TITLE: Laplace Distribution Constant
QUESTION [1 upvotes]: Given the Laplace distribution PDF as $$f_X(x)=ke^{-\lambda|x|} \> \forall x \in \mathbb{R}$$ How may we determine what k is in terms of $\lambda$. I have a hunch that k is equal to $\frac{\lambda}{2}$ due to the fact that our x is calculated in absolute terms and therefore our limits of integration may be split up into two integrals where $x \leq \mu$ and $x > \mu$ but is there a more rigorous/intuitive argument?

REPLY [2 votes]: Your intuition is exactly correct, but you need to formalize it.  Since $f_X$ is a density, we must have $$\int_{x = -\infty}^\infty f_X(x) \, dx = 1,$$ or in your case, $$\int_{x=-\infty}^\infty ke^{-\lambda|x|} \, dx = 1.$$  Since $k$ is a positive constant that does not depend on the variable of integration $x$, and since $$|x| = \begin{cases} x, & x \ge 0 \\ -x, & x < 0, \end{cases}$$ our equation becomes $$\int_{x = -\infty}^0 e^{-\lambda(-x)} \, dx + \int_{x=0}^\infty e^{-\lambda x} \, dx = \frac{1}{k}.$$  Now evaluate the two integrals and solve for $k$ in terms of $\lambda$.
Alternatively, we can recognize the symmetry of $f_X$, namely for all $x \in \mathbb R$, we have $$f_X(-x) = k e^{-\lambda|-x|} = ke^{-\lambda|x|} = f_X(x).$$  Therefore, $$\int_{x=-\infty}^\infty f_X(x) \, dx = 2\int_{x=0}^\infty f_X(x) \, dx.$$
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