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For a continuous random variable, the probability of any particular value is zero. Only by integrating over some range is a non-zero probability obtained.

The components of the Bayes theorem are random variables.

Thus, in the case of Bayes written for continuous data, say D is the height mass of something, why isn't the dedenominator $P(D)$ zero? The probability density of any particular value is zero (yes)?

$$ P(M|D) = \frac{ P(D|M) P(M) }{ P(D) } $$

I believe that in the Bayes formula we can assign D as the value of a single observation OR as a collection of observations (yes?). If so, in the case where the collection is a continuous interval, then $P(D)$ would be non-zero.

I am also aware that $P(D) = \int P(D|M) P(M) \, dM$. However this does not answer the question for me. Again consider the case where the data is the mass (or height). You can measure a particular mass, but the probability of that exact mass is zero.

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In the case where $D$ is a continuous random variable, the expression $P(D)$ refers to the probability density of $D$, instead of the probability mass.

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  • $\begingroup$ Right, but that is the source of the problejm (I believe). The probability density for a single value from a continuous distribution is zero. Am I wrong? $\endgroup$
    – basicidea
    Commented Apr 19, 2019 at 4:14
  • $\begingroup$ please distinguish probability mass from "mass" in the question. I am using mass (related to weight) as an example of a continuous quantity. $\endgroup$
    – basicidea
    Commented Apr 19, 2019 at 4:16
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    $\begingroup$ @basicidea "The probability density for a single value from a continuous distribution is $0$." No, that's not correct. The key word here is density. For example, if $X$ is uniformly distributed on $[0,1]$, the probability density of $X$ at $1/3$ is equal to $1$. $\endgroup$
    – littleO
    Commented Apr 19, 2019 at 6:23

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