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# Probability density function

A probability density function serves to represent a probability distribution in terms of integrals. If a probability distribution has density f(x), then intuitively the infinitesimal interval [x, x + dx] has probability f(x) dx. A probability density function can be seen as a "smoothed out" version of a histogram: if one empirically measures values of a random variable repeatedly and produces a histogram depicting relative frequencies of output ranges, then this histogram will resemble the random variable's probability density (assuming that the variable is sampled sufficiently often and the output ranges are sufficiently narrow).

Formally, a probability distribution has density f(x) if f(x) is a non-negative Lebesgue integrable function RR such that the probability of the interval [a, b] is given by

$\int_a^b f(x)\,dx$
for any two numbers a and b. This implies that the total integral of f must be 1. Conversely, any non-negative Lebesgue integrable function with total integral 1 is the probability density of a suitably defined probability distribution.

For example, the uniform distribution on the interval [0,1] has probability density f(x) = 1 for 0 ≤ x ≤ 1 and zero elsewhere. The standard normal distribution has probability density

$f(x)={e^{-{x^2\over 2}}\over \sqrt{2\pi}}$.

If a random variable X is given and its distribution admits a probability density function f(x), then the expected value of X (if it exists) can be calculated as

$E(X)=\int x\,f(x)dx$

Not every probability distribution has a density function; for instance the distributions of discrete random variables do not. A distribution has a density function if and only if its cumulative distribution function F(x) is absolutely continuous. In this case, F is almost everywhere differentiable, and its derivative can be used as probability density. If a probability distribution admits a density, then the probability of every one-point set {a} is zero. (It is a common mistake to think of f(a) as the probability of {a}, but this is incorrect; in fact, f(a) will often be bigger than 1.)

Two densities f and g for the same distribution can only differ on a set of Lebesgue measure zero.

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