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Tags probability , randomness

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Old 18th June 2008, 12:35 PM   #1
mijopaalmc
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"A measurable function of a random variable is a random variable"?

The title of the thread is a frequently stated (often without a proof) theorem in probability theory an statitistics.

In another thread, someone pointed out to me that functions such as f(X)=0X+c and f(X)=X^0+c, where X is a random variable and c is a real number, are not random variables because they yield the same value for ever value of the random variable X.

Now, this seems sensible; however, it does seem strange that such a proof of such an important theorem in probability theory and statistics would have such an often overlooked, obvious counterexample.

Could someone help me reconcile the proof with the apparent counterexample?

Note: I would be happy to provide a more thorough explanation of my understanding of the basic concepts of probability theory. I did not want to make this post needlessly long though.
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Old 18th June 2008, 12:52 PM   #2
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Non-trivially, certainly functions gain a degree of predictability that random numbers do not have. For instance, X is a random real number

f(x) = x^2 is a random, non-negative real number
f(x) = x^(1/2) will significantly influence the distribution of random numbers in a predictable manner (i.e. it will look 'less random').

So I'm not so sure that the results of these functions have the same degree of randomness as the original function.
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Old 18th June 2008, 01:24 PM   #3
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Conventionally, if a function is declared as a function of another variable, e.g. f(X) in your example, it means the result of f(X) is dependent on the value of X.

Look at your examples. Neither of them are dependent on the value of X; they are both independent of X. Therefore, these functions are not functions of X as stated.
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Old 18th June 2008, 01:38 PM   #4
Steven Howard
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Originally Posted by mijopaalmc View Post
In another thread, someone pointed out to me that functions such as f(X)=0X+c and f(X)=X^0+c, where X is a random variable and c is a real number, are not random variables because they yield the same value for ever value of the random variable X.
When you talk about "measurable functions" and "random variables" you're into set theory, which is way over my head. But just from a naive, natural language standpoint, it seems to me that when people say "a function of a random variable" that's probably just shorthand for "a function whose value is dependent on the value of a random variable."

Last edited by Steven Howard; 18th June 2008 at 01:39 PM.
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Old 18th June 2008, 01:46 PM   #5
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A random variable is just a measurable function on a probability space. Even though a function roughly means something whose value depends on its input, functions taking constant values are allowed as functions. So you can have a random variable that is constant.

This sort of inclusion of trivial cases is almost universal in mathematics: without it you'd need exceptions all over the place. For example, if X and Y are two random variables on the same space, then X-Y is a random variable. But of course it might be constant even if X and Y are not.

ETA: To CoolSceptic - for just the same reason the function from the real numbers to the real numbers which assigns the value 27 to any input is regarded as a function by all mathematicians.

Last edited by Meridian; 18th June 2008 at 01:49 PM.
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Old 18th June 2008, 01:49 PM   #6
mijopaalmc
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Originally Posted by CoolSceptic View Post
Conventionally, if a function is declared as a function of another variable, e.g. f(X) in your example, it means the result of f(X) is dependent on the value of X.

Look at your examples. Neither of them are dependent on the value of X; they are both independent of X. Therefore, these functions are not functions of X as stated.
Except, as I understand it, there is a constant function that has the same value regardless of the independent. In this case, f maps every point in its domain to the same point in its codomain and is still a function because no more than one point in its codomain is mapped to by any single point in its domain.

Last edited by mijopaalmc; 18th June 2008 at 01:58 PM.
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Old 18th June 2008, 01:51 PM   #7
69dodge
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Originally Posted by mijopaalmc View Post
In another thread, someone pointed out to me that functions such as f(X)=0X+c and f(X)=X^0+c, where X is a random variable and c is a real number, are not random variables because they yield the same value for ever value of the random variable X.

Now, this seems sensible; however, it does seem strange that such a proof of such an important theorem in probability theory and statistics would have such an often overlooked, obvious counterexample.

Could someone help me reconcile the proof with the apparent counterexample?
The proof is using a definition of "random variable" according to which a constant is a random variable: it simply happens to take on one particular value with probability 1, and every other value with probability 0.

