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I have a variable x. I generate variables u=f1(x), v=f2(x), w=f3(x), z=f4(x). I have the marginal probability density functions of u,v,w,z. I wish to find the joint probability density function. The variables u,v,w,z are correlated as they are all generated from x.

If it helps:

  1. u,v,w,z can be approximated as Beta distributions and this will be used as their PDFs.
  2. x is an exponential random variable

Thanks in advance

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  • $\begingroup$ It rather depends on what you know about $f_1,f_2,f_3,f_4$. Are they deterministic? Are they all increasing functions of $x$? If so, you can answer your question via the cumulative distribution functions of $u,v,w,z$. $\endgroup$
    – Henry
    Jun 26, 2012 at 6:39
  • $\begingroup$ Margin distributions can be determined from joint distributions but usually not the other way around because the correspondec is not unique. That is for example you can find more than one joint distribution that generates the same pair of marginals. Form example suppose you have a bivariate normal with mean vector (0,0) both variances equal to 1 and correlation coefficient r. Then each marginal distribution is N(0,1) but there are infinitely many such bivariate distributions (one for every choice of -1<=r<=1. $\endgroup$
    – Michael R. Chernick
    Jun 26, 2012 at 22:19
  • $\begingroup$ @ Henry: Yes the functions are deterministic and yes they are increasing functions of x. Can you please elaborate or direct me to a site where I can read up on how to find the Joint Distribution? $\endgroup$
    – user992654
    Jun 27, 2012 at 3:45
  • $\begingroup$ @MichaelChernick True but off-topic. $\endgroup$
    – Did
    Jun 27, 2012 at 7:37
  • $\begingroup$ Sorry I missed some of the specifics of the question that make the joint distribution unique namely the specific functions f1, f2, f3 and f4. So I thought I was addressing the question. $\endgroup$
    – Michael R. Chernick
    Jun 27, 2012 at 14:55

1 Answer 1

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There is no joint density function since the random variable $(U,V,W,Z)$ takes values on a subset $D=\{(f_1^{-1}(x),f_2^{-1}(x),f_3^{-1}(x),f_4^{-1}(x))\mid x\in\mathbb R\}$ of $\mathbb R^4$ which has Lebesgue measure zero.

Informally, $D$ has co-dimension $3$, hence one can compare $D$ to a line in $\mathbb R^4$. Formally, for every measurable function $\varphi$ on $\mathbb R^4$, $$ \mathrm E(\varphi(U,V,W,Z))=\int\varphi(f_1(x),f_2(x),f_3(x),f_4(x))\,g(x)\mathrm dx, $$ where $g$ is the density of the distribution of $X$ hence $\mathrm E(\varphi(U,V,W,Z))$ is an integral on (a subset of) $\mathbb R$ instead of $\mathbb R^4$.

The simplest analogue is when $U=V=X$ with $X$ uniformly distributed on $[0,1]$. Then $(U,V)$ is uniformly distributed on the diagonal $\Delta=\{(x,x)\mid x\in[0,1]\}$ hence the distribution of $(U,V)$ is $$ \mathrm dP_{(U,V)}(u,v)=\mathbf 1_{u\in[0,1]}\,\delta_u(\mathrm dv)\,\mathrm du, $$ where, for every $u$, $\delta_u$ is the Dirac distribution at $u$. One sees that $\mathrm dP_{(U,V)}(u,v)$ has no density with respect to Lebesgue measure $\mathrm du\mathrm dv$.

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