# What is a constant?

Discussion in 'Mathematics and Physics' started by Papabravo, Apr 19, 2007.

1. ### PapabravoWell-Known Member

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I'm getting bored with the .99...~=1 thread so I thought I'd start another one with the potential for controversy. Here goes.
Code (text):

What is the difference between a constant k, and a random variable [B]K[/B]
with mean k and variance zero.

2. ### PommieWell-Known MemberMost Helpful Member

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I'm going to guess that they are the same unless the fact that a compiler would still calculate the random K is counted as a difference.

Mike.

3. ### PapabravoWell-Known Member

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Great, except this is the "Math & Physics" category, not the compiler construction and implementation category.

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5. ### RadioRonWell-Known Member

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if K happens to always have a value of k as you describe, then K-k=0. So the difference between them is 0. Isn't this kind of obvious?

6. ### PapabravoWell-Known Member

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I was using the word "difference" in a broader sense than "arithmetic difference". Look at the formal definition of a random variable and the formal definition of a constant. Can you still claim that there is no difference between them, despite arithmetic difference being zero?

7. ### RadioRonWell-Known Member

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Well, in general terms, and after reading the formal definitions, I find that a constant is simply an unchanging value, whereas a random variable is a function (wikipedia says "a random variable is a measurable function from a probability space to the real numbers"). So in general its a bit like comparing apples and oranges.

In any case, since this is the Math and Physics category, as you point out, what more need be said than: if the mean of K=k and variance=0 then K-k=0?

8. ### PapabravoWell-Known Member

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So does the random variable K, with mean k and variance zero belong to the set of all random variables or does it belong to the set of all constants? Which way you decide implies the answer to weather there is a difference between the two. Set membership does require a distinction. What do ya'll think?

Last edited: Apr 22, 2007
9. ### PommieWell-Known MemberMost Helpful Member

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I'll guess it belongs to constants because, randoms have to have variance!=0. Do I get a distinction and therefore become one of the members of the set?

10. ### tkbitsMember

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Does "mean" and "variance" have any meanings associated only with single values?

Because of the formulas, I've only known of their use to describe distributions, or samplings. These are collections of numbers, some of which may have the same value, but each number has a separate identity. The uniqueness of each instance of a value is usually associated with a set such as position, individual creature, or time.

Isn't a "random variable" a mathematical abstraction of distributions?

11. ### PapabravoWell-Known Member

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Well random variable is actually a misnomer. It is actually a mapping or function. It maps elements of a proabability space onto the real numbers. If the mapping maps every element of the space onto the same real number k, then we have mean k, and variance zero. There is nothing in the formal definition of a random variable that excludes this situation.

And thus the nature of the original question. With respect to actual physical constants as opposed to mathematical ones there is some debate about the nature of their constancy. The fine structure constant may have take different values in different epochs in the evolution of the universe.

Last edited: Apr 22, 2007
12. ### 3v0Coop Build CoordinatorForum Supporter

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If physical laws as we know them do not hold inside a black hole, then within the black hole constants would be different if they exist at all. If you include the interior of black holes (at least the super massive types) to be a part of our universe then you do not have to travel in time to see constants change.

Gets even more convoluted if you toss string theory into the mix. Do you suppose a traveler could use a constant such as the speed of light as an address to locate a specific universe/time ?

All very wishy washy.

13. ### PapabravoWell-Known Member

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The more you try to get your arms around questions like this the more the two things seem to morph into each other. Each having aspects of the other.

14. ### 3iMaJNew Member

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First you need to understand the definition of a random variable. I'll state it and then answer your question.

A random variable is a variable K whose values are random but have an associated probability distribution function. So, generally speaking K has a range of -inf < K < inf (for some types of distributions). This is true for continuous random values, but discrete random values have a finite set of K values that they can take on each of which have a probability associated with them. Drawing a card K from a deck of cards would be considered a discrete random variable.

Now on to your question, you're asking about a continuous random variable K with mean of k and variance of 0. Since our continous random variable has a variance of 0 and a mean of k means we have an impulse at location k with an associated probability of 1. Which means there is no difference between a constant k and a random variable K with mean k and variance 0.

However, I'm not convinced you actually meant variance 0, so I'll assume that you meant variance V and answer your question the way I intended. I'd like to point out when you constrain the variance to 0 that takes all the "randomness" away from the problem. It's technically not defined, but physically thats what it relates to.

Lets assume variance V and mean k. Lets lay out the 3 axioms of probability theory. Lets quickly define our entire sample space (the space that contains all possible values that K can take on as :delta: )

1. Pr(K) => 0
2. Prdelta: ) = 1
Lets quickly define A1UA2UA3U....UAn = :delta:

3. sum(Ai) over all i = 1

now the equivalent of continuous random variables

1. fk(k) = 0 (which means the probability of k is actually 0 )
2. sum(fk(k)) = 1
3. F(inf) = 1
4. F(-inf) = 0
5. Pr(k1 < K < k2) = F(k2) - F(k1), where F is called the cumulative distribution function.

So, whats the difference between constant k and random variable K with variance V? It means that Pr(k) = 0 which means its probabilistically impossible to get a random number generator (or anything else for that matter) to generate EXACTLY k, but as average an infinite number of random variables together (called an expectation) you'll get your constant k.

15. ### PapabravoWell-Known Member

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Problem is, a random variable is actually a function, not a variable. Regardless of any properties a function or a constant may have, they are different things.

16. ### 3iMaJNew Member

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No. A random variable is not a function, it is in fact a variable. A function maps a variable from the domain space to the range space. So in the context of our problem it maps our random variable in the sample space to the probability space.

A random variable isn't special just because it's random, its essentially the same as y = mx + b, you wouldn't call x the function would you? No. The function is f(x) it maps x > y.

A function is merely a bridge between two spaces. In what I said earlier, the probability density function mapped our random variable K to its associated probabilities. With out the function we wouldn't be able to analyze the random variable.

17. ### PapabravoWell-Known Member

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I think you're wrong. Check the following from the wikipedia article:

http://en.wikipedia.org/wiki/Random_variable

In probability theory, a random variable is a quantity whose values are random and to which a probability distribution is assigned. Formally, a random variable is a measurable function from a sample space to the measurable space of possible values of the variable. The formal definition of random variables places experiments involving real-valued outcomes firmly within the measure-theoretic framework and allows us to construct distribution functions of real-valued random variables.

So tell me Herr Doktor 3iMaJ, from which institution of higer learning did you receive your degree?

18. ### SceadwianBanned

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In the real world EVERY value has a random variable associated with it, especially in electronics, even if it's something as basic as the thermal noise floor. I think there is WAY too much patting of oneself on the back and searching for questions which other people are easy to mistake for simple problems which in fact are complex ones going on in this thread which make it USELESS because it won't result in a real world result aside from much stone throwing and glutting of various individuals degrees and intelligence quotients. Which serves no purpose other than reminding us that all humans have ego's which need padding every now and again.

19. ### 3iMaJNew Member

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Probability theory is used constantly in the "real world," so having a good understanding in the building blocks of it is very relevant. Debating is not always egotistical stone throwing as you point out, rather sometimes good for furthering the participant's understanding of the subject matter.

Without probability theory most if your "real world" things that you point out wouldn't work. Consider cell phones, routers, digital televisions, I dare to say you can even apply probability theory to analog filtering... (Random Processes, but that'll be for another thread).

It's just someone else's flavor of mathemathics, don't be so quick to disregard something that is a pillar in modern engineering.