$X$ ~ $uniform(0,1)$, $f_Y (y | X=x) = I( x<y<x+1 )$ ( for $0<x<1$ ). Find....
a) What is the distribution of $Y$, given $X = x$?
b) What is $f(x,y)$? Distribution of $(X,Y)$?
c) $f_Y (y) =$ ?, Distribution of $Y$?
My attempts:
a) Is the uniform distribution just $(x,x+1)$?
b) $f(x,y) = \left\{ \begin{array}{ll} 1 & : 0<x<1 \,\,and \,\,x<y<x+1 \\ 0 & : else \end{array} \right.$
c) $f_Y(y) = \left\{ \begin{array}{ll} y & : 0<y<1 \\ 2-y &: 1<y<2 \\ 0 & : else \end{array} \right.$
Does my work seem to be on the correct path? Thanks!
$\color{red}{\checkmark} \quad [Y\mid X=x] \sim {\cal U}(x,x+1), \forall x\in (0,1)$
$\begin{align}\color{red}{\checkmark} \quad f_{X,Y}(x, y) & = f_X(x)f_{Y\mid X}(y \mid x) \\ & = \operatorname{\bf 1}_{(0,1)}(x)\cdot \operatorname{\bf 1}_{(x, x+1)}(y) \operatorname{\bf 1}_{(0,1)}(x) \\ & = \operatorname{\bf 1}_{(0,1)}(x)\operatorname{\bf 1}_{(x, x+1)}(y) \\ & = \begin{cases}1 & : x\in (0,1)\cap y\in (x, x+1) \\ 0 & : \text{elsewise}\end{cases} \\ & = \begin{cases}1 & : y\in(0,1)\cap x\in (0,y) \\ 1 & : y\in(1,2)\cap x\in (y-1,1) \\ 0 & : \text{elsewise} \end{cases} \end{align}$
$\begin{align}\color{red}{\checkmark} \quad f_Y(y) & = \int_{\bf X} f_{X,Y}(x,y)\operatorname{d}x \\ & = \int_{\bf X} \operatorname{\bf 1}_{(0,1)}(x)\operatorname{\bf 1}_{(x, x+1)}(y) \operatorname{d}x \\ & = \int_{\bf X} \operatorname{\bf 1}_{(0,1]}(y)\operatorname{\bf 1}_{(0,y)}(x)+ \operatorname{\bf 1}_{(1, 2)}(y)\operatorname{\bf 1}_{(y-1, 1)}(x) \operatorname{d}x \\ & = \operatorname{\bf 1}_{(0,1]}(y)\int_{0}^y \operatorname{d}x+ \operatorname{\bf 1}_{(1, 2)}(y)\int_{y-1}^1 \operatorname{d}x \\ & = \operatorname{\bf 1}_{(0,1]}(y)\cdot (y) + \operatorname{\bf 1}_{(1,2)}(y)\cdot(2-y) \\ & = \begin{cases}y & : y\in (0, 1] \\ 2-y & : y\in (1,2) \\ 0 & : \text{else}\end{cases} \end{align}$