$P(X+Y>Z)$ given $X, Y, Z$ are i.i.d random variables

Question

Given $X, Y, Z$ i.i.d random variables the probability $P(X+Y>Z)$ can be found by the following three approaches:

1. $X+Y-Z > 0$ region cuts the $X Y Z$ volume into two equal volumes as $X+Y-Z=0$ plane passes through origin. Thus, $P(X+Y>Z)=\frac{1}{2}$ which seems doubtful as $X+Y$ is mostly more than $Z$.
2. $$P(X+Y>Z) = E(I_{X+Y>Z}) = E(E(I_{X+Y>Z}|Z))$$ $$E(E(I_{X+Y>Z}|Z)) = E(P(X+Y>z)) = E((1-F_{X+Y}(z))) = \int_z (1-F_{X+Y}(z))f_{Z}(z)dz$$ where, $$F_{X+Y}(z) = \int_{-\infty} ^{z} \int_{-\infty} ^{\infty} f_{X}(k)f_{Y}(h-k)dk$$

It is very complicated to substitute $F_{X+Y}(z)$ in the final expression of $P(X+Y>Z)$. Thus, it's quite difficult to comment on $P(X+Y>Z)$. Please help in this approach to solve it.

3. $E[X] = E[Y] = E[Z]$, because $X, Y, Z$ follow identical, independent distributions.

$E[X + Y] = 2\times E[Z]$,

Intuition says that $P(X+Y>Z)$ thus ought to be 2/3, but I do not know if this is correct, nor how to make this reasoning rigorous.

Help in figuring out which approaches are correct, wrong and why ?

Answer 1

Using Bayes theorem: \begin{align} P(X+Y>Z) & =\int_{-\infty}^\infty dx\int_{-\infty}^\infty dy~f(x,y)P(X+Y>Z|X=x,Y=y)\\ &=\int_{-\infty}^\infty dx\int_{-\infty}^\infty dy~f(x)~f(y)P(Z<x+y)\\ &=\int_{-\infty}^\infty dx\int_{-\infty}^\infty dy~f(x)~f(y)\int_{-\infty}^{x+y}dz~f(z) \end{align}

Answer 2

Let's take the following random variables: $X,Y,Z$ all take $0$ or $1$ with probability $\frac12$ and are independent.

Consider the possible cases which are of the same probability $\left(\frac18, \frac18, ...\right)$:

$$\begin{matrix} \color{white}{\checkmark}XYZ\\ \color{white}{\checkmark}\ 0\ 0\ 0\ \\ \color{white}{\checkmark}\ 0\ 0\ 1\\ \color{green}{\checkmark}0\ 1\ 0\\ \color{white}{\checkmark}0\ 1\ 1\\ \color{green}{\checkmark}1\ 0\ 0\\ \color{white}{\checkmark}1\ 0\ 1\\ \color{green}{\checkmark}1\ 1\ 0\\ \color{green}{\checkmark}1\ 1\ 1\end{matrix}$$

$X+Y>Z$ takes place where the check marks are. The probability of the event is $$\frac12$$

and not $$\frac23.$$