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Here in the 4 pictures in the last answer, is the vertical axe the probability? I.e. it seems to me that it is somewhat unnormalized : it has the value 2 in the 2nd picture and 3 in the 3rd picture. In the l.h.s. of each picture they write that this vertical axe really should be a probability. So why it is not in the range $[0,1]$? Also, they write

Note that as soon as you see your first Tail after the 3rd flip, the prior probability of p is now 0 at p=1 - ie there is SOME probability of seeing Tail.

So the value at each point should be in the range $[0,1]$, but it is not!

user2925716
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2 Answers2

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These plots are PDFs, i.e. Probability Density Functions, which is used for dealing with continuous random variables. Specific values don't represent probabilities (e.g. $f(x_0)\neq P(X=x_0)$); instead they represent some kind of a measure of probability.

A final note: Instead of summing, you need to integrate the PDFs, in which you get $1$. Probability of a point is approximated as $P(X=x_0) \approx f(x_0)dx$, which is infinitely small, and need to be integrated for obtaining a proper sum.

gunes
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  • But they write "Note that as soon as you see your first Tail after the 3rd flip, the prior probability of p is now 0 at p=1 - ie there is SOME probability of seeing Tail." So they have a *value* (0) at a point (p=1)! Moreover they labeled the vertical axe as "prior prob (p)". I'm confused.If I integrate how do I obtain a value 0 at p=1 as they write in the displayed sentence in my OP? – user2925716 Dec 21 '18 at 13:53
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    If PDF is $0$ at some point, then that point has no probability volume, using the approximation above. But, this doesn't mean that the PDF is probability, right? – gunes Dec 21 '18 at 13:55
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These 4 plots show the probability densitiy functions of some distributions, therefore they do not need to be in the range of $[0,1]$. They are not cumulative distribution functions.

nope
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