High School Can PDF values be equal to zero at some given points?

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A function can be a probability density function (PDF) even if it reaches zero at certain points, as long as it meets the criteria of being non-negative and integrating to one over its entire range. Many well-known distributions, such as the gamma and exponential distributions, have density functions that are zero over sets of infinite measure. The existence of a Borel-measurable map that satisfies these conditions ensures that a corresponding probability measure can be defined. Additionally, while a density function may not be unique, it is unique almost everywhere. Therefore, a function that meets these criteria can indeed be a valid PDF.
Schwann
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Suppose we have a function which looks like this:
probability.jpg

It seems like it meets criteria of probability density functions: this function is asymptotic to zero as x approaches infinity and also it is not negative. My question is: if at some points this function reaches zero (as I have shown above), does that mean that in cannot be PDF?
 
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Many pdf functions have zeros.
 
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The only conditions for a function ##f(x)## to be a pdf are ##f(x)\ge 0## for all ##x## and ##\int_{-\infty}^\infty f(x)dx=1##.
 
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An example of a well-known absolute continuous distribution with density function that is zero on a set of infinite measure is the gamma-distribution and the exponential distribution (which is a special case of the former).
 
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Schwann said:
Suppose we have a function which looks like this:
View attachment 252880
It seems like it meets criteria of probability density functions: this function is asymptotic to zero as x approaches infinity and also it is not negative. My question is: if at some points this function reaches zero (as I have shown above), does that mean that in cannot be PDF?

If you have a Borel-measurable map ##f:\mathbb{R}\to [0,\infty[## such that ##\int_\mathbb{R} f =1 ##, then we get a measure

$$\mu(A) =\int_A f d\lambda, A \in \mathcal{B}(\mathbb{R})$$

and this is a probability distribution. One can even show that there is a probability space on which there exists a random variable with this distribution. More formally, there exists a probability space ##(\Omega, \mathcal{F},\mathbb{P})## and a random variable ##X: (\Omega, \mathcal{F})\to (\mathbb{R},\mathcal{B}(\mathbb{R}))## such that ##\mu=\mathbb{P}_X##.

So to anwer your question: such a function is certainly a density function of some random variable on some probability space.

Last remark: a density function ##f## need not be unique, but it is unique almost everywhere.
 
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The standard _A " operator" maps a Null Hypothesis Ho into a decision set { Do not reject:=1 and reject :=0}. In this sense ( HA)_A , makes no sense. Since H0, HA aren't exhaustive, can we find an alternative operator, _A' , so that ( H_A)_A' makes sense? Isn't Pearson Neyman related to this? Hope I'm making sense. Edit: I was motivated by a superficial similarity of the idea with double transposition of matrices M, with ## (M^{T})^{T}=M##, and just wanted to see if it made sense to talk...

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