Suppose that $X$ and $Y$ are independent and identically distributed random variables with probability density function $f(x)$ that is symmetric about the origin.

We have $P[|X+Y|≤k] \ge a$. Can I show that there exist constants $k_1,k_2,a_1,a_2$ so that $P[|X| \le k_1] \ge a_1$
and $P[|Y| \le k_2] \ge a_2$?

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    $\begingroup$ Please use math typsetting to improve your question's readability. Readable questions are more likely to be answered. More information is available here: meta.math.stackexchange.com/questions/5020/… $\endgroup$ – Sycorax Jan 11 '16 at 16:08
  • $\begingroup$ Since $a_1=a_2=0$ will always work regardless of $K$, $a$, $K_1$, or $K_2$, this question must be missing some essential condition, constraint, or assumption. Could you please edit this post to expand on the missing information? $\endgroup$ – whuber Jan 11 '16 at 19:41
  • $\begingroup$ Also, $X$ and $Y$ are i.i.d. Thus if you can prove $\mathbb{P}(|X|\leq K_1) \geq a_1$ holds then the same inequality holds for $Y$ (and vice versa). I.e. $a_2$ and $K_2$ are distractions here $\endgroup$ – P.Windridge Jan 11 '16 at 21:23
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    $\begingroup$ Another point is that for any random variable we have $\mathbb{P}(|X|\leq K) = \mathbb{P}(X\leq K) - \mathbb{P}(X < -K) \to 1$ as $K \to \infty$. I.e., for any $a < 1$, there exists $K$ large enough so that $\mathbb{P}(|X|\leq K) > a$. So I guess you want some relationship to hold between $a_1$ and $a$, etc. $\endgroup$ – P.Windridge Jan 11 '16 at 21:40

Consider the pdf of X+Y. If X and Y have stdev $\sigma$, then X+Y has stdev $\sqrt(2) * \sigma$

Furthermore, drawing a picture of the distribution along X and Y axes, you can see that the limits create diagonal lines, one running through (0,K) and (K,0), the other through (-K,0) and (0,-K). The integral of the joint pdf in this area is >=a.

Now slice this area with the X=Y line. Letting X=Y does not diminish the fact that the area integrates to >=a. But note that now we can integrate a zero-mean gaussian with standard deviation of $\sqrt(2)*\sigma$ from $-\frac{\sqrt(2)}{2}*K$ to $+\frac{\sqrt(2)}{2}*K$ and get >= a

Thus K1 and K2 can both be $\frac{\sqrt(2)}{2}*K$ and a1 and a2 are both a

I've checked this with $\sigma = 1$ and $\sigma = 2$ and k = 1, 2, and 3


imagine the limit lines are blue (for the original problem). the new gaussian with sqrt(2) * sigma lies along the red line. Note that the distance to the limit is now $K*\frac{\sqrt(2)}{2}$

enter image description here

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  • $\begingroup$ You lost me where you introduced the Gaussian: could you be more explicit about how that is related to the arbitrary symmetric distribution described in the question? $\endgroup$ – whuber Jan 11 '16 at 19:37
  • $\begingroup$ I appreciate the picture, but what does a Gaussian have to do with the question? We don't even know that $X$ and $Y$ have a finite standard deviation! $\endgroup$ – whuber Jan 11 '16 at 20:24
  • $\begingroup$ Sorry, I guess I just assumed that they did. So, in that case, this new gaussian should be just like X and Y (they are identical, independent) except for the sqrt(2) larger standard deviation. Thus we can use the characteristics of this one with its stdev and probability limits to imply those of X and Y. ... or at least I think that we can $\endgroup$ – MikeP Jan 11 '16 at 20:32
  • $\begingroup$ Also how does the answer use that $X$ and $Y$ are independent? Perhaps I've misunderstood what you mean by "Letting $X=Y$ does not diminish the fact that the area integrates to $\leq a$", but allowing dependence we could take $X=-Y$, so that $|X+Y| = 0\leq K$, and I can't see how you can deduce much from that :) $\endgroup$ – P.Windridge Jan 11 '16 at 21:41
  • $\begingroup$ Were they not independent, then the standard deviation of the sum would not just the sqrt(2) * the standard deviation either. $\endgroup$ – MikeP Jan 11 '16 at 22:41

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