tight bounds on probability of sum of laplace random variables. - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T13:17:49Z http://mathoverflow.net/feeds/question/66763 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/66763/tight-bounds-on-probability-of-sum-of-laplace-random-variables tight bounds on probability of sum of laplace random variables. Vedarun 2011-06-02T19:34:54Z 2012-07-13T18:17:21Z <p>Are there tight upper and lower bounds on the density of the sum of $n$ i.i.d laplace random variables that depend on $n$ and the individual laplacian densities?</p> http://mathoverflow.net/questions/66763/tight-bounds-on-probability-of-sum-of-laplace-random-variables/67302#67302 Answer by Douglas Zare for tight bounds on probability of sum of laplace random variables. Douglas Zare 2011-06-08T19:55:00Z 2011-06-08T19:55:00Z <p>You can read off the density function as follows: Since a Laplace distribution with mean $0$ is the difference of two IID exponential distributions, the sum of $n$ IID Laplace distributions with mean $0$ is the difference of two IID gamma (or Erlang) distributions. Although gamma distributions are not memoryless, the difference between two independent gamma distributions can be computed directly by convolution.</p> <p>For simplicity, rescale so that the parameter $\lambda=1$. The probability density function for the appropriate gamma distribution with $\lambda=1$ is $\frac{1}{(n-1)!} x^{n-1} e^{-x}$ for $x\gt 0$. So, the difference between two has probability density function</p> <p>$$\frac{1}{(n-1)!^2} \int_{y=0}^\infty (y+|x|)^{n-1} e^{-y-|x|} y^{n-1} e^{-y}~dy$$</p> <p>$$=\frac{1}{e^{|x|} (n-1)!^2}\int_{y=0}^\infty (y^2+y|x|)^{n-1} e^{-2y}~dy$$</p> <p>$$=\frac{1}{e^{|x|} (n-1)!^2} \sum_{i=0}^{n-1} \int_{y=0}^\infty {n-1 \choose i}y^{2(n-1)-i}|x|^i e^{-2y}~dy$$</p> <p>$$=\frac{1}{e^{|x|} (n-1)!^2} \sum_{i=0}^{n-1} {n-1 \choose i}\frac{(2n-i-2)! |x|^i}{2^{2n-i-1}}$$</p> <p>which is just a polynomial in $|x|$ divided by $e^{|x|}$.</p> <p>$n=1: \frac{1}{2 \exp(|x|)}$.</p> <p>$n=2: \frac{|x|+1}{4 \exp(|x|)}$.</p> <p>$n=3: \frac{|x|^2 + 3|x| + 3}{16 \exp(|x|)}$.</p> <p>$n=4: \frac{|x|^3 + 6|x|^2+ 15|x| + 15}{96 \exp(|x|)}$.</p> http://mathoverflow.net/questions/66763/tight-bounds-on-probability-of-sum-of-laplace-random-variables/102157#102157 Answer by Yisong Yue for tight bounds on probability of sum of laplace random variables. Yisong Yue 2012-07-13T18:17:21Z 2012-07-13T18:17:21Z <p>Can one write a general formula that depends on arbitrary n? For example, the gamma function can take as input arbitrary positive n (and corresponds to the sum of n exponentially distributed RVs when n is integral). </p>