I'm working on a random walk problem with an two-sided exponential distribution (i.e., a Laplace distribution).

From [1], I know that "The sum of $n$ independent $\operatorname{Exp}(\lambda)$ exponential random variables is a $\operatorname{Gamma}(n, \lambda)$ gamma distributed". The sketch of a proof of this proposition given by @heropup in answer to a question on StackExchange [2].

In my problem, I do not have an exponential distrubution, I have Laplace distribution [3].


Q.1 What type of distribution is the sum of $n$ independent Laplace random variables?

Q.2 In an analogous method to @heropup in [2], can you prove that your answer to Q.1 is correct by comparing the moment generating functions, by using induction, or by other means.

My solution

Solution to Question 1

From what I understand from Christophe Leuridan's accepted solution in [4], and from further reading into the matter, there is not an accepted convention for the distribution in question.

Solution to Question 2

Adapting from the framework provided by Christophe Leuridan in [4], let $X_1,\ldots,X_n,Y_1,\ldots,Y_n$ be identical independent random variables each with distribution Exp$(\lambda)$. Then, from \url{https://en.wikipedia.org/wiki/Laplace_distribution} $X_1-Y_1,\ldots,X_n-Y_n$ are $n$ independent random variables with distribution Laplace$(0,b)$, where $b = \lambda^{-1} $; whereas $X_1+\cdots+X_n$ and $Y_1+\cdots+Y_n$ are both independent random variables with distribution Gamma$(n,\lambda)$. By $(\lambda)^{*n,b}$ I denote the distribution formed by the sum of $n$ independent $\operatorname{Laplace}(n,b)$ Laplace random variables. By $Z_n$, where $Z_n := (X_1+\cdots+X_n)-(Y_1+\cdots+Y_n)$, I denote a random variable with that is the difference of two independent random variables with distribution Gamma$(n,\lambda)$.

From \url{https://math.stackexchange.com/posts/4737796/edit} I know that the probability density function of the sum of two independent random variables $U$ and $V$, each of which has a probability density function, is the convolution of their separate density functions: $$ f_{U+V}(z) = \int_{-\infty}^z f_U(x) f_V(z - x)\,dx. $$ Thus, $$ f_{Z_n}(z) = \int_{-\infty}^z f_{\left(X_1+\cdots+X_n\right)}(x) f_{-\left(X_1+\cdots+X_n\right)}(z - x)\,dx. $$

Now, as $F_\xi(y)$ I denote the cumulative density function of the random variable $\xi$. In particular, $$F_{-U}(y)=\mathbb{P}(-U\leq y)=\mathbb{P}(U \geq -y)=1-\mathbb{P}(U\le -y)=\\=1-F_{U}(-y).$$ To find the density of $-U$, we differentiate $F_{-U}(u)$ : $$f_{-U}(u)=\frac{d}{du}\left(F_{-U}(u)\right)=\frac{d}{du}(1-F_{U}(-u))=\\=f_{U}(-u).$$ Thus, $$ f_{Z_n}(z) = \int_{-\infty}^z f_{\left(X_1+\cdots+X_n\right)}(x) f_{\left(X_1+\cdots+X_n\right)}(x - z)\,dy. $$ With the aid of the indicator function, $\mathbf{1}_{u>0} $, I write that the probability-density distribution is given by \begin{align*} f_{Z_n}(z) &= \int_\mathbb{R} f_{X_1+\cdots+X_n}(x)f_{Y_1+\cdots+Y_n}(x-z) dx \\ &= \int_\mathbb{R} \frac{\lambda^n}{\Gamma(n)}x^{n-1}e^{-\lambda x}\mathbf{1}_{x>0} \quad\frac{\lambda^n}{\Gamma(n)}(x-z)^{n-1}e^{-\lambda (x-z)}\mathbf{1}_{x-z>0} dx \\ &= \frac{\lambda^{2n}e^{\lambda z}}{\left(\Gamma(n)\right)^2}\int_{\max(z,0)}^\infty x^{n-1}(x-z)^{n-1}e^{-2\lambda x}dx. \end{align*} With the binomial theorem we obtain that \begin{align*} f_{Z_n}(z) &= \frac{\lambda^{2n}e^{\lambda z}}{\left(\Gamma(n)\right)^2}\int_{\max(z,0)}^\infty x^{n-1} \sum_{k=0}^{n-1} {{n-1} \choose k}x^{{n-1}-k}(-z)^k e^{-2\lambda x}dx. \\ &= \frac{\lambda^{2n}e^{\lambda z} }{\left(\Gamma(n)\right)^2}\sum_{k=0}^{n-1} (-z)^k {{n-1} \choose k} \int_{\max(z,0)}^\infty x^{{2n-2}-k} e^{-2\lambda x}dx. \end{align*} From a table of integrals we have that \begin{align*} \int_{\max(z,0)}^\infty x^{ 2n-2 -k} e^{-2\lambda x}dx &= \left[ e^{-2\lambda x} \sum_{r=0}^{2n-2 -k}(-1)^r \frac{(2n-2 -k)! {x }^{2n-2 -k-r}}{(2n-2 -k-r)! (-2\lambda)^{r+1}} \right]_{\max\left( z ,0\right)}^\infty \\ &= -e^{-2\lambda \max\left( z ,0\right)} \sum_{r=0}^{2n-2 -k}(-1)^r \frac{(2n-2 -k)! {\max\left( z ,0\right) }^{2n-2 -k-r}}{(2n-2 -k-r)! (-2\lambda)^{r+1}} . \end{align*} Thus, \begin{align*} f_{Z_n}(z) &= \frac{\lambda^{2n}e^{\lambda z} }{\left(\Gamma(n)\right)^2}\sum_{k=0}^{n-1} (-z)^k {{n-1} \choose k} e^{-2\lambda \max\left( z ,0\right)} \sum_{r=0}^{2n-2 -k}(-1)^{r+1} \frac{(2n-2 -k)! {\max\left( z ,0\right) }^{2n-2 -k-r}}{(2n-2 -k-r)! (-2\lambda)^{r+1}} . \end{align*} and \begin{align*} \boxed{ f_{Z_n}(z) = \frac{\lambda^{2n}e^{\lambda \left[z -2 \max\left( z ,0\right)\right]} }{\left(\Gamma(n)\right)^2}\sum_{k=0}^{n-1} (-z)^k {{n-1} \choose k}\Gamma(2n- 1-k) \sum_{r=0}^{2n-2 -k} \frac{ {\max\left( z ,0\right) }^{2n-2 -k-r}}{(2n-2 -k-r)! ( 2\lambda)^{r+1}} .} \end{align*}

