Assume that for $i\in\{1,2\}$ $\theta_i$ is distributed uniformly in the interval $[1,2]$, while $\epsilon_i$ is distributed uniformly in the interval $[-1/2,1/2]$ (both random variables are independent of each other and also across $i$). I am interested in computing:
\begin{align*} R^*= \int\int \max_i\left\{\theta_i+\epsilon_i-1,0\right\}dF(\theta_i)dG(\epsilon_i) \end{align*}
in the supports of each random variable.
Is there a way to compute this to obtain a closed-formed expression?
Observe that $\max (x+y-1,0)>0$ iff $x>1-y$ so if $y>0$, this condition is satisfied for $x \in ]1,2]$, and if $y<0$, you want $x \in (1-y,2]$. Your integral becomes then equal to: $$\int_0^\frac{1}{2} \int_1^2 (x+y-1) dx dy+ \int_{-\frac{1}{2}}^{0} \int_{1-y}^2 (x+y-1) dx dy.$$ This you can integrate easily and obtain (if I haven't done any mistake in the calculations) $\frac{25}{48}$.
Interpretation 1.
One interpretation of your question, given your comment that the integrand is $$\ \max\{\ \max\left\{\theta_1+\epsilon_1-1,0\right\}, \max\left\{\theta_2+\epsilon_2-1,0\right\}\ \}\ ,$$ is that you want the following expected value: $$\begin{align}R^*&= \mathrm{E}\,[\max(\ \max(\theta_1+\epsilon_1-1,0),\ \max(\theta_2+\epsilon_2-1,0)\ )]\\ &=\mathrm{E}\,[\max(\ \max(X_1,0),\ \max(X_2,0)\ )] \end{align}$$ where $X_1=\theta_1+\epsilon_1-1$ and $X_2=\theta_2+\epsilon_2-1$. Here $\theta_1,\epsilon_1,\theta_2,\epsilon_2$ are mutually independent, with $\theta_i\sim \text{Uniform}[1,2],\ \epsilon_i\sim\text{Uniform}[-\frac{1}{2},\frac{1}{2}]$.
Letting $Y_i=\max(X_i,0)$, we use the following fact (with $n=2$) to compute $R^*$:
Fact 1: If $Y_1,...,Y_n$ are iid nonnegative continuous random variables with CDF $F_Y$, then $$R^*=\mathrm{E}\,[\max(Y_1,...,Y_n)\,]= \int_0^{\infty}\left(1-F_Y(y)^n\right) \, \mathrm{d}y.$$ Now $F_Y(y)=\mathrm{P}[\max(X_i,0)\le y]=\mathrm{P}[X_i\le y]\cdot 1_{[0.\infty)}(y)$, so in the region of integration $(y\ge 0)$ we have simply $F_Y(y)=\mathrm{P}[X_i\le y]=F_{X_i}(y)$.
To obtain the CDF $F_{X_i}$, we use the following facts:
Fact 2: If $U,V$ are iid Uniform$[0,1]$, then $T=U+V$ has a symmetric triangular distribution with CDF $$F_T(t)=\frac{t^2}{2}\cdot 1_{[0,1]}(t) + \left(1-\frac{(2-t)^2}{2}\right)\cdot 1_{[1,2]}(t) + 1\cdot 1_{[2,\infty)}(t).$$
Fact 3: If $W\sim\text{Uniform}[0,1]$, then $a+(b-a)\cdot W\sim \text{Uniform}[a,b]$.
We have then the following: $$\begin{align}X_i&=\theta_i+\epsilon_i-1\ \ \sim\ \ \left(1+U_i\right)+\left(-\frac{1}{2}+V_i\right)-1=U_i+V_i-\frac{1}{2}=T_i-\frac{1}{2}\end{align}$$
where the $U_i,V_i$ are iid Uniform$[0,1]$, and by Fact 2, the $T_i$ are iid with the triangular distribution CDF $F_T.$
Therefore, when $y\ge 0$, $$F_Y(y)=\mathrm{P}[X_i\le y]=\mathrm{P}\left[T_i\le y+\frac{1}{2}\right]=F_T\left(y+\frac{1}{2}\right),$$ where $F_T$ is the piecewise function given in Fact 2.
Letting $H(y)=1-F_Y(y)^2$, we finally obtain (using SageMath) $$R^*=\mathrm{E}\,[\max(Y_1,Y_2)\,]= \int_0^{\infty}H(y) \, \mathrm{d}y=\color{blue}{\frac{1411}{1920}}= 0.734895833333333... $$
var("y")
H = piecewise( [ [(-1/2,1/2),1-( (y+1/2)^2/2 )^2], [[1/2,3/2],1-( 1-(3/2-y)^2/2 )^2] ] )
ans = integrate(H, y, 0, 3/2)
print ans, ans.n()
1411/1920 0.734895833333333
An easy Monte Carlo simulation confirms this result:
def a(n):
tot = 0
for _ in xrange(n):
x = random() + random() - 1/2
y = random() + random() - 1/2
tot += max(max(x,0), max(y,0))
return tot/n
for n in [1..8]:
print n, a(10^n)
1 0.729610490768
2 0.721972793888
3 0.729694176852
4 0.739766983435
5 0.735572807136
6 0.734985659743
7 0.734830221675
8 0.734934516699
Interpretation 2.
If there were no subscript $i$ on the $\max$ in your integrand, contrary to your statement that the integrand is $\ \max\{\ \max\left\{\theta_1+\epsilon_1-1,0\right\}, \max\left\{\theta_2+\epsilon_2-1,0\right\}\ \},\ $ then we would have the following (calling this new and different object $R^*_i$, although $R^*_1=R^*_2$ because the $X_i$ are iid): $$\begin{align}R^*_i&=\int\int \max\left\{\theta_i+\epsilon_i-1,0\right\}dF(\theta_i)dG(\epsilon_i)\\ \\ &=\mathrm{E}\,[\max(X_i,0)]\\ \\ &=\mathrm{E}\,[Y_i]\\ \\ &=\int_0^{\infty}\left(1-F_Y(y)\right) \, \mathrm{d}y\tag{by Fact 1}\\ \\ &=\int_0^{\infty}\left(1-F_T\left(y+\frac{1}{2}\right)\right)\, \mathrm{d}y\tag{by Facts 2 & 3}\\ \\ &= \color{blue}{\frac{25}{48}}=0.520833333333... \end{align}$$
Using SageMath again, where now the integrand is $H(y)=1-F_Y(y)$:
var("y")
H = piecewise( [ [(-1/2,1/2),1-( (y+1/2)^2/2 )], [[1/2,3/2],1-( 1-(3/2-y)^2/2 )] ] )
ans = integrate(H, y, 0, 3/2)
print ans, ans.n()
25/48 0.520833333333333
and again confirming by an easy Monte Carlo simulation:
def b(n):
tot = 0
for _ in xrange(n):
x = random() + random() - 1/2
if x > 0: tot += x
return tot/n
for n in [1..8]:
print n, b(10^n)
1 0.687577778529
2 0.591626977083
3 0.524521271125
4 0.523581803279
5 0.522742615831
6 0.520354088095
7 0.520978341799
8 0.520873563125
(This also confirms that @KarloKik's answer would be correct in this interpretation that's contrary to your comment.)
