Let $X$ and $Y$ are uniformly (independently) distributed RV's such that $X \sim U[0,a]$ and $Y \sim U[0,b]$, where $a,b >0$. Further, let $Z$ is a RV (which is independent of the above two RV's) with CDF $F(z)$ and support $(\underline{z}, \bar{z})$, where $\underline{z}, \bar{z} > 0$. How do we find



If $a=b=1$ then the probability $\text{Pr}(X+Y\leq t)$ is found as the area below the line $x+y=t$, so that $X+Y$ has probability distribution $G$ given by

$$G(t)=\begin{cases} \frac{t^2}{2}& t\in[0,1]\\ 1-\frac{(2-t)^2}{2} & t\in[1,2] \end{cases}$$

with density

$$g(t)=\begin{cases} t& t\in[0,1]\\ 2-t & t\in[1,2] \end{cases}.$$

(This is the triangular density on $[0,2]$ with mode at $1$.)

For any given realization of $z$ we have

$$Pr(X+Y\leq z)=G(z).$$

Integrating over all possible realizations:

$$\text{Pr}(X+Y\leq Z)=\int_{\underline{z}}^{\overline{z}}G(z)f(z)dz=\int_{\underline{z}}^{1}zf(z)dz+\int_{1}^{\overline{z}}(2-z)f(z)dz.$$

You could rewrite this in any number of ways. For example, since:


we have (denoting the mean of $Z$ by $\mu$)

$$\text{Pr}(X+Y\leq Z)=\mu-2\int_{1}^{\overline{z}}(z-1)f(z)dz.$$

This could be further manipulated using integration by parts.

When $a\neq 1$ and $b\neq1$ the expression for $G$ is messier, but you can find it in a similar way.


You should calculate the PDF $f_{X+Y}(t)$ for $X+Y$ e.g. by looking at the rectangle $[0,a] \times [0,b]$, calculate the area below $X+Y<t$ and take the derivative (it's a piece wise linear 'bump'). From then the easiest approach is seemingly (?) to use conditional expectation (and independence): $$ P(X+Y<Z) = E(1_{X+Y<Z}) = E( E(1_{X+Y<Z}|X+Y=t)) = E( E(1_{t<Z} |X+Y=t)) = E(1-F(t)|X+Y=t)) = \int (1-F(t)) f_{X+Y}(t) dt .$$


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