Question

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?

Answer 1

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.

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

$$\frac{1}{(n-1)!^2} \int_{y=0}^\infty (y+|x|)^{n-1} e^{-y-|x|} y^{n-1} e^{-y}~dy$$

$$=\frac{1}{e^{|x|} (n-1)!^2}\int_{y=0}^\infty (y^2+y|x|)^{n-1} e^{-2y}~dy$$

$$=\frac{1}{e^{|x|} (n-1)!^2} \sum_{i=0}^{n-1} \int_{y=0}^\infty {n-1 \choose i}y^{2(n-1)-i}|x|^i e^{-2y}~dy$$

$$=\frac{1}{e^{|x|} (n-1)!^2} \sum_{i=0}^{n-1} {n-1 \choose i}\frac{(2n-i-2)! |x|^i}{2^{2n-i-1}}$$

which is just a polynomial in $|x|$ divided by $e^{|x|}$.

$n=1: \frac{1}{2 \exp(|x|)}$.

$n=2: \frac{|x|+1}{4 \exp(|x|)} $.

$n=3: \frac{|x|^2 + 3|x| + 3}{16 \exp(|x|)}$.

$n=4: \frac{|x|^3 + 6|x|^2+ 15|x| + 15}{96 \exp(|x|)}$.

Answer 2

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).