We are given independent and identically distributed random variables $X,Y$ with probability density function $f(\cdot)$ that is symmetrical about $0$.

We need to prove that $P(|X+Y|\leq 2|X|) > \dfrac{1}{2}$

Starting with $P(|X+Y|\leq 2|X|)=P((X+Y)^2\leq 4X^2)=P(-3X^2+Y^2+2XY\leq 0)$

Now, $-3X^2+Y^2+2XY=0$ represents a pair of distinct real lines : $X=Y$ and $3X+Y=0$.

So, $P(|X+Y|\leq 2|X|)=P(-3X^2+Y^2+2XY\leq 0)=P[(Y-X)(3X+Y)\leq 0]$.

So two cases arise,

(i) $Y-X\geq0$ and $3X+Y\leq0$


(ii) $Y-X\leq0$ and $3X+Y\geq0$

So, we have $P(|X+Y|\leq 2|X|)=\int_0^{\infty}\int_{(\frac{-y}{3})}^{y}f_{X,Y}(x,y)\:dxdy\:+\: \int_{-\infty}^{0}\int_{y}^{(\frac{-y}{3})}f_{X,Y}(x,y)\:dxdy$

Since $X,Y$ are independent $f_{X,Y}(x,y)=f_X(x)f_Y(y)$.

I am not able to move forward from here, in the solution the integral on the far right side is straight away written to be $(\dfrac{1}{2})$ and hence the sum becomes $\geq(\frac{1}{2})$.

Can anyone explain what just happened there ?

  • $\begingroup$ Are you assuming that $x$ and $y$ are continuous r.v.s? $\endgroup$
    – jbowman
    Mar 23, 2018 at 19:12
  • $\begingroup$ The two cases have equal probability, since $(X,Y)$ has the same distribution as $(-X,-Y)$. The corresponding iterated integrals are therefore equal, so there's no way the far right one can be equal to $1/2$. $\endgroup$
    – grand_chat
    Mar 23, 2018 at 19:26
  • $\begingroup$ So, there are more than just these two cases ? @grand_chat $\endgroup$
    – User9523
    Mar 23, 2018 at 19:54

4 Answers 4


I think it's much easier to solve this problem using the triangle inequality rather than using a squaring approach. Since $|X+Y| \le |X| + |Y|$, we have that

$$P(|X+Y|\le2|X|) \ge P(|X|+|Y|≤2|X|)=P(|Y|≤|X|)=1/2$$

Do you specifically need to show that the probability is greater than 1/2?

  • 2
    $\begingroup$ Far better than my answer, which I shall now delete so as to avoid taking away well-deserved upvotes. But a counterexample to the $> 1/2$ exists, namely, $x, y \sim \text{U}(-1,1)$ and $x=1/4 \implies y \in [-3/4, 1/4]$, which evidently has probability equal to $1/2$. $\endgroup$
    – jbowman
    Mar 23, 2018 at 20:05
  • $\begingroup$ I'm not sure I follow you're counterexample. Can you elaborate. Based on my math, I have that if $X$ and $Y$ are uniform on $(-1, 1)$, then the probability of interest is 2/3. Are you sure that strict inequality doesn't hold? $\endgroup$
    – jjet
    Mar 23, 2018 at 20:41
  • 1
    $\begingroup$ The inequality does not have to be strict if the distributions are discrete. If I calculated correctly, if A = probability the point (X,Y) lies between the lines Y= -X and Y = -3X or the lines Y=X and Y = X/3, then P(|X+Y|<2|X|) =(1+A)/2. So if A = 0, P(|X+Y|<2|X|) is exactly 1/2. $\endgroup$ Mar 23, 2018 at 22:26
  • $\begingroup$ When you said "Whoops, this is incorrect", what were you including in "this"? $\endgroup$ Mar 23, 2018 at 22:30
  • $\begingroup$ I had meant the solution provided in the first edit. I just deleted the edits for clarity since they served no purpose. $\endgroup$
    – jjet
    Mar 23, 2018 at 22:35

If you draw the lines Y = X and Y = -3X, the first has slope 1, and the second has slope -3. The two lines divide the plane into four quadrants, with the solution set being the left and right quadrants. So you just have to show that more than half of the probability mass is in those two quadrants. Call the two quadrants together figure1. Now rotate this by 90, and call this figure2. Rotating by 90 degrees is the same as Y' = X, X' = -Y. The distributions are identical, so exchanging X and Y leaves the probability the same. They are symmetric about zero, so multiplying by -1 leave the probability the same. Thus, this transformation leaves the probability the same. Since every point in the plane is covered by either figure1 or figure2, and some points are covered by both, and this transformation doesn't affect the probability mass, it follows that figure1 contains at least half of the probability mass, and if you can proved that the overlapped areas contain a positive amount of probability mass, then it follows that figure1 contains more than half of the probability mass.


