# What does it exactly mean if a random variable follows a distribution

Imagine there's a random variable such as $$ε$$. Then we say that $$ε$$ is i.i.d and follows a normal distribution with mean $$0$$ and variance $$σ^2$$.

What does this mean? Is this not a variable anymore? Is this a function now? I see this in most books and such but I'm still unclear what exactly it means or what it does and etc.

In terms of regression, I know this variable is basically the random errors, but what does it mean if this vector of random errors follows a normal distribution?

I.I.D. means independent and identically distributed, so $$\epsilon$$ is a vector of component random variables with the same distribution.

The meaning of "A follows an X distribution" is equivalent to saying that it "has a distribution," which is to say that it is a random quantity that can be determined only in probability.

In the example of regression that you refer to, $$Y=f(X) + \epsilon; \epsilon \stackrel{i.i.d.}{\sim} N(0,\sigma^2)$$, so the response variable $$Y$$ is equal to some function of the independent $$X$$ on average, and errors are normally distributed with mean zero, i.e. the observed $$Y$$ is not exactly $$f(X)$$.

A random variable $$\varepsilon \sim \mathrm{N}(0,\sigma^2)$$ is not the kind of variable considered when thinking about function arguments or solving equations, but actually represents the outcome of a random experiment. (Mathematically rigorously, but not so important, one would say: it is a function mapping from a sample space into the space in which the random variable lives.)

How can this be understood? A probability measure, like $$\mathrm{N}(0,\sigma^2)$$ assigns values to sets, so-called events. In this case, the probability of $$\varepsilon$$ ending up in a set $$A$$ has probability $$\mathrm{N}(0,\sigma^2)(A) = \int_A \frac{1}{\sqrt{2\pi\sigma^2}}\exp\left(-\frac{1}{2\sigma^2} \|x \|^2 \right) \mathrm{d}x.$$ That means, if you repeatedly saw i.i.d. (independent and identically distributed) $$\varepsilon$$'s, they would (in the large data limit) on average end up in $$A$$, precisely $$\mathrm{N}(0,\sigma^2)(A)\cdot 100 \%$$ of the time.

• How is it not a random variable? It has a distribution, so it is a random variable. – Tim Apr 8 '19 at 21:28
• It is a random variable, but not a „variable“ how we typically understand it. – Jonas Apr 9 '19 at 4:31
• That is..? What do you mean by variable? – Tim Apr 9 '19 at 4:51
• Like an unknown value with respect to which we want to solve an equation, or the argument of a function taking values in a given set. A random variable is a measurable function from a probability space to a measurable space. To me, those are completely different concepts; even if they appear similar. – Jonas Apr 9 '19 at 5:31
• When you see "variable" mentioned in probability theory text, they usually mean "random variable". Also "variable" in statistics means something else then in algebra. So it would be best if you could edit and make it more precise what kind of "variable" the random variable is not. – Tim Apr 9 '19 at 6:53