Probability of absolute value of a sum of two symmetric random variables - Cross Validated most recent 30 from stats.stackexchange.com 2019-09-15T16:29:21Z https://stats.stackexchange.com/feeds/question/190176 https://creativecommons.org/licenses/by-sa/4.0/rdf https://stats.stackexchange.com/q/190176 -1 Probability of absolute value of a sum of two symmetric random variables user100358 https://stats.stackexchange.com/users/100358 2016-01-11T15:08:23Z 2017-02-26T11:43:18Z <p>Suppose that $X$ and $Y$ are independent and identically distributed random variables with probability density function $f(x)$ that is symmetric about the origin.</p> <p>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 $P[|X| \le k_1] \ge a_1$<br> and $P[|Y| \le k_2] \ge a_2$? </p> https://stats.stackexchange.com/questions/190176/-/190209#190209 -1 Answer by MikeP for Probability of absolute value of a sum of two symmetric random variables MikeP https://stats.stackexchange.com/users/36115 2016-01-11T18:52:10Z 2016-01-12T14:26:23Z <p>Consider the pdf of X+Y. If X and Y have stdev $\sigma$, then X+Y has stdev $\sqrt(2) * \sigma$</p> <p>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.</p> <p>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</p> <p>Thus K1 and K2 can both be $\frac{\sqrt(2)}{2}*K$ and a1 and a2 are both a</p> <p>I've checked this with $\sigma = 1$ and $\sigma = 2$ and k = 1, 2, and 3</p> <p>added:</p> <p>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 $K*\frac{\sqrt(2)}{2}$</p> <p><a href="https://i.stack.imgur.com/NApnt.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/NApnt.jpg" alt="enter image description here"></a></p>