Compute $P(X+Y>Z)$ where $X, Y$ and $Z$ are independent, uniform random variables in the interval $[0,1]$.

Question

I know that if $W=X+Y$ then $$f_{W}(t) = t\mathbf{1}_{[0,1]}(t)+(2-t)\mathbf{1}_{(1,2]}(t).$$ Thus, we want to find $P(W>Z)=P(W-Z>0).$ Using Bayes Theorem we get that $$P(W-Z>0) = P(W-Z>0|W\leq 1)P(W\leq 1)+P(W-Z>0|W> 1)P(W> 1)$$ $$=\int_{0}^{1} \int_{0}^{w} dz \cdot P(W\leq 1)+P(Z\leq 1)P(W>1)$$ $$=\frac{1}{2}\cdot \frac{1}{2}+1\cdot \frac{1}{2}=\frac{3}{4}.$$

Is this calculation correct?

Edit

Based on the suggestion made in the comment we consider the following:

$$P(W>Z)=\int_{0}^{2}P(Z<w)f_{W}(w)dw$$ $$=\int_{0}^{1}w^2\cdot dw+\int_{1}^{2}(2-w)\cdot dw=\frac{1}{3}+2-\frac{3}{2}=5/6.$$ Does this make sense?

Answer 1

$x+y > z\\ 0\le z \le 1$

Either $x +y > 1$ and we can ignore $z$ or $x+y\le 1$ and $z < x+y$

$\int_0^1\int_0^{1-x}\int_0^{x+y} dz\ dy\ dx + \int_0^1\int_{1-x}^{1} dy\ dx\\ \frac 13 + \frac 12$

Answer 2

Using the Irwin–Hall distribution

$$W = X + Y,\qquad f_W(w)= w 1_{(0,1]}(w) + (2 - w)1_{(1,2)}(w)$$

then you get

$$ \begin{align} P(W > Z) &= P(W > Z\mid W > 1)P(W > 1)+ P(W > Z, W \leq 1) \\ &= 1 \cdot \int_1^2(2-w)dw + \int_0^1\int_0^wwdzdw \\ &= \frac{1}{2} + \frac{1}{3} \end{align}$$

as Doug M points out. You forgot a $w$ and $dw$ in $\int_{0}^{1} \int_{0}^{w} dz$.