# Probability involving sum of uniform distributions and bounds

Let $X_1 = \min(\max(\eta_1 +0.5 + a,0),1)$

Let $X_2 = \min(\max(\eta_2+0.5+b,0),1)$

where ${a,b} \in \mathbb{R}$ and $\eta_i\sim U[-d,d]$ for $i=1,2$ (where U represents the Uniform distribution) and $X_i$ must be between 0 and 1.

I need to find: $P(X_1+X_2\geq 1)$

My idea is to break it down into pieces. I.e. start with $P(X_1=1)\cup P(X_2=1)$ and then add to that $P(X_1+X_2\geq 1) | X_i\in[0,1]$

Finding the first term should be straight forward (I think, since they are independent distributions).

As for the second term, using an answer to a previous question of mine, I get it looks something like:

$$\int_{-0.5-b}^{0.5-b}\left(\int_{-a-b-\eta_2}^{0.5-a}\frac{1}{4d^2} \, d\eta_1\right) \, d\eta_2$$

Before I attempt to solve this, can someone let me know if I'm on the right track

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You have a typo in the definition of $X_2$: there's something missing after the comma. – Robert Israel May 6 '12 at 18:15

For every random variables $\xi_1$ and $\xi_2$ and for $\zeta=\min(\max(\xi_1,0),1)+\min(\max(\xi_2,0),1)$ $$[\zeta\geqslant1]=[\xi_1\geqslant1\ \text{or}\ \xi_2\geqslant1\ \text{or}\ \xi_1+\xi_2\geqslant1],$$ hence $$\mathrm P(\zeta\geqslant1)=1-\mathrm P(\xi_1\leqslant1,\xi_2\leqslant1,\xi_1+\xi_2\leqslant1).$$ Using this for $\xi_1=d\cdot U+a+\frac12$, $\xi_2=dV+b+\frac12$ and $(U,V)$ i.i.d. uniform on $[-1,1]$, one gets $$\mathrm P(\zeta\geqslant1)=1-\mathrm P(U\leqslant u,V\leqslant v,U+V\leqslant w),$$ with $$u=(\tfrac12-a)/d,\quad v=(\tfrac12-b)/d,\quad w=-(a+b)/d.$$ From this point, the computations depend very much on the relative values of $a$, $b$ and $d$. For example, if $a=b=\frac12$ and $d\geqslant1$, $$\mathrm P(\zeta\geqslant1)=(3/4)+1/(8d^2).$$