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What does it mean to use the probability density function of the gaussian distribution as a "probability"? the probability of any x value in a continuous range

What does it mean to use the probability density function of the gaussian distribution as a "probability"? the probability of any x value in a continuous range is infinitely small. Is it really using the probability of a value "as extreme or more" as x?

bstamour 9y ago
It doesn't make sense, in my opinion (though I could be ignorant). For classification, wouldn't it be better to stick to discrete probability mass functions?
- ilyashabeeb 9y ago
  The attributes are real-valued. You cannot choose to use discrete probability functions. You have to make use of probability density functions.
amallia 9y ago
If you have a bag of [1.0 3.1 5.2 7.8 7.8 7.9 8.1 8.2 9.9 10.1] what is the probability of picking 8.0? It is definitely higher then 1.1, isn't it? I hope this clarifies the pdf usage.
- nerfhammer 9y ago
  The probability is zero. pdf = slope of the cdf. The value of the pdf at a given point is not a probability, it's the instantaneous rate of change of a probability. You need to integrate the pdf over a range to get a probability. You could take the area under the pdf (i.e., integrate) for a window around a given x or use the area under the tail of the pdf past x (i.e., p-value).
- nerdponx 9y ago
  You have an interval of real numbers [0, 1]. What's the probability of picking 0.3639401?
  - nightcracker 9y ago
    If we're using a uniform random number generator in [0, 1] with 32-bit IEEE754 numbers, the closest to 0.3639401 is 2^-2 * (1 + 3823195*2^-23) = 0.3639400899410247802734375 The numbers directly before and after it are: 2^-2 * (1 + 3823194*2^-23) = 0.363940060138702392578125 2^-2 * (1 + 3823196*2^-23) = 0.36394011974334716796875 Giving a range of size 2.980232238769531250 * 10^-8 in which all numbers compare equal to 0.3639401 in IEEE754 32-bit floating point. And since we're looking at a domain of [0, 1], that's also immediately our probability. (I'm fully aware this isn't what you were asking, but I found it fun either way)
pedrosorio 9y ago
It is not a probability. https://en.wikipedia.org/wiki/Joint_probability_distribution#Mixed_case https://en.wikipedia.org/wiki/Joint_probability_distribution...
- nerfhammer 9y ago
  P(Y=y|X=x)fX(x) isn't that like p-value?
  - pedrosorio 9y ago
    It's a probability (conditional probability of the discrete rv Y given the continuous X=x) multiplied by the density of the continuous rv at X=x. It is the value of the joint density at X=x, Y=y. p-values are unrelated to this but are computed by looking at the cumulative distribution function (the integral of the density).

bstamour 9y ago

It doesn't make sense, in my opinion (though I could be ignorant). For classification, wouldn't it be better to stick to discrete probability mass functions?

ilyashabeeb 9y ago
The attributes are real-valued. You cannot choose to use discrete probability functions. You have to make use of probability density functions.

amallia 9y ago

If you have a bag of [1.0 3.1 5.2 7.8 7.8 7.9 8.1 8.2 9.9 10.1] what is the probability of picking 8.0? It is definitely higher then 1.1, isn't it? I hope this clarifies the pdf usage.

nerfhammer 9y ago
The probability is zero. pdf = slope of the cdf. The value of the pdf at a given point is not a probability, it's the instantaneous rate of change of a probability. You need to integrate the pdf over a range to get a probability. You could take the area under the pdf (i.e., integrate) for a window around a given x or use the area under the tail of the pdf past x (i.e., p-value).
nerdponx 9y ago
You have an interval of real numbers [0, 1]. What's the probability of picking 0.3639401?
- nightcracker 9y ago
  If we're using a uniform random number generator in [0, 1] with 32-bit IEEE754 numbers, the closest to 0.3639401 is 2^-2 * (1 + 3823195*2^-23) = 0.3639400899410247802734375 The numbers directly before and after it are: 2^-2 * (1 + 3823194*2^-23) = 0.363940060138702392578125 2^-2 * (1 + 3823196*2^-23) = 0.36394011974334716796875 Giving a range of size 2.980232238769531250 * 10^-8 in which all numbers compare equal to 0.3639401 in IEEE754 32-bit floating point. And since we're looking at a domain of [0, 1], that's also immediately our probability. (I'm fully aware this isn't what you were asking, but I found it fun either way)

pedrosorio 9y ago

It is not a probability. https://en.wikipedia.org/wiki/Joint_probability_distribution#Mixed_case https://en.wikipedia.org/wiki/Joint_probability_distribution...

nerfhammer 9y ago
P(Y=y|X=x)fX(x) isn't that like p-value?
- pedrosorio 9y ago
  It's a probability (conditional probability of the discrete rv Y given the continuous X=x) multiplied by the density of the continuous rv at X=x. It is the value of the joint density at X=x, Y=y. p-values are unrelated to this but are computed by looking at the cumulative distribution function (the integral of the density).