The question is as follows:
Consider a PDF that is positive only within an interval $[a, b]$ and is symmetric around the mean $(a + b)/2$. Let $X$ and $Y$ be independent random variables that both have this PDF. Suppose that you have calculated the PDF of $X + Y$. How can you easily obtain the PDF of $X - Y$?
This is in the convolution section of the book which introduces the theorem that the PDF of $Z_1$ given $Z_1 = X + Y$ is: $$f_{Z_1}(x) = \int_{-\infty}^\infty f_X(x)f_Y(z-x)dx$$
I initially tried to do this by representing $Z_2=X-Y$ as a convolution of the two PDFs: $$f_{Z_2}(x) = \int_{-\infty}^\infty f_X(x)f_Y(x-z)dx$$
However, I was not sure where to go from here. In the book they give this as the solution:
We have $X −Y = X +Z −(a+b)$, where $Z = a+b−Y$ is distributed identically with $X$ and $Y$ . Thus, the PDF of $X + Z$ is the same as the PDF of $X + Y$ , and the PDF of $X − Y$ is obtained by shifting the PDF of $X + Y$ to the left by $a + b$.
I don't quite understand the solution. How did they get $Z = a+b−Y$ and why is $X + Z$ the same as $X + Y$?