Difference Uniform rv's

Question

Let $U_{1}\sim U(0,1)$ be a standard uniform random variable. Is $U_{1}-U_{1}$ uniformly distributed?

I've been trying to work this out as follows: Let $A,B$ be rv's $$P(A-B\leq x)=\int_{0}^{1}P(A\leq u,B \geq u-x)du$$ Take $A=B=U_{1}$ then $$P(U_{1}-U_{1}\leq x)=\int_{0}^{1}P(U_{1}\leq u,U_{1} \geq u-x)du$$ $$=\int_{0}^{1}P(u-x\leq U_{1}\leq u)du=\int_{0}^{1}xdu=x$$ Which would mean that indeed $U_{1}-U_{1}$ is uniformly distributed.

Am I approaching this in the right way and is my statement correct or am I overlooking something?

Answer 1

I would approach the question this way:

If $-1 \lt x \lt 0$ then $$P(A-B \lt x)= \int_{a=0}^{1+x} \int_{b=a-x}^{1} \,db \,da=\int_{a=0}^{1+x} (1-a+x) \,da =\frac{(1+x)^2}{2}$$ while by symmetry if $0 \lt x \lt 1$ then $$P(A-B \lt x)=1-\frac{(1-x)^2}{2}.$$

So the density of the difference of these two independent random variables uniformly distributed on $[0,1]$ is the deriviative with respect to $x$ of these cumulative distribution functions:

• $1+x$ when $-1 \lt x \lt 0$
• $1-x$ when $0 \lt x \lt 1$

Since the density varies with $x$, it is clearly not uniform, and is in fact triangular centred at $0$.

Answer 2

This is essentially the same as Henry's approach but with a bit more detail.

You can use conditional probabilities to solve this. First, note that $X = U_1 - U_2$ can take values in $[-1,1]$. Then, compute the distribution of $X$ conditioned on $U_2$. Finally, uncondition on $U_2$ to get the result.
\begin{align*} P(U_1 - U_2 \leq x) &= \int_{u_2=0}^1 P(U_1 \leq x + u_2|U_2 = u_2)f_{U_2}(u_2)du_2\\ &=\int_{u_2=0}^1 \max(\min(x+u_2,1),0)f_{U_2}(u_2)du_2\\ \end{align*} The integral can be computed to obtain the distribution of $X$ derived in Henry's answer.