Probability of absolute value of a sum of two symmetric random variables Suppose that $X$ and $Y$ are independent and identically distributed random variables with probability density function $f(x)$ that is symmetric about the origin.
Show that $\mathbb{P}[|X+Y|\le2|X|]>0.5$
 A: Assuming only that $X$ and $Y$ are exchangeable (which holds if $X$ and $Y$ are i.i.d.), note that, almost surely,
$$|X+Y|\leqslant|X|+|Y|\leqslant2\max(|X|,|Y|),
$$ hence the events $A=[|X+Y|\leqslant2|X|]$ and $B=[|X+Y|\leqslant2|Y|]$ are such that $A\cup B=\Omega$ and $P[A]=P[B]$. In particular, 
$$P[A]\geqslant\tfrac12.
$$
If $P[A]=\frac12$, then $P[A\cap B]=0$. For every $x\geqslant0$, 
$$
[X\in[x,2x],Y\in[-2x,-x]]\subseteq A\cap B.
$$
If $X$ and $Y$ are i.i.d. and symmetric, then $P[X\in[x,2x]]=P[Y\in[-2x,-x]]$ hence $P[X\in[x,2x]]=0$ for every $x\geqslant0$. This is impossible, hence 
$$P[A]\gt\tfrac12.$$
A: Let $P$ be the given probability.  Note that for a given $x \geq 0$, we have $|x+y|\leq 2|x|$ just in case $-3x\leq y\leq x$.  Exploiting the symmetry of the density, we thus obtain that $\displaystyle P = 2\int_0^\infty \int_{-3x}^x f(x,y)\ dy\ dx$, where $f(x,y)= f(x)f(y)$ is the joint density function.  Then $\displaystyle P > 2\int_0^\infty \int_{-x}^x f(x,y)\ dy\ dx$, since the integrand is nonnegative (strict inequality because the omitted part is positive somewhere).  Since $f(x,y) = f(y,x) = f(-x,y) = f(y, -x)$, the integral of $f(x,y)$ over the region $R = \{(x,y)|\ x\geq 0, -x\leq y \leq x\}$ is exactly $\frac{1}{4}$ of the integral of $f(x,y)$ over the whole plane (because it's equal to the integrals over the regions given by applying the three transformations $(x,y)\rightarrow (y,x)$, $(x,y)\rightarrow (-x,y)$, and $(x,y)\rightarrow (y,-x)$ to $R$, and those four regions together cover the whole plane). The integral of $f(x,y)$ over the whole plane is $1$; hence $P > 2\cdot\frac{1}{4} = \frac{1}{2}$.
A: Due to symmetry, independence, and equality in distribution, we have
\begin{align}
P\bigl[|X+Y| \leq 2|X|]\bigr] &= P\bigl[|X| \geq |Y|]\bigr] +
\frac{1}{2}P\bigl[3|X| \geq |Y| > |X|\bigr]\\
&= \frac12 + 
\frac{1}{2}P\bigl[3|X| \geq |Y| \geq |X|\bigr].
\end{align}
It thus suffices to prove that $P\bigl[3|X| \geq |Y| \geq |X|\bigr] > 0$.  This can be done by contradiction.  For any $z\geq 0$, note that
$$P\bigl[3|X| \geq |Y| \geq |X|\bigr]\geq 
\frac{1}{2}P\bigl[|X|,|Y| \in [z,2z]\bigr] =
\frac{1}{2}P\bigl[|X|\in [z,2z]\bigr]^2.$$
Hence, if the first probability is zero, it must be the case that $P\bigl[|X|\in [z,2z]\bigr]=0$ for all $z\geq 0$, and since the positive real line can be covered with countably many such intervals, it follows that $P[|X|\geq 0]=0$, which is impossible.