(And, on previewing my post, I see Meridian just said the same thing.)
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Old 18th June 2008, 02:05 PM   #8
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The proof states that the function has to be a Borel measurable function. So we need a mathematician to tell us whether a maping to a constant is a Borel measurable function.
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Old 18th June 2008, 02:07 PM   #9
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Originally Posted by Reality Check View Post
The proof states that the function has to be a Borel measurable function. So we need a mathematician to tell us whether a maping to a constant is a Borel measurable function.
One already did
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Old 18th June 2008, 02:54 PM   #10
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Your posting states that the function taking constant values is a trivial case and I agree.
But it would be nice if the proof in the OP was addressed.
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Old 18th June 2008, 03:04 PM   #11
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Ah, I knew there was a reason I always avoided set theory like the plague.

RealityCheck, I think the point is not just that a topological mapping from some set of numbers to a constant is a valid (albeit trivial) function, but that a constant is a valid (albeit trivial) random variable.

It's all in the definitions, of course
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Old 18th June 2008, 03:40 PM   #12
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Originally Posted by Reality Check View Post
Your posting states that the function taking constant values is a trivial case and I agree.
But it would be nice if the proof in the OP was addressed.
I'm not quite sure what you mean: the question in the OP has been addressed. If you'd like a proof of the theorem - it proves itself once you know the definitions. Roughly speaking, a function $f:X\to Y$ is measurable if for every measurable subset $B$ of $Y$ its preimage under $f$, i.e., $f^{-1}(B)=\{x\in X:f(x)\in B\}$ is a measurable subset of $X$. What measurable means may be different for the two spaces: more formally you have to specify in advance a collection of measurable sets satisfying certain rules. If the spaces are the real numbers, the default collection is the Borel measurable sets (or the Lebesgue measurable sets).

Anyway, a constant function is always measurable, since the preimage of any set under a constant function is either the empty set or the whole space $X$, both of which are always measurable. For the theorem: a random variable $W$ simply means a measurable function from some probability space $\Omega$ to some other space $X$, usually the real numbers. Then as a function, the random variable $f(W)$ is just the composition $f\circ W$ of $W$ and$f$. Since the composition of two measurable functions is measurable (easy from the definition), $f$ measurable implies $f(W)$ measurable.
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Old 18th June 2008, 03:48 PM   #13
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Originally Posted by Meridian View Post
I'm not quite sure what you mean: the question in the OP has been addressed. If you'd like a proof of the theorem - it proves itself once you know the definitions. Roughly speaking, a function $f:X\to Y$ is measurable if for every measurable subset $B$ of $Y$ its preimage under $f$, i.e., $f^{-1}(B)=\{x\in X:f(x)\in B\}$ is a measurable subset of $X$. What measurable means may be different for the two spaces: more formally you have to specify in advance a collection of measurable sets satisfying certain rules. If the spaces are the real numbers, the default collection is the Borel measurable sets (or the Lebesgue measurable sets).

Anyway, a constant function is always measurable, since the preimage of any set under a constant function is either the empty set or the whole space $X$, both of which are always measurable. For the theorem: a random variable $W$ simply means a measurable function from some probability space $\Omega$ to some other space $X$, usually the real numbers. Then as a function, the random variable $f(W)$ is just the composition $f\circ W$ of $W$ and$f$. Since the composition of two measurable functions is measurable (easy from the definition), $f$ measurable implies $f(W)$ measurable.
<-------------- Your explanation



[my poor, aching head]
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Old 18th June 2008, 04:10 PM   #14
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For a function not to be measurable, it needs to be "very complicated". The reason why a constant might intuitively seem not to qualify as a random variable is that it's "too simple". But too simple is ok; too complicated is what causes trouble.

Regarding the definition of "y is a function of x" as "y depends on x", it's probably better to think of it as meaning "y doesn't depend on anything other than x". That way, it's clearer that a constant y satisfies the definition.
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Old 18th June 2008, 04:11 PM   #15
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Originally Posted by Meridian View Post
I'm not quite sure what you mean: the question in the OP has been addressed. If you'd like a proof of the theorem - it proves itself once you know the definitions. Roughly speaking, a function is measurable if for every measurable subset of its preimage under , i.e., is a measurable subset of . What measurable means may be different for the two spaces: more formally you have to specify in advance a collection of measurable sets satisfying certain rules. If the spaces are the real numbers, the default collection is the Borel measurable sets (or the Lebesgue measurable sets).

