Difference of uniform random variables Note:- The question might have been asked several times before I present here. I am trying to derive it using some new approach.  
$X,Y$ are i.i.d $U(0,1)$. Find the distribution of $X-Y$.  
I have tried to do it using distribution function technique, and then breaking it into two parts. But I got stuck in the following steps...  
$G(t)=P(X-Y\le t)=P(X\le t+Y) $ 
$=P(X\le t+Y \cap t+Y\le 1)+P(X\le t+Y \cap t+Y>1)$ 
Then I am stuck what to do. Will this method give success?
 A: Let $W=-Y$, then the density of $Z=X+W$ is given by the convolution:
\begin{align}
f_Z(z) &= (f_X\star f_W)(z)\\
&= \int_{\mathbb R} f_X(x)f_W(z-x)\ \mathsf dx\\
&= \int_{\mathbb R} \mathsf 1_{(0,1)}(x)\cdot\mathsf 1_{(-1,0)}(z-x)\ \mathsf dx\\
&= \begin{cases}
\int_0^{1+z}\ \mathsf dx = 1+z,& -1<z<0\\
\int_z^1\ \mathsf dx = 1-z,& 0<z<1.
\end{cases}
\end{align}
A: A slight variant on @Math1000's strategy is to use $1-Y\sim U(0,1)$ so $X-Y+1$ has an $n=2$ Irwin–Hall distribution. Another approach is to note $X$ has MGF $\frac{e^t-1}{t}$, so $X-Y$ has MGF $$\frac{e^t-1}{t}\frac{e^{-t}-1}{-t}=\frac{e^t-1-t}{t^2}+t\leftrightarrow -t.$$But $\frac{e^t-1-t}{t^2}$ is the MGF of the distribution with pdf $2(1-x)$ on $[0,\,1]$, since $$\int_0^12(1-x)e^{tx}dx=\frac{1}{t^2}[(2t+2-2tx)e^{tx}]_0^1=\frac{2(e^t-1-t)}{t^2}.$$Therefore, $X-Y$ has pdf $1-|x|$ on $[-1,\,1]$.
A: Just to illustrate this other post, in which the point was made that the distribution of two independent random variables $X$ and $Y$ is obtained by the cross-correlation of the original distributions, i.e. $\int_{-\infty}^\infty f(\tau) g(t+\tau) \rm d\tau,$ this is how the pdf (and cdf) above can be obtained without the change of sign:
With $Z=Y-X,$ and both $Y$ and $X\sim \text{Unif}[0,1],$ i.e. $f_X(x)=f_Y(y)=\mathbf 1_{x,y\in [0,1]},$
$$f_{Z}(z)=\int_{0}^1 f_X(x)f_Y(z+x) \rm dx$$
A quick sketch suggests breaking the integration in two parts:

If $-1\leq z < 0,$
$$f_Z(z) =\int_{-z}^1 \rm dx= 1+z$$
If $0\leq z \leq 1$
$$f_Z(z) =\int_0^{1-z}\rm dx=1-z$$
The cdf is for $-1\leq z < 0,$
$$F_Z(z)=\int_{-1}^z (1+z)\rm dz= \frac 1 2 (z^2 + 2z +1)$$
and if $0\leq z \leq 1,$
$$F_Z(z) = \frac 1 2 + \int_0^z (1-z)\rm dz= -\frac 1 2 (z^2-2z - 1).$$
