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I think I don't get the probability density function. At least on uniform distribution.

There are infinitely many numbers between $0$ and $1$, in probability, I understand this means that the weight it assigns to individual points must necessarily be zero. For this reason, we represent a continuous distribution with a probability density function (pdf) such that the probability of seeing a value in a certain interval equals the integral of the density function over the interval.

Then I don't get why, in this book page 74, the density function for the uniform distribution is just:

def uniform_pdf(x):
    return 1 if x >= 0 and x < 1 else 0

Shouldn't it be:

def uniform_pdf(x):
    return x if x >= 0 and x < 1 else 0
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  • $\begingroup$ Using the second distribution definition violates the nature of the uniform distribution; any two intervals $(a,b)$ and $(c,d)$ of equal length inside $[0,1]$ must receive the same probability. $\endgroup$
    – WaveX
    Oct 24, 2019 at 23:40
  • $\begingroup$ With your pdf, is the probability of being below $\frac12$ equal to $\int\limits_0^{1/2} x \, dx = \frac18$ ? Or $\int\limits_0^{1/2} 1 \, dx = \frac12$ ? $\endgroup$
    – Henry
    Oct 24, 2019 at 23:41
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    $\begingroup$ You seem to be thinking of the distribution, rather than the density. $\endgroup$
    – saulspatz
    Oct 25, 2019 at 0:05

1 Answer 1

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As you write, " the probability of seeing a value in a certain interval equals the integral of the density function over the interval." For the uniform distribution, the probability of seeing a value in the interval $[a,b]$ ($0\le a\le b\le1$) is $b-a$, and this agrees with $\int_a^b1\,dx$, not with $\int_a^bx\,dx$.

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