Integral of Probability Density Function over the Reals is Equal to One
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Theorem
Let $\struct {\Omega, \Sigma, \Pr}$ be a probability space.
Let $X$ be an absolutely continuous random variable.
Let $f_X$ be a probability density function for $X$.
Let $\map \BB \R$ be the Borel $\sigma$-algebra on $\R$.
Let $\lambda$ be the Lebesgue measure on $\struct {\R, \map \BB \R}$.
Then:
- $\ds \int_\R f_X \rd \lambda = 1$
where $\ds \int_\R f_X \rd \lambda$ denotes the Lebesgue integral of $f_X$.
Proof
Let $P_X$ be the probability distribution of $X$.
From the definition of a probability density function, $f_X$ is a Radon-Nikodym derivative of $P_X$ with respect to $\lambda$.
We then have:
\(\ds 1\) | \(=\) | \(\ds \map {P_X} \R\) | Definition of Probability Measure, Probability Distribution is Probability Measure | |||||||||||
\(\ds \) | \(=\) | \(\ds \int_\R f_X \rd \lambda\) | Definition of Radon-Nikodym Derivative |
$\blacksquare$