Sanity check Let's do a sanity check \begin{align*} f_{Z_1}(0) &= \frac{ 1}{ ( 2\lambda) } . \end{align*} This is exactly the value of the Laplace distribution's probability-distribution function.


[1] https://en.wikipedia.org/wiki/Exponential_distribution

[2] Gamma Distribution out of sum of exponential random variables

[3] https://en.wikipedia.org/wiki/Laplace_distribution

[4] Christophe Leuridan (https://math.stackexchange.com/users/421890/christophe-leuridan), gamma distribution is to exponential distribution what ---- distribution is to Laplace distribution?, URL (version: 2024-02-06): https://math.stackexchange.com/q/4858203


1 Answer 1


Let $X_1,\ldots,X_n,Y_1,\ldots,Y_n$ be independent random variables with distribution Exp$(\lambda)$. Then $X_1-Y_1,\ldots,X_n-Y_n$ are independent random variables with distribution Laplace$(\lambda)$, whereas $X_1+\cdots+X_n$ and $Y_1+\cdots+Y_n$ are independent random variables with distribution Gamma$(n,\lambda)$.

Hence $Z_n := (X_1+\cdots+X_n)-(Y_1+\cdots+Y_n)$ has distribution Laplace$(\lambda)^{*n}$, and is the difference of two independent random variables with distribution Gamma$(n,\lambda)$.

If I do not make mistakes, the density is given by \begin{eqnarray*} f_{Z_n}(z) &=& \int_\mathbb{R} f_{X_1+\cdots+X_n}(x)f_{Y_1+\cdots+Y_n}(x-z)dx\\ &=& \int_\mathbb{R} \frac{\lambda^n}{n!}x^{n-1}e^{-\lambda x}\mathbb{1}_{x>0} \frac{\lambda^n}{n!}(x-z)^{n-1}e^{-\lambda (x-z)}\mathbb{1}_{x-z>0} dx \\ &=& \frac{\lambda^{2n}}{n!^2}e^{\lambda z}\int_{\max(z,0)}^\infty x^{n-1}(x-z)^{n-1}e^{-2\lambda x}dx. \end{eqnarray*} The integral can be computed using the Newton binomial formula and the equalities $$2\lambda\int_{z}^\infty \frac{(2\lambda x)^k}{k!}e^{-2\lambda x}dx = \sum_{j=0}^k \frac{(2\lambda z)^j}{j!}e^{-2\lambda z}$$ This provides an heavy expression.

  • $\begingroup$ You mean Laplace(0,$\lambda$), correct? $\endgroup$ Commented Feb 6 at 21:20
  • $\begingroup$ Your answer would be improved if you gave the density explicitly. $\endgroup$ Commented Feb 6 at 21:21
  • $\begingroup$ I now give hints, but I let you complete the computation. $\endgroup$ Commented Feb 6 at 21:55
  • 1
    $\begingroup$ If $X$ and $Y$ are independent and have densities $f_X$ and $f_Y$, then $X+Y$ has density $f_X*f_Y$. And $-Y$ has density $f_{-Y} : z \mapsto f_Y(-z)$. This is the two things I use to get the density of a difference of independent random variables. $\endgroup$ Commented Feb 7 at 14:03

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