Multi-variate uniform distribution Suppose that $(X,Y,Z)$ is uniformly distributed on $\{ (x,y,z) : 0 \leq x \leq y \leq z \leq 1 \}$
a. I want to find out joint density function of $(X,Y,Z)$.
b. I want to find out probability density function of each pair of variables.
c. I want to find out probability density function of each variable.
d. I want to determine the dependency relationships between the variables.
Answers
a. Joint density of $(X,Y,Z)$ is $\int\limits_0^y\int\limits_x^z\int\limits_y^1 dzdydx$ $ 0 \leq x \leq 1,0 \leq y \leq 1,0 \leq z \leq 1$.
$=\frac16$ So its inverse 6 is the joint density function because total probability must be equal to one.
b. I calculated the PDf of each pair of variable f(x,y)=6(1-y),f(x,z)=6(z-x),f(y,z)=6y.
c. I also computed PDF of each variable $f(x)=3(1-x)^2$, $f(y)=6y(1-y)$and $f(z)=3z^2-3$.
d. Each pair of variables is dependent.
But I am doubtful about the way,by which I have computed these answers. would any member explain me the steps involved in the computation of aforesaid answers?
 A: a. The integral $$\int\limits_0^y\int\limits_x^z\int\limits_y^1 dzdydx$$is meaningless, because of its bounds, while the result density 3!=6 is correct. 
When integrating a function with multiple arguments, like $$f(x,y,z)=\mathbb{I}_{0\le x\le y\le z\le 1}$$ in the current setting, you start with one argument, $z$ say, keeping the other fixed, which means here that you integrate $f$ as a function of $z$ only, that is
$$f(z|x,y)=\mathbb{I}_{y\le z\le 1}$$ You then integrate the resulting function of $(x,y)$, $$\mathbb{I}_{0\le x\le y\le 1}\,\int_y^1 \text{d}z=\mathbb{I}_{0\le x\le y\le 1}\,(1-y)$$ as a function of $y$ say, keeping $x$ fixed, which leads to$$\mathbb{I}_{0\le x\le 1}\,\int_x^1(1-y)\text{d}y=\mathbb{I}_{0\le x\le 1}\,\frac{(1-x)^2}{2}$$
And this function of $x$ and only $x$ is then to be integrated in $x$:
$$\int  \mathbb{I}_{0\le x\le 1}\,\frac{(1-x)^2}{2} \text{d}x=
\int_0^1 \frac{(1-x)^2}{2} \text{d}x=\frac{1}{3\times 2}$$.
b. The densities $$ f_1(x,y)=6(1-y),\ f_2(x,z)=6(z-x),\ f_3(y,z)=6y$$ are missing proper indicator functions to represent the constraints on the support.
Namely, it should be$$ f_1(x,y)=6(1-y)\mathbb{I}_{0\le x\le y\le 1},\ f_2(x,z)=6(z-x)\mathbb{I}_{0\le x\le z\le 1},\ f_3(y,z)=6y\mathbb{I}_{0\le y\le z\le 1}$$
c. The density of $Z$ is incorrect. 
Namely, when considering $f_3(y,z)=6y\mathbb{I}_{0\le y\le z\le 1}$, you have to derive$$\int_0^z 6y\,\text{d}y=3z^2\,.$$It exhibits the right symmetry with the distribution of $X$.
d. You should compute the covariances between the pairs.
This means deriving
$$\begin{align*}
\int xy\,f_1(x,y)\,\text{d}x\text{d}y&=6\int_0^1\int_x^1 xy\,(1-y)\,\text{d}x\text{d}y\\
\int xz\,f_2(x,z)\,\text{d}x\text{d}z&=6\int_0^1\int_x^1 xz(z-x)\,\text{d}x\text{d}z\\
\int zy\,f_3(y,z)\,\text{d}z\text{d}y&=6\int_0^1\int_y^1 zy\,y\,\text{d}z\text{d}y\\
\end{align*}$$and the means of $X$, $Y$, and $Z$, with $\mathbb{E}[Z]=1-\mathbb{E}[X]$, since $$\text{cov}(X,Y)=\mathbb{E}[XY]-\mathbb{E}[X]\mathbb{E}[Y]$$
