-1
$\begingroup$

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 $P[|Y| \le k_2] \ge a_2$?

$\endgroup$
  • 4
    $\begingroup$ 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
  • 1
    $\begingroup$ 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
-1
$\begingroup$

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

added:

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

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.