Convolution of weighted uniformly distributed random variables

Question

I have problems with deriving the CDF of a weighted difference of iid uniformly distributed random variables.

Assume that $X_1$ and $X_2$ $\stackrel{iid}{\sim} U[0,1]$. Define Z = $a\cdot X_1 - X_2$, where $a > 0$. How can I derive $P(Z<z)$? I was struggling with applying the general definition of a convolution on my particular case...

Thanks in advance!

Answer 1

Hint:

For fixed $z$:$$\mathsf P(Z<z)=\int_0^1\int_0^1[ax-y<z]dydx$$ where $[ax-y<z]$ is the function $\mathbb R^2\to\mathbb R$ taking value $1$ if $ax-y<z$ and taking value $0$ otherwise.

Answer 2

On the one hand we have $$P(A) = E[1_A]$$ on the other hand we have $$E[g(X_1,X_2)] = \int g(x_1,x_2) f_{(X_1,X_2)}(x_1,x_2) dz$$ for each measurable function $g$ where $f_{(X_1,X_2)}$ denotes the joint density of $X_1$ and $X_2$

Now take $$A = \{Z < z \}$$ and consider that for independent random variables $X_1,X_2$ it holds $$f_{(X_1,X_2)}(x_1,x_2) = f_{X_1}(x_1)f_{X_2}(x_2)$$ to get

$$P(Z < z) = \int_{\Bbb R^2} 1_{\{aX_1 - X_2 < z\}}\, dx_1dx_2$$

Solving the integral will give you the wanted result.