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"Doing Math with Python: Use Programming to Explore Algebra, Statistics, Calculus, and More!" in chapter 7 says

A probability density function, P(x), expresses the probability of the value of a random variable being close to x, an arbitrary value.

Is it true? the pdf is just close to?

Is it reasonable to view this close as some kind of approximation? Is probability density function is precise or approximation?

I didn't see any indication of close in wikipedia on probability density function.

Consider the PDF of the standard normal distribution, $\varphi(x) = \frac{1}{\sqrt{2\pi}} e^{-x^2/2}$ at point $x = 0$, is equal to $\varphi(x) = \frac{1}{\sqrt{2\pi}}$.

Is the value of $\frac{1}{\sqrt{2\pi}}$ a precise or approximate value of the PDF of the standard normal distribution at point $x = 0$?

JJJohn
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2 Answers2

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The probability density function specified just that: the probability a given value arises in a random process governed by the probability density function. Nevertheless, the probability of any specific value $x$ arising (where that value is specified to arbitrary precision) is infinitesimal. Thus, in practice one has a measuring device with a finite resolution, or specifies a value selected to finite precision, and the smaller that range, the closer the measurements are to revealing the density at $x$.

So the ideal function describes infinitesimally small ranges (and thus infinitesimally improbable events), but in any experiment there is a finite range.

Just look at the pdf of a normal distribution and think about the above explanation.

enter image description here

Clear?

David G. Stork
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  • Thank you so much. Is the value of $\frac{1}{\sqrt{2\pi}}$ a precise or approximate value of the PDF of the standard normal distribution at point $x = 0$? – JJJohn Oct 01 '19 at 04:46
  • $\frac{1}{\sqrt{2}}$ is the precise value of the probability density at $x=0$. But the finite probability is $$\lim_{\Delta \to 0} \int\limits_{x - \Delta}^{x+\Delta} p(x)\ dx$$ – David G. Stork Oct 01 '19 at 05:25
  • Thanks for your comment! What is does the "finite probability" mean? Is it a canonical term? – JJJohn Oct 01 '19 at 07:28
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    "Finite probability" here colloquially means "non-infinitessimal probability." Given a normal distribution ${\cal N}(0,1)$, the probability of selecting the exact value $x = 1/3$ is infinitesimally small. The probability you select an $x$ in the range $[1/3 - \Delta, 1/3 + \Delta]$ for $\Delta > 0$ is a finite probability. Clear? – David G. Stork Oct 01 '19 at 17:51
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As you say, the pdf of the standard normal is $\varphi(x) = \frac{1}{\sqrt{2\pi}} e^{-x^2/2}$. When we evaluate this at $x=0$ the value is exactly $\frac 1{\sqrt{2\pi}}$. It comes from integrating $e^{-x^2/2}$ from $-\infty$ to $\infty$, getting $\sqrt{2 \pi}$ and using that as a normalizing factor. There is nothing approximate about it.

The chance of a random variable having the value exactly $0$ or any other number is $0$. We normally consider the chance the random variable is in some small range around some value. If we consider the range from $-\epsilon$ to $+\epsilon$, the chance a random variable is within that range is approximately $2\epsilon \cdot \frac 1{\sqrt {2 \pi}}$. This is approximate because $\varphi(x)$ is not constant over the interval. It decreases away from $0$. If $\epsilon$ is small, the decrease is small, and the approximation is good.

Ross Millikan
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