Density Functions

Q: Consider the function f(x) = (e^-|x|)/2 for x belonging to Real numbers. Prove that f is a density function.

Approach:
For f(x) to be classified as a density function there are 2 things required:
1. f(x) >= 0 for all x belonging to Real numbers.
2. Indefinite Integral f(x)dx = 1.

Property 1 is simple enough to prove as f(x) >= 0 for all values of x. However, I am stumped on proving property 2. As far as I can tell, the integral is undefined. Am I going about the solution of this question all wrong or is there something else that proves a density function?

For functions involving absolute values, you can break them up into two different functions and then work through as two separate cases.

So, if you are to integrate the function you have written, then:

For x > 0, you can replace |x| by x , that is, integrate (e^-x)/2 dx.

For x < 0, you can replace |x|by -x, that is, integrate (e^x)/2 dx

$\displaystyle \displaystyle \int_{ - \infty }^\infty {\frac{{e^{ - |x|} }}
{2}dx} = 1$

Notice that $\displaystyle f(x) = \dfrac{{e^{ - |x|} }}{2}$ is an even function.

Quote:

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Originally Posted by ipacheco

Q: Consider the function f(x) = (e^-|x|)/2 for x belonging to Real numbers. Prove that f is a density function.

Approach:
For f(x) to be classified as a density function there are 2 things required:
1. f(x) >= 0 for all x belonging to Real numbers.
2. Indefinite Integral f(x)dx = 1.

Property 1 is simple enough to prove as f(x) >= 0 for all values of x. However, I am stumped on proving property 2. As far as I can tell, the integral is undefined. Am I going about the solution of this question all wrong or is there something else that proves a density function?

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Thread of related interest: http://www.mathhelpforum.com/math-he...tml#post566813

Thanks everyone. I used the two case method where I split up the integral based on whether x was positive or negative. It became simple enough after that. Thanks for the help.