the centre of a circle of radius $r$, and $B$ is a point at distance $d$ from $A$ ($d < r$). Suppose a large number of points, indexed by $i$, are uniformly distributed by area within the circle. Define random variables $X$ and $Y$ as the distances from $A$ and $B$ respectively to $i$. What is the correlation coefficient $\rho$ between $X$ and $Y$? Either an exact formula or a good approximation would be useful.

What I Have Done

Starting from the formula:

$$\rho = \frac{cov(X,Y)}{\sigma_X\sigma_Y}$$

I calculated $\mu_X$ and hence $\sigma_X$ as follows:

$$\mu_X = \int_0^r z.(2\pi z/\pi r^2) dz = \int_0^r 2z^2/r^2 dz = \bigg[ 2z^3/3r^2 \bigg]_0^r = 2r/3$$

$$\sigma^2_X = \int_0^r [z - (2r/3)]^2(2\pi z/\pi r^2) dz = \int_0^r [z^2 - 4rz/3 + 4r^2/9](2z/r^2) dz$$ $$\sigma^2_X = \frac{1}{r^2} \int_0^r [2z^3 - 8rz^2/3 + 8r^2z/9] dz = \frac{1}{r^2} \bigg[\frac{z^4}{2} - \frac{8rz^3}{9} + \frac{4r^2z^2}{9} \bigg]_0^r$$ $$\sigma^2_X = \frac{1}{r^2}\bigg[\frac{r^4}{2} - \frac{8r^4}{9} + \frac{4r^4}{9}\bigg] = \frac{r^2}{18}$$ $$\sigma_X = \frac{r}{3\sqrt{2}} \approx 0.236r$$

However, I'm not sure how to proceed to find $cov(X,Y)$ and $\sigma_Y$, which appear to need more sophisticated applications of calculus.

Intuitively, I can see that there will be a positive correlation when $d$ is much smaller than $r$, so that A and B are relatively close together, but I'm not sure whether the correlation will become negative for larger $d$.

Update 14/6/2017

Prompted by whuber's comment I estimated $\rho$ for the case with $B$ on the edge of the circle by an approximate method. A distribution of 120 points within the circle was defined as follows. The circle was imagined divided into 10 concentric rings of equal width. Numbering the rings from the centre, their areas are in the ratio $1:3:5:...:2n-1:...:19$. Along the middle of each ring, 12 points were positioned at intervals of 30 degrees relative to the line $AB$. The distance of each point from $B$ was found using the cosine rule. In calculating $\rho$, values associated with points were weighted by the areas of their respective rings (this is equivalent to having the number of points in each ring proportional to its area). This gave the following results (rounded to 3 significant figures):

$$\sigma_X = 0.235r$$

$$\sigma_Y = 0.470r$$

$$cov(X,Y) = 0.0175r^2$$

$$\rho = 0.158$$

This estimate of $\rho$ is reasonably close to whuber's 0.164.

  • 2
    $\begingroup$ The calculus is messy: you will likely need to perform numerical integration. The correlation remains positive, however. To see that, convince yourself that it is least when $B$ lies on the circumference of the circle, where the correlation is close to $0.164$. Alternatively, argue that the correlation monotonically decreases with the distance from $A$ to $B$, then remark that it must be $1$ when $A=B$ and asymptotically $0$ as $B$ moves arbitrarily far from $A$ (it needn't be in the unit circle for the calculation to make sense). $\endgroup$ – whuber Jun 12 '17 at 17:31

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