Conditional probability of two continuous random variables

Question

"EDIT: Changed my $k'=k$ to $k'=1$, hope the question does not change alot, due to comments"

I have:
$X$~$U[0,1]$, as $k$ is just a number, could be $1,2,3$ or even a million.
$Y$~$U[0,X]$
$Z$~$U[0,Y]$
I have to find $f_{\left(Z|Y,X\right)},f_{\left(Y|X\right)},f_{\left(Z|X\right)}$
Problem it, I really do not know how to start finding them.
To find $f_{\left(Y|X\right)}$, I thought of maybe finding $f_{\left(X,Y\right)}$ will make me be able to find it, but I still have to find $f_Y$ afterwards and its also problematic ( And I do not really know how to find $f_{\left(X,Y\right)}$ also, sadly.
since $Y$ and $Z$ are dependent on random variable, I do not know how to look at it.
Any help will be welcomed, how to see a uniform random variable that depend on another random variable.

Answer 1

since $Y$ and $Z$ are dependent on random variable, I do not know how to look at it.

Exactly. $~Y\sim\mathcal U[1, X]$ , which is more properly written as $(Y\mid X)\sim\mathcal U[0,X]$, means that $Y$ is uniformly distributed dependent on the value given for $X$.  Thus it is telling you exactly that the conditional probability density function is:

$$f_{\small (Y\mid X)}(y\mid x) = (1/x)\,\mathbf 1_{y\in[0,x]}$$

Likewise you have $f_{\small(Z\mid Y)}(z\mid y)$ so you may use this to find $f_{\small(Z\mid X)}(z\mid x)$ .

Answer 2

The simplest solution would be $f(x,y,z)=1/xy\quad 0\leq z\leq y\leq x\leq 1$, in that case: \begin{gather*} f(x,y)=\int_0^y\frac{1}{xy}dz=\frac{1}{x},\quad f(x,z)=\int_z^x\frac{1}{xy}dy=\frac{1}{x}ln\frac{x}{z},\quad f(y,z)=\int_y^1\frac{1}{xy}dx=\frac{1}{y}ln\frac{1}{y}\\ f(x)=\int_0^x\frac{1}{x}dy=1,\quad f(y)=\int_y^1\frac{1}{x}dx=ln\frac{1}{y},\quad f(z)=\int_z^1\frac{1}{y}ln\frac{1}{y}dy=\frac{1}{2}ln^2\frac{1}{z} \end{gather*} From where: \begin{gather*} f_{Y/X}(x,y)=\frac{1}{x},\quad f_{Z/X}(x,z)=\frac{1}{x}ln\frac{x}{z},\quad f_{Z/Y}(y,z)=\frac{1}{y},\quad f_{Z/X,Y}(x,y,z)=\frac{1}{y} \end{gather*} But there could be other solutions where $f_{Z/X}$ and $f_{Z/X,Y}$ were different.