most recent 30 from http://mathoverflow.net 2013-05-23T04:27:56Z http://mathoverflow.net/feeds/question/95719 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/95719/on-the-difference-of-exponential-random-variables Mensarguens 2012-05-02T04:38:57Z 2012-07-24T00:28:35Z

<p>Assume two random variables X,Y are exponentially distributed with rates p and q respectively, and we know that the r.v. X-Y is distributed like X'-Y' where X',Y'are exponential random variables, independent among themselves and independent of X andY, with rates p and q.</p> <p>Does it follow that X and Y are independent?</p>

http://mathoverflow.net/questions/95719/on-the-difference-of-exponential-random-variables/102969#102969 Douglas Zare 2012-07-24T00:28:35Z 2012-07-24T00:28:35Z

<p>No.</p> <p>Consider an analogous question for the random variables $Z_1$, $Z_2$ uniformly distributed on $\lbrace0,1,2,3\rbrace$ whose sum has the probabilities $1/16, 2/16, 3/16, 4/16, 3/16, 2/16, 1/16$ of taking the values $0, 1, 2, 3, 4, 5, 6$, respectively. Must $Z_1$ and $Z_2$ be independent? No, here is a possible joint distribution:</p> <p>$$\frac 1{16}\begin{pmatrix}1 &amp; 0 &amp; 2 &amp; 1 \\ 2 &amp; 1 &amp; 1 &amp; 0 \\ 0 &amp; 1 &amp; 1 &amp; 2 \\ 1 &amp; 2 &amp; 0 &amp; 1\end{pmatrix}.$$</p> <p>The sum of each row or column is $1/4$, and the sums of the diagonals are $1/16, 2/16, 3/16, 4/16, 3/16, 2/16, 1/16$. </p> <p>Using this, we can construct random variables uniformly distributed on $[0,1]$ whose sum is a uniform sum distribution, but which are not independent. Let the density of the joint distribution on the unit square be $0$, $1$, or $2$ according to the $16$th of the square and the pattern above.</p> <p>Using this, for any distributions which have densities greater than a positive constant on some intervals, we can write the distributions as mixtures of uniform distributions and something else, and we use the above on the uniform distributions to construct nonindependent copies whose sum has the same distribution as independent copies. Doing this for $X'$ and $-Y'$ produces exponentially distributed random variables which are not independent, but whose difference is the same as the difference of independent exponential random variables.</p> <p>There are some discrete distributions which do have the property that if the sum looks like the variables are independent, then they must be independent.</p>