# Can a probability density function be used directly as probability function?

This might be something basic but it confuses me greatly.

I am reading a literature, where they use the probability density function of a Gaussian distribution, that is

$$f(x)=\frac{1}{\sigma\sqrt{2\pi}} e^{ -\frac{(x-\mu)^2}{2\sigma^2} }$$

directly as a probability function - that means,

$$p(x\mid\sigma^2,\mu)=\frac{1}{\sigma\sqrt{2\pi}} e^{ -\frac{(x-\mu)^2}{2\sigma^2} }\;.$$

However, from what I read elsewhere, probability density function cannot be used like that, because it can be bigger than 1.

So I am confused.

• It looks as if that is just their notation for the density function. – André Nicolas Jun 26 '13 at 1:12
• oh. so it's not actual probability, it just... looks like probability? – Karel Jun 26 '13 at 1:15
• I think $p$ is not a good name for a density function, the default if $f$ possibly with subscript, but I have seen $p$ before. – André Nicolas Jun 26 '13 at 1:17
• hm. no, further down, they specifically say that it's the probability (given those parameters). – Karel Jun 26 '13 at 1:18
• Well, then it is wrong. Maybe they mean that $p(x)\,dx$ is a probability. – André Nicolas Jun 26 '13 at 1:21

Unless the writer is making a mistake, the symbol $p$ is being used where it is more common to use $f$, possibly with subscript, as in $f_X$.
Some people use $p$, in the discrete case, for the probability mass function, and, in the continuous case, for the probability density function. In the continuous case, $p(x)$ is not a probability.
You are describing a likelihood function. So $L(\theta|x) = P(x| \theta)$ instead of $P(\theta|x)$. The integral of the likelihood function over its domain could be greater than $1$.