Sum of random variables decomposition Let $0< \delta<\frac{1}{2}$. Why do we have for $X_{1},X_{2}$ iid:
\begin{equation}
\{X_{1}+X_{2}>x\}\subset\{X_{1}>(1-\delta)x\}\cup\{X_{2}>(1-\delta)x\}\cup\{X_{1}> \delta x,\ X_{2}>\delta x\}?
\end{equation}
 A: As a visual demonstration

The yellow area (to the right of the rightmost green line), 
the blue area (above the higher green horizontal line), 
and the green area (above and to the right of the other two lines) 
together more than cover the dark area.
It will not work if $x \lt 0$.
A: Just consider the case when $X_1<(1-\delta)x$ and $X_2<(1-\delta)x$, which means $(X_1,X_2)$ is not in the union of the two first sets. Then, if $X_1+X_2>x$, $X_1>x-X_2$ and, since $X_2<(1-\delta)x$,
$$
X_1 > x - (1-\delta)x = \delta x
$$
The same reasoning applies to $X_2$.
A: Sometimes it is easier to show $B^c\subseteq A^c$ instead of $A\subseteq B$, where $^c$ denotes the complement. In your case it suffices to show that
$$
\{X_1\leq (1-\delta)x\}\cap \{X_2\leq (1-\delta x)\}\cap \left(\{X_1\leq \delta x\}\cup \{X_2\leq \delta x\}\right)\subseteq \{X_1+X_2\leq x\}.
$$
Now take an $\omega$ belonging to the left hand side. Then $X_1(\omega)\leq (1-\delta)x$ and $X_2(\omega)\leq (1-\delta)x$ and either $X_1(\omega)\leq \delta x$ or $X_2(\omega)\leq \delta x$. Let us assume that $X_1(\omega)\leq \delta x$. Then
$$
X_1(\omega)+X_2(\omega)\leq \delta x + (1-\delta)x = x,
$$
and hence $\omega$ belongs to the right hand side.
