Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It only takes a minute to sign up.
Sign up to join this community
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:
Thanks in advance
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 $\mathrm du\mathrm dv$.