The image below demonstrates how a partitioning of the area $|x+y| \leq 2|x|$ helps to proof $P[|X+Y| \leq 2|X|] > \frac{1}{2}$.

The hatched region (region 2) corresponds to your area $$|x+y| \leq 2|x| \qquad \text{or} \qquad (y-x)(3x+y) \leq 0 $$

Part of this region (the pink colored hatched region, region 2a) is a mirror image of the complement of region 2 (the pink coloured region, region 1). From this you can deduce that:

$$P[|X+Y| \leq 2|X|] = \frac{1 + P[(X-Y)(3Y-X) \geq 0] + P[(-X-Y)(3X+Y) \geq 0]}{2}$$

where the extra terms $P[(X-Y)(3Y-X) \geq 0]$ and $P[(-X-Y)(3X+Y) \geq 0]$ relate to the probability that X,Y are in the gray coloured hatched regions.


Note that the lines are not included in the complement region 1. Thus the discrete uniform example $x,y \sim U(−1,1)$ by jbowman does not work. The points (1,1), (-1,-1), (-1,1), (1,-1) are inside the gray hatched region and eventually inside the region for which $|x+y| \leq 2|x|$ is true.

In the same way for any distribution with finite pdf or probability mass at least somewhere. there will be some positive contribution near the lines Y=X such that the inequality is more strict $>$ instead of $\geq$.

For instance in the discrete case $P(X=x,Y=y)$ for $x=y$ is equal to $P(X=x)^2$ which is due to the property that $f_X=f_Y$

In the continuous case you could evaluate:

$$\begin{array} \\P[|X+Y| \leq 2|X|] &= \frac{1 + 4 \int_{x=0}^{x=\infty} \left( \int_{t=\frac{1}{3}x}^{t=x} f_X(t) dt \right) f_X(x) dx}{2}\\ & = \frac{1 + 4 \int_{x=0}^{x=\infty} \left( F(x)-F(\frac{1}{3}x)\right) f_X(x) dx}{2} \\ \end{array}$$

where the integral must be non-zero if f_X(x) is non zero in at least some continuous region of non zero size (such that $F(x)-F(\frac{1}{3}x)$ is non-zero in a region with non-zero probability).

Some other equality is

$$P[|X+Y| \leq 2|X|] = \frac{1}{2} + P[\frac{1}{3} |Y| \leq |X| \leq |Y| ]$$


Changing this probability into an expectation is the key to solving this problem easily. As noted by @Accumulation and @Martijn Weterings, the absolute value function divides the (x,y)-plane into conic regions. Note that

$$ P(|X+Y|<2|X|)=E[I(|X+Y|<2|X|)] $$

where $I()$ is the usual zero-one indicator function. (Note that we can use "$<$" in place of "$\le$" by assuming $X$ and $Y$ are continuous - a requirement that's needed to ensure the strict inequality you want.) Then, the indicator can be rewritten as



$$U=I(X < 0, -X < Y < -3X) + I(X > 0, -3X < Y < -X)$$ and $$V=I(X < 0, X < Y < -X) + I(X > 0, -X < Y < X)$$

Based on the definitions of $U$ and $V$, it is clear that the regions on which each random variable equals one are non-overlapping (i.e. $P(V=1,U=1)=0$). Additionally, we can rewrite $V$ as

$$V=I(|X|+|Y| \le 2|X|)=I(|Y| \le |X|)$$


$$P(|X+Y|<2|X|)=E[I(|X+Y|<2|X|)]=E[U+V]=E[U]+E[V]=P(U=1)+P(|Y| \le |X|)=P(U=1)+1/2$$

All that remains is to show that $P(U=1)>0$. Let $A\times B\in \textbf{R}^2$ be a rectangle centered about the origin. Then, $P(X\in A, Y\in B)=P(X\in A)P(X\in B)$. Necessarily, there exist intervals $A$ and $B$ over which $f_X(x)$ integrates to some number greater than zero. Also, any such $A$ and $B$ will contain area over which $U=1$. This follows from the fact that the area over which $U=1$ takes the form of two cones extending from the origin. Therefore, $P(U=1)>0$ and the result follows.


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