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?


2 Answers 2


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 $[x_1,x_2] \times [y_1,y_2]$.

  • $\begingroup$ 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
    Commented Dec 24, 2016 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.
    Commented Dec 24, 2016 at 18:01

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:

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


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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.