9
$\begingroup$

This post says

A PDF is used to specify the probability of the random variable falling within a particular range of values, as opposed to taking on any one value.

Is it true?

this is the PDF of the standard normal distribution.

$$\varphi(x) = \frac{1}{\sqrt{2\pi}} e^{-x^2/2}$$

plug in x=0 into the formula above, I can get the probability of taking on one value.

Does that post mean the PDF could be used both for point and interval?

Improve this question
$\endgroup$
3
  • 2
    $\begingroup$ This answer by whuber goes into more detail of how to interpret the value of the PDF at a certain point. $\endgroup$
    – COOLSerdash
    Oct 5, 2019 at 13:11
  • $\begingroup$ Welcome to CV yaojp. I suspect notation may be playing a role in your perplexity: the $\varphi(x)$ function is the cumulative distribution function of the standard normal distribution which is explicitly an integral from $-\infty$ to $x$ so its non-zero probabilities must come from intervals. $\endgroup$
    – Alexis
    Oct 5, 2019 at 16:49
  • 1
    $\begingroup$ You can also interprete the density the following way: for a very small interval $[x-\epsilon, x+\epsilon]$ it holds: $P(X \in [x-\epsilon,x+\epsilon]) \approx 2\epsilon f(x)$ $\endgroup$
    – Sebastian
    Oct 6, 2019 at 9:21

2 Answers 2

Reset to default
24
$\begingroup$

The citation is true. When you plug $x=0$ to the PDF function, you do NOT get the probability of taking this particular value. The resulting number is probability density which is not a probability. The probability of taking exactly $x=0$ is zero (consider the infinite number of similarly-likely values in the tiny interval $x\in[0,10^{-100}]$).

To further convince yourself that this $\varphi(x)$ cannot be a probability, consider decreasing the standard deviation of your normal distribution from $\sigma = 1$ to $\sigma = \frac{1}{100}$. Now, $\varphi(0)=\frac{100}{\sqrt{2\pi}}$ - much more than one. Not a probability.

Improve this answer
$\endgroup$
4
  • $\begingroup$ Thanks for your answer. What is $x\in[0..10^{-100}]$ for? Is it discrete? Would you please give a set notation, something like $\{0, ..., 10^{-99}, 10^{-100}\}$? $\endgroup$
    – JJJohn
    Oct 5, 2019 at 4:07
  • $\begingroup$ Sorry about the sloppy notation. Continous, not discrete - all numbers between 0 and $10^{-100}$. The point is that if $\varphi(x)$ was the probability of taking one particular value, the total probability for this tiny interval would be infinity. $\endgroup$
    – Trisoloriansunscreen
    Oct 5, 2019 at 12:04
  • $\begingroup$ +1 Welcome to CV, @Trisoloriansunscreen. I am amused by your name, assuming as I do, that it references Cixin Liu's trilogy? $\endgroup$
    – Alexis
    Oct 5, 2019 at 16:47
  • 1
    $\begingroup$ @Alexis, correct :) $\endgroup$
    – Trisoloriansunscreen
    Oct 6, 2019 at 1:14
4
$\begingroup$

Elaborating a bit on Trisoloriansunscreen's answer: it's very much true that you only got a probability density function. I'd like to draw an analogy for you. Imagine you have a 3D object, say some complex spaceship, and you know the mass density at every point.

For example, some parts of the spaceship might contain water, which has a mass density of $997 \frac{\text{g}}{\text{l}}$. Does this already tell you anything about the mass of the whole spaceship? No, it does not! Precisely because you only know this value at a specific point. You've got no information on how much water there actually is. It might be $1\ \text{ml}$ or $1\ \text{l}$.

Now suppose you know the amount of water, let's say $2\ \text{l}$. By simple multiplication $997 \frac{\text{g}}{\text{l}} \cdot 2\ \text{l}$, you get roughly $1994\ \text{g}$. I would like to make the point that you just did integration in disguise! Consider the following picture:

2D Cartesian coordinate system showing "Amount of water" (x-axis) vs "Mass Density" (y-axis) with a horizontal line at 997 g/l and a designated interval of 2l at the x-axis. The area of the "curve" (=horizontal line) is shadowed and amounts to the multiplication given before.

The mass you calculated is just the greenly shaded rectangular area. This was only doable as a simple multiplication because the mass density was constant for the amount of water considered and thus yielded a rectangular area.

What if you had mixed forms of water, e.g. some gaseous, some liquid, some in varying temperatures and so on? It could look like this:

Similar picture as before, but now mass density is not a constant function, but some arbitrary continuous function "going up and going down".

Now for computing the mass you would need to integrate that mass density function over the amount of water. Do you see the parallel to probability density functions now? To get an actual probability (cf. mass) you need to integrate the probability density (cf. mass density) over some domain.

Improve this answer
$\endgroup$
1
  • 3
    $\begingroup$ @Downvoter: Could you please include constructive feedback in a comment? :) $\endgroup$
    – ComFreek
    Oct 6, 2019 at 12:28

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.