# Showing that Laplace distribution is a probability density

• Jun 9th 2010, 12:58 AM
acevipa
Showing that Laplace distribution is a probability density
How would you show that the laplace distribution is a probability density, i.e.

$\displaystyle f(x)\geq 0$ and $\displaystyle \int_{-\infty}^{\infty}f(x)dx=1$

Laplace Distribution:

$\displaystyle f(x)=\frac{1}{2}e^{-|x|}$ for $\displaystyle -\infty<x<\infty$

Proving that $\displaystyle f(x) \geq 0$ is quite obvious

To prove that the integral equals 1, would you consider the 2 cases where $\displaystyle x\geq 0$ and $\displaystyle x<0$? And then from that, the sum of the integrals should equal 1?
• Jun 9th 2010, 01:01 AM
mr fantastic
Quote:

Originally Posted by acevipa
How would you show that the laplace distribution is a probability density, i.e.

$\displaystyle f(x)\geq 0$ and $\displaystyle \int_{-\infty}^{\infty}f(x)dx=1$

Laplace Distribution:

$\displaystyle f(x)=\frac{1}{2}e^{-|x|}$ for $\displaystyle -\infty<x<\infty$

Proving that $\displaystyle f(x) \geq 0$ is quite obvious

To prove that the integral equals 1, would you consider the 2 cases where $\displaystyle x\geq 0$ and $\displaystyle x<0$? And then from that, the sum of the integrals should equal 1?

Yes.