# I Difficulty with distribution functions

1. Sep 17, 2016

### matthew9

My undergrad probability theory course just got to random variables and distribution functions. Up until this point, the material was very straightforward and I could understand what was being done, but I feel that I am just not seeing the jump between probability with sets and probability with random variables/distribution functions. From what I understand, the RV will replace the set allowing us to have a sort of shortcut in the calculations, but after that I am not sure. Any help would be greatly appreciated!

2. Sep 17, 2016

### Krylov

The set is always there. It is usually called the "sample space" and denoted by $\Omega$. The other elements of what is called a "probability space" are the collection of events, often denoted by $\mathcal{A}$, and the probability measure, often denoted by $\mathbb{P}$. In the simplest of cases (your course probably started there), the collection $\mathcal{A}$ is just the power set of $\Omega$ (i.e. the collection of all subsets of $\Omega$).

In this simplest case, a random variable is nothing but a function $X : \Omega \to \mathbb{R}$ (in which case $X$ is called a "continuous random variable") or a function $X : \Omega \to \mathbb{Z}$ (in which case $X$ is called a "discrete random variable"). You can now obtain events by taking inverse images of sets in the codomain of $X$, say $\mathbb{R}$. Indeed, for any $B \in \mathbb{R}$ the inverse image $X^{-1}(B)$ is a subset of $\Omega$ or, equivalently, an element of $\mathcal{A}$ and henceforth an event.

This way you can define a function $F : \mathbb{R} \to [0,1]$ by putting
$$F(x) := \mathbb{P}[X^{-1}((-\infty,x])] = \mathbb{P}(\{\omega \in \Omega\,:\, X(\omega) \le x\})$$
for all $x \in \mathbb{R}$. This function is conventionally called the "distribution function" of the random variable $X$. As you can see, the underlying set $\Omega$, its collection of events $\mathcal{A}$ and the probability measure $\mathbb{P}$ are all still very much present, but now in addition there is the random variable $X$ with its distribution function $F$.

3. Sep 17, 2016

### EnumaElish

A random variable is neither. It's a function, not a variable. And it is deterministic, not random.

4. Sep 17, 2016

### Krylov

Undoubtedly this will clarify a lot to the OP.

5. Sep 17, 2016

### EnumaElish

I think you explained it well in your reply. I'm just saying it's a total misnomer. Don't go looking for a variable that is random. Look for a deterministic function.

6. Sep 18, 2016

### Stephen Tashi

It isn't accurate to think that a random variable "replaces" "the set".

In most general scenario for a probability model, there is a set of elements Omega. We think of an single element in that set as a "point" or an "outcome" that cannot be further subdivided into in other points or outcomes. There is a "probability measure" mu defined on Omega that tells us the probability of certain subsets of Omega. It may not tell us the probability of each possible subset of Omega, but it tells us the probability of some subsets of Omega, including the probability of entire set and the probability of the null set. In most practical examples, mu can tell us the probability of a given single outcome.

The probability measure mu is an abstract concept. It doesn't specify any formula or procedure for find the probability of a given subset of Omega. Abstractly we can only think of mu as being specified by a perhaps infinite list of subsets and their associated probabilities.

When we have a probability model that uses a "random variable" , the random variable has a distribution function. The distribution function allows us to use a computational procedure to find the probability of certain subsets of Omega. The distribution function doesn't "replace" Omega. If you want to say the distribution function "replaces" something, you could say that it replaces the abstract implementation of mu as a big list with an implementation of mu by specific computations.

7. Sep 23, 2016

### Zafa Pi

Your absolutely right that the term random variable is a misnomer and is a function. But saying it isn't random is a bit misleading because random is not a term in probability theory and in fact is an undefined vague idea about reality. Same goes for deterministic.

Also Krylov's bit of sarcasm is uncalled for since his reply is very likely over the head of the OP. For an intro to probability Ω should be finite with proscribed probabilities. This is fact completely general (see Nelson's "Radical Elementary Probability Theory"),

8. Sep 23, 2016

### Krylov

Oh dear, do you think that is why he left after starting the thread and did not return so far?

9. Sep 23, 2016

### Zafa Pi

Look at post #1, what do you think. <<Mentor note: comment removed.>>
And BTW you said "In this simplest case, a random variable is nothing but a function X:Ω→R (in which case X is called a "continuous random variable")" which is false. A continuous r.v. is one that gives points probability 0.

Last edited by a moderator: Sep 23, 2016
10. Sep 23, 2016

### Krylov

<<Mentor note: reply to deleted comment removed>>
No this terminology is not false, see here, where this is explicitly stated as the second sentence in the second paragraph. Note that I was not talking about an absolutely continuous random variable.

Last edited by a moderator: Sep 23, 2016
11. Sep 23, 2016

### Orodruin

Staff Emeritus
The OP has not been back to since before any of the replies in this thread were posted so clearly anything the replies state have nothing to do with it.

12. Sep 23, 2016

### Zafa Pi

Two points: If one writes X: Ω→R that allows X to be constant, i.e. take on one value, whereas the wiki article requires the the range to be uncountable. However I really don't think this is an important issue The main thing is that I think the wiki article got it wrong. If the range of X is {0} U [1,2] where the probability of 0 is 1/2 and the probability of [1,2] is 1/2, but otherwise uniform, would you call X continuous in spite of it having uncountable range? Most sources don't call it continuous.
When I googled "continuous random variable probability" I found my definition. The idea behind a continuous r.v. is that the PDF is continuous (need not be absolutely continuous) which implies a point has probability 0.

13. Sep 26, 2016

### pwsnafu

It's a mixed random variable.

The existence of the Radon-Nikodym derivative requires that the distribution function to be absolutely continuous, this is a classic result in measure theory. Hence the term "absolutely continuous random variable".
For an example of a continuous random variable with no pdf, take the Cantor distribution.