# Finding Expectation of a probability density function Z = Y - X problem from MIT 6.041?

Problem is here 2(c)
Solution is here 2(c)
$$\quad \ Z\ =\ Y\ -\ X\quad\quad; 0 $$\quad\quad\ => Z\ =\ Y\ -\ X\quad\quad; -x $$=> Z\ =\ Y\ -\ X\quad\quad; -x $$f_{X,Z}(x,z)= f_X(x)f_{Z|X}(z|x)\quad; -x We are given that Y is uniformly distributed, if we fix X then Z will also have the same distribution as Y but shifted by a constant.
Y has a Uniform distribution of $$1/2x$$.
$$f_{Y|X}(y |x)= \frac{1}{2x}\quad; 0 So Z should also have the same distridution but shifted by a constant say 'c'
$$f_{Z|X}(z |x)= \frac{1}{2x} - c\quad; -x

But the solution says: $$f_{Z|X}(z |x)= \frac{1}{2x}\quad; -x Whats am missing?