# gamma distribution is to exponential distribution what ---- distribution is to Laplace distribution?

Context

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].

Questions

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.

Bibliography

[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

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.
• You mean Laplace(0,$\lambda$), correct? Commented Feb 6 at 21:20
• 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. Commented Feb 7 at 14:03