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Probability density function

In mathematics, a probability density function (pdf) 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. Informally, a probability density function can be seen as a "smoothed out" version of a histogram: if one empirically measures values of a continuous 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 R → R 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.

Simplified explanation

In simple English, the probability density function is any function f(x) that describes the probability density in terms of the input variable x in a manner described below.

f(x) is greater than or equal to zero for all values of x
The total area under the graph is 1. Refer to equation below.

$\int_{-\infty}^\infty \,f(x)\,dx = 1$

The actual probability can then be calculated by taking the integral of the function f(x) by the integration interval of the input variable x.

For example: the variable x being within the interval 4.3 < x < 7.8 would have the actual probability of

$\Pr(4.3<x<7.8) = \int_{4.3}^{7.8} f(x)\,dx.$

Further details

For example, the continuous 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

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

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

$\operatorname{E}(X)=\int_{-\infty}^\infty x\,f(x)\,dx.$

Not every probability distribution has a density function: the distributions of discrete random variables do not; nor does the Cantor distribution, even though it has no discrete component, i.e., does not assign positive probability to any individual point.

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:

$\frac{d}{dx}F(x) = f(x).$

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 - consider a random variable with a uniform distribution between 0 and 1/2.

Two probability densities f and g represent the same probability distribution precisely if they differ only on a set of Lebesgue measure zero.

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