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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.

Brian M. Scott
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Karel
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2 Answers2

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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 $1$.

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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.

André Nicolas
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