This is a book example, but I'm trying to understand it more fully.
So let X $\sim U(0,1)$ and $Z \sim U(0,\frac{1}{10})$. let X,Z be independent uniform distributions. Consider Y=X+Z where the joint distribution $f_{XY}$ can be derived from the joint distribution of (X,Z). So I think this is the joint (X,Z) given $f_{XZ}$ = xz, =(1)(10)=10 for X is uniform, and Z is uniform on $[0,\frac{1}{10}]$
My attempt to work this out is given: Let X=U, Let Y=U+Z so For $f_{XY}(x,y) = f_{XZ}(u,y-u)=u(y-u) = 10$ because y-u is in Z (?). so $f_{XY}=10$ ?
The book explains that by examining the marginal $Y|X=x$ you have Y=x+Z which is uniform on (x,x+$\frac{1}{10})$ and results the $f_{XY}=10$.
my question is how can I show $f_{X+Y}$ using convolution?
How I can solve this using convolution is as follows: Let Y=X+Z
$f_{X+Y}=\int_{0}^{1} f_{X}(y-t)f_{Z}(t)dt$
$f_{X+Y}=10 \int_{0}^{1} f_{X}(y-t)dt$
where the latter is uniform under [0,1] and is equal to 1
so $f_{X+Y}=10$ not sure if this is correct but reasonable.
any help showing that f$_{XY}=10$ using a convolution approach would be greatly appreciated.