I have a question about variable substitution for a multivariate integral. Let me first pose the question and then provide some context, since I think the context is not necessarily very important.

Say we have (for example) the following integral $$\int_0^1 \int_0^{1-\varepsilon_1} \int_0^{1-\varepsilon_1 -\varepsilon_2} 24 (1- \varepsilon_1 - \varepsilon_2 - \varepsilon_3) d\varepsilon_3 d\varepsilon_2 d\varepsilon_1= 1$$

Furthermore, let $z = \varepsilon_1 + \varepsilon_2 + \varepsilon_3$. I'm trying to find $f(z)$ satisfying $$\int_0^1 f(z) dz = 1$$ corresponding to the integral above. At first glance, this seems like an easy problem, but whatever I do, I cannot seem to eliminate all the epsilon!

I suppose there are three cases:

$$\text{substituting } \varepsilon_1 = z - \varepsilon_2 - \varepsilon_3: \qquad \int_{\varepsilon_2 + \varepsilon_3}^{1+\varepsilon_2 + \varepsilon_3} \int_0^{1-\varepsilon_1} \int_{0}^{1- \varepsilon_1 - \varepsilon_2} 24 (1- z) d\varepsilon_3 d\varepsilon_2 dz$$

$$\text{substituting } \varepsilon_2 = z - \varepsilon_1 - \varepsilon_3: \qquad \int_0^1 \int_{\varepsilon_1 + \varepsilon_3}^{1+\varepsilon_3} \int_{0}^{1- \varepsilon_1 - \varepsilon_2} 24 (1- z) d\varepsilon_3 dz d\varepsilon_1$$

$$\text{substituting } \varepsilon_3 = z - \varepsilon_1 - \varepsilon_2: \qquad \int_0^1 \int_0^{1-\varepsilon_1} \int_{\varepsilon_1 + \varepsilon_2}^{1} 24 (1- z) dz d\varepsilon_2 d\varepsilon_1$$

The first two integrals will actually not evaluate to a numerical value. Only the last one will, but $z$ is eliminated in the first step (which is not what I want of course). So my question is: can I partially solve this integral in a way that leaves me with a function $f(z)$?

I am trying to compute the convolution of a linear combination of three dependent variables $\varepsilon_1$, $\varepsilon_2$ and $\varepsilon_3$. So I am trying to determine the distribution of $\alpha_1 \varepsilon_1 + \alpha_2 \varepsilon_2 + \alpha_3 \varepsilon_3 + \cdots + \alpha_n \varepsilon_n$. I was able to determine the joint distribution $f(\alpha_1 \varepsilon_1, \ldots, \alpha_n \varepsilon_n)$, so I really only have the integration problem left to solve.

  • $\begingroup$ Really confusing. What means "corresponding to the integral above"? $\endgroup$ – Martín-Blas Pérez Pinilla May 12 '18 at 11:11

The expression $$ \int_{\varepsilon_2+\varepsilon_3}^{1+\varepsilon_2+\varepsilon_3} \int_0^{1-\varepsilon_1}\int_0^{1-\varepsilon_1-\varepsilon_2} 24(1 - z)\,d\varepsilon_3\,d\varepsilon_2\,dz $$ is utterly nonsensical. The external integration limits must be constant.

If you do the change of variable $$\epsilon_1 = z - \epsilon_2 - \epsilon_3,$$ $$\epsilon_2 = \epsilon_2,$$ $$\epsilon_3 = \epsilon_3,$$ in the original integral, as the Jacobian is 1, after the change the new integral in the order $d\varepsilon_3\,d\varepsilon_2\,dz$ is $$ \int_0^1\int_0^z\int_0^{z-\varepsilon_2}24(1 - z)\,d\varepsilon_3\, d\varepsilon_2\,dz = \int_0^1 12(z^2 - z^3)\,dz = 1. $$

  • $\begingroup$ Thanks for your answer! I understand that the expression with non-constant external limits is nonsense, but I hoped it made it easier to spot my (faulty) rationale. The solution you provide makes sense, but I still don't understand how you obtained the integration bounds. For example, if I use your change of variable, when I compute the new bounds for $\varepsilon_1$ then I get: (lower bound) $z = 0 + \varepsilon_2 + \varepsilon_3 = \varepsilon_2 + \varepsilon_3$ and (upper bound) $z = 0 + \varepsilon_2 + \varepsilon_3 = \varepsilon_2 + \varepsilon_3$. $\endgroup$ – Lerak May 14 '18 at 9:36
  • $\begingroup$ I understand that this cannot be right, since the external integration limits are non-constant, but then how did you get your integration limits using your change of variables? $\endgroup$ – Lerak May 14 '18 at 9:38
  • $\begingroup$ @Karel, $0\le z = (\cdots)\le 1$ is obvious. Fixed $z$, use that $0\le\epsilon_2\le z$. Finally, $\epsilon_ 2 + \epsilon_3\le z$. $\endgroup$ – Martín-Blas Pérez Pinilla May 14 '18 at 10:02
  • $\begingroup$ @Karel, $\epsilon_1$ isn't among the new variables. $\endgroup$ – Martín-Blas Pérez Pinilla May 14 '18 at 10:04

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