2

Given $k$ independent, uniform random variables $X_i \sim U(-a,a)$, $i=0,\dots k$, and two sets of coefficients $\{\alpha_i\}$ and $\{ \beta_i\}$, let $U = \sum_{i=1}^k \alpha_i X_i$ and $V = \sum_{i=1}^k \beta_i X_i$.

What is the analytic expression for joint distribution of $(U, V)$?

The closest to a starting point I could get was potentially using the Irwin-Hall distribution to derive a pdf for one of the sums, but even this I am not sure about, and I have no ideas on how to calculate the joint pdf described above.

EDITED to include amended bounds for RVs and requirement for analytic expression.

whuber
  • 281,159
  • 54
  • 637
  • 1,101
h2learn
  • 53
  • 6
  • 3
    If you run a change of variables from $(X_1,\ldots,X_n)$ to $(U,V,X_3,\ldots,X_n)$ and integrate $(X_3,\ldots,X_n)$ out, what are the difficulties? – Xi'an Jul 08 '19 at 13:11
  • @Xi'an Thanks for the comment. I may be mistaken, but I thought that when changing variables e.g. $U = g(X_1, \dots, X_k)$ that $g$ is required to be one-to-one - which is not the case here. Would you mind elaborating a bit more on how to do this please? – h2learn Jul 09 '19 at 12:32
  • 2
    You are free to reorder the indexes so that the transformation proposed by @Xi'an is one-to-one. Could you exhibit the coefficients $\alpha_i$ and $\beta_j$ you are using that do not make the transformation one-to-one for *any* possible ordering of the indexes? – whuber Jul 09 '19 at 13:27
  • 1
    @Xi'an @ whuber Apologies - my mistake there. I had misunderstood the requirement to apply to each $g_i$ in $(g_1, g_2, \dots, g_k)$. Yes, the function $g(X_1, \dots, X_k) = (U,V,X_3, \dots, X_k)$ is one to one, and the change of variables idea is a good one. Thank you. – h2learn Jul 09 '19 at 17:11
  • As a follow up, I am looking for an analytic closed form expression though, and it is looking to me like the resulting bounds of integration after the change of variables, needed to integrate out the other X's, will be complicated and thus the integration difficult to perform analytically for general $\alpha_i$, $\beta_i$, $k$ and $a$. Is there something simple I am missing there? – h2learn Jul 09 '19 at 17:12
  • 2
    I don't think you're missing anything simple. See my post at https://stats.stackexchange.com/a/43075/919 for some insight. BTW, changing the bounds of your variables doesn't change the problem at all, because those changes can be incorporated within the $\alpha_i,$ $\beta_i,$ and a constant. Thus, you can just as well assume each $X_i$ is supported on $[-1,1]$ or even $[0,1].$ – whuber Jul 09 '19 at 17:15
  • @whuber Thanks for the link to that answer - which is excellent by the way. I think I understand that if my constants $\alpha_i$ and $\beta_i$ are arbitrary, then I can incorporate changed integration bounds in those constants. In my case, though, $a$, $\alpha_i$ and $\beta_i$ are fixed constants, and it would not be helpful if I end up with a pdf $f_{U,V}(u,v)$ whose support is defined in terms of new constants which differ in an unknown way from the original $a$, $\alpha_i$ and $\beta_i$. Am I wrong here in thinking that in this case changing the integration bound explicitly is necessary? – h2learn Jul 10 '19 at 15:40
  • 1
    I think it's simpler than that. Unless your coefficients have special values (related to sums of subsets of the distribution support endpoints), you might as well solve the general problem; and the solution of the general problem is simplified by either standardizing the coefficients or standardizing the supports. Using $3k$ variables to express a problem that can be described in full generality with just $2k-2$ variables is not terribly useful. – whuber Jul 10 '19 at 15:48
  • Let us [continue this discussion in chat](https://chat.stackexchange.com/rooms/95987/discussion-between-price-and-whuber). – h2learn Jul 10 '19 at 16:03

0 Answers0