Probability of $A+C>B+D$ for $A\sim U(x,1)$, $B\sim U(y,1)$, and $C,D\sim U(0,1)$

Question

Let $A$ be uniformly distributed on $[x,1]$ and $B$ be uniformly distributed on $[y,1]$, where $0\leq x,y\leq 1$.

If $C$ and $D$ are uniformly distributed on $[0,1]$, what is the probability that $A+C > B+D$?

I know that both $A+C$ and $B+D$ will be between $x$ and $2$ and $y$ and $2$, respectively, but definitely not uniformly. I'd imagine it's some type of normalized distribution but I haven't been able to find anything on how to do something like this. I know what the distribution of adding two numbers from the same uniform distribution is, but this is different.

Answer 1

I will setup the integral you need to compute this probability and then provide a Mathematica script for evaluating it.

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We have $A\sim\mathcal U(x,1)$ and $B\sim\mathcal U(y,1)$ with $0\leq x,y\leq 1$ as well as $C,D\sim\mathcal U(0,1)$. Assuming all random variables are mutually independent we define $U=A+C$ and $V=B+D$. The goal is then to find $$ P=\mathsf P(U>V). $$

We first find the density of $U$. By the convolution of probability densities $$ \begin{aligned} f_U(u) &=\int_{-\infty}^\infty f_A(t)f_C(u-t)\,\mathrm dt\\ &=\int_{-\infty}^\infty \frac{1}{1-x} \mathbf 1_{t\in(x,1)}\mathbf 1_{u-t\in(0,1)}\,\mathrm dt\\ &=\frac{1}{1-x}\int_{-\infty}^\infty \mathbf 1_{t\in(x,1)}\mathbf 1_{t\in(u-1,u)}\,\mathrm dt\\ &=\frac{1}{1-x}\int_{-\infty}^\infty \mathbf 1_{t\in(x,1)\cap t\in(u-1,u)}\,\mathrm dt\\ &=\frac{1}{1-x}\int_{\max(x,u-1)}^{\min(1,u)}\mathrm dt. \end{aligned} $$ Thus, $$ f_U(u)=\frac{1}{1-x}(\min(1,u)-\max(x,u-1)),\quad u\in(x,2) $$ and $$ f_V(v)=\frac{1}{1-y}(\min(1,v)-\max(y,v-1)),\quad v\in(y,2). $$

To find $P$ we need to integrate $f_{U,V}(u,v)=f_U(u)f_V(v)$ over the region $u>v$ that also coincides with the support of $(U,V)$. The region is complicated because it is described differently depending on if $x\geq y$ or $x<y$. Combining each case we can write the probability $P$ as $$ P=\int_{\max(x,y)}^2\int_y^u f_U(u)f_V(v)\,\mathrm dv\mathrm du. $$ For the special case $x=y$ we can show $P=1/2$ by symmetry.

Finally, here is a Mathematica script to evaluate $P$ for specific choices of $x$ and $y$:

x = 0.4;
y = 0.3;
fU[u_] := 1/(1 - x) (Min[1, u] - Max[x, u - 1]);
fV[v_] := 1/(1 - y) (Min[1, v] - Max[y, v - 1]);
Integrate[fU[u, x] fV[v, y], {u, Max[x, y], 2}, {v, y, u}]