# Why does Uniform distribution make sense?

This might be a dumb question, but I am suddenly confused on how to understand the PDF of a uniform distribution.

For instance, the PDF of standard uniform is always equal to 1... How is that possible? Shouldn't a PDF always integrate to 1? I'm lost on how to understand the uniform distribution intuitively (besides just solving for the PDF using the area of the rectangle).

Sorry if this is a silly question, but any explanation would be appreciated!

• Hint: what is the domain over which the PDF of the standard uniform distribution takes the value of 1? What would happen to the value of the PDF if a uniform distribution were to be defined over a different domain, given that it must integrate to a value of ` over its domain? – EdM May 25 '20 at 19:57
• @EdM The domain for standard uniform is [0,1], and anything outside the domain takes zero probability. But isn't the uniform distribution a continuous distribution? So wouldn't the standard uniform's PDF mean that infinitely many numbers (e.g. 0.21, 0.7, 0.999) occur with probability 1? – J. Doe May 25 '20 at 20:07
• The value of probability density function $f(x)$ can be considered the probability per unit measure around $x$. So for a continuous distribution the probability in a region $dx$ around $x$ is $f(x) dx$. The answer from @Tim works that through. You seem to be thinking about discrete distributions, with counting measure providing a finite probability at a single value of $x$. Introductory presentations usually use the terminology probability mass function for discrete distributions. – EdM May 25 '20 at 21:47

One of the simplest ways to numerically approximate integral (calculate the "area under the curve")

$$\int_a^b f(x) \, dx$$

is the quadrature, where you approximate the smooth function with a number of rectangular "bins", calculate their areas and sum them up, as shown on the image below taken from the linked Wikipedia article. Now recall that uniform distribution on $$(a, b)$$ is a kind of rectangle, so the "approximation" is in fact the exact solution. The quadrature approximates integral by

$$\int_a^b f(x) \, dx \approx (b-a) \, f(\tfrac{b-a}{2})$$

The area of the rectangle is width $$b-a$$ times height $$f(x)$$ for $$x = \tfrac{b-a}{2}$$, where in case of uniform random variable, $$f(x)$$ would be the same for any $$x$$ within the domain. For standard uniform this is $$1 \times 1 = 1$$.