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For a 2-place real function $H$, H-volume of $[x_1,x_2]\times[y_1,y_2]$ is $H(x_2,y_2)-H(x_2,y_1)-H(x_1,y_2)+H(x_1,y_1)$. What is really the intuition of the H-volume?

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The H-Volume is the volume contained by the rectangle $[x_1,x_2] \times [y_1,y_2]$ of a 3-dimensional function $H(x,y)$. To visualize this, see the Figure H-Volume of the $\Pi$ copula

which is the contour plot of the independence copula (which is simply a 3-dimensional function with some special properties that make it a copula function).

The H-Volume is the volume contained within the box labeled $R3$. However, remember that the Copula function is defined as the H-Volume of the copula function $H$ from $[0,u] \times [0,v]$. Thus, $H(x_2,y_2)$ in reference to the figure would be the volume contained by $R1+R2+R3+R4$. To get the region of interest, which is just $[x_1,x_2] \times [y_1,y_2]$, we must subtract out $R2$ and $R4$. However, by subtracting out $R2$ and $R4$, we have also subtracted out $R1$ twice. We thus add $R1$ back into the equation (recall that $R1$ is included when computing $H(x_2,y_2)$.

To think about it in 3-D terms, see the Figure H-Volume of the $\Pi$ copula 3-D Visualization. The H-Volume of this 3-D function, which happens to be the independence copula density, is the volume enclosed under the blue shaded area, where the points are given by the rectangle $[x_1,x_2] \times [y_1,y_2]$.

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  • $\begingroup$ Regarding your last statement? Does it mean that $H-$volume is $\int\limits_{y1}^{y2} \int\limits_{x1}^{x2} H(u,v) \text{d}u\text{d}v$? $\endgroup$
    – DEVA
    Dec 24 '16 at 16:30
  • $\begingroup$ While in general, the H-volume may be defined as you put it, for copulas, the integral you have provided is not correct interpretation, if we view $H$ as the distribution function and $h$ as the density. The H-Volume in reference to the copula is the integral of the copula density. Recall that $C(u,v) = \int_0^v \int_0^u c(u,v) du dv = C(u,v)-C(0,v)-C(u,0)+C(0,0)$. Similarly, for arbitrary bounds, it would be $\int_{v_1}^{v_2} \int_{u_1}^{u_2} c(u,v) du dv = C(u_2,v_2)-C(u_2,v_1)-C(u_1,v_2)+C(u_1,v_1)$. I updated the figure above to show the copula density rather than the copula function. $\endgroup$
    – Kiran K.
    Dec 24 '16 at 18:01
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The H-volume gives the the volume of some n-box (or hyper-rectangle) from some measure. Volume being generalised here for higher dimensions. When the measure is the Lebesgue measure, then the H-volume is our usual notion of length, area and volume in Euclidian space:

  • $V_{Lebesgue}([0,0.5]) = 0.5$ (length)
  • $V_{Lebesgue}([0,0.5]\times[0,0.5]) = 0.25$ (area)
  • $V_{Lebesgue}([0,0.5]\times[0,0.5]\times[0,0.5]) = 0.125$ (volume)
  • $V_{Lebesgue}([0,0.5]\times[0,0.5]\times[0,0.5]\times[0,0.5]) = 0.0625$ (volume)

It's useful in probability theory because it can be used to compute the probability mass in some n-box given a multidimensional cdf H.

The two dimensional calculation is the one you mentioned above:

$V_{H}([x_1,x_2] \times [y_1, y_2]) = H(x_2, y_2) - H(x_2, y_1) - H(x_1, y_2) + H(x_1, y_1)$

Pictorially:

enter image description here

Where the contoured lines are a bivariate cdf H.

The general formula is:

$V_{H}(J) = \sum_{\textbf{c}\; \in \; \text{vertices}(J)} \text{sign}_J(\textbf{c}) H(\textbf{c})$

where the sum is performed over all the vertices of the hyper-rectangle J, and with the sign of the sum element being:

$\text{sign}_J(\textbf{c}) = \begin{cases} \;\;1, & \text{if c has an even number of lower bounds}.\\ -1, & \text{if c has an odd number of lower bounds}. \end{cases}$

The independence or $\pi$ copula ($\pi(u,v) = uv$) is the same as the Lebesgue measure.

There is a matlab function for performing this calculation in any dimension: https://github.com/AnderGray/Hvolume-Matlab

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