Anyway, a constant function is always measurable, since the preimage of any set under a constant function is either the empty set or the whole space , both of which are always measurable. For the theorem: a random variable simply means a measurable function from some probability space to some other space , usually the real numbers. Then as a function, the random variable is just the composition of and. Since the composition of two measurable functions is measurable (easy from the definition), measurable implies measurable.
That is what I was looking for - a proof that a constant function is always measurable. Thus a constant function is a Borel measurable function, the proof in the OP includes it and it is not a counterexample.

That should satisfy mijopaalmc.
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Old 18th June 2008, 04:18 PM   #16
Meridian
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Originally Posted by GreyICE View Post
[my poor, aching head]
You're not missing much! I'm not sure why I posted it, since the whole thing is one of those things that's almost certainly only interesting if you are a mathematician. The whole business with measurable sets and measurable functions arises because when you try to make some common sense stuff work mathematically, you run into contradictions, so it turns out you need some restrictions on the definitions. It's very much like needing the axioms of set theory to avoid the `set of all sets not belonging to themselves' paradox.

In fact, any function you can possibly conceive of is measurable. In practice, a random variable is just a function on a probability space, assigning a number, e.g., number of heads, to each possible outcome. So a function of a random variable is still a random variable. A constant function is a function, or a constant random variable (e.g., number of heads + number of tails, if you toss 3 coins) is a random variable, for the same reason that 0 is an integer or a square is a rectangle.
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Old 18th June 2008, 04:18 PM   #17
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Originally Posted by Reality Check View Post
Your posting states that the function taking constant values is a trivial case and I agree.
But it would be nice if the proof in the OP was addressed.
When I read this thread, I think what's needed is to take a step back and explain the ideas behind the whole stuff. It's 20 years since I've looked at either probability theory or measure theory, so please correct me if I goof up on some of the stuff.

In probability theory, the basic notion is a stochast X which "records" your events. If your event is something like the throw of a dice, you can speak of the chance you throw a 5: P[X = 5], or another value. However, when your event can have an arbitrary real value - say, you take someone's length - then it's senseless to speak of P[X = 1.83] as that chance is 0. Then you have a density function f(x) to capture the relative chance of getting that value - relative to all other values - and to really calculate the chance for some set of values to be hit, you take the integral. The usual distribution function taken is its primitive, so

$$F(x) = \int_{-\infty}^{x} f(t) dt$$

which gives the chance P[X < x] that your event records a smaller value than x. So the issue of being able to work with a stochast is that the density function is integrable. And given that it is integrable, you may then next pose the question for which sets S you want to know the chance that your stochast "hits" the set:

$$P[X \in S] = \int_{t \in S} f(t) dt$$

That's the kind of look at integrals you're probably not used to from high school mathematics, and that's where measure theory comes in. Measure theory is about setting up a system of subsets -- called "measurable sets -- of a given set for which it makes sense to define integrals. So the above formula makes sense if S is a measurable set. Think in the first place for S of sets like intervals [1.5, 3.14] -- that's where measure theory on real numbers starts with -- but also of (arbitrary) unions and intersections of those.

In particular, measure theory assigns to every measurable set a "size" of the set (just integrate the constant function 1 over the set).

When you look at the definition of random variable, you see it is in effect already a function from a probability space S1 to a measurable space S2. A measurable space is a set that is endowed with a system of measurable subsets; a probability space is a measurable space where the whole set has size 1: which is logical as the probability of any result at all is per definition 1.

Think of the space S1 of the possible events ("taking someone's length"), and S2 as their translation into some number; when you nitpick, they're two different spaces. The translation has to be a measurable function which means that the function behaves nice w.r.t. to the measure systems on S1 and S2, so the question P[X \in T], where T is a measurable subset of S2, can be calculated. The word "random" is maybe misleading; you could already assign to a random variable always the value 3, whatever the outcome of the event.

The theorem says that "taking a measurable function g of a random variable is again a random variable". If you look at the definitions, that just means that you compose the transformation function from the random variable with another (measurable) function, you again get a measurable function mapping from the original event space S1 now to a new space S3, and again you can calculate P[g(x) \in T] for any measurable subset T of S3.

So, in short, random variables are not really about randomness but more about being able to ask "what's the chance the event has a value in some reasonable set?", and measurable functions are about behaving nicely w.r.t. those reasonable sets.
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Old 18th June 2008, 04:23 PM   #18
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Originally Posted by Meridian View Post
In fact, any function you can possibly conceive of is measurable.
Such a statement is just an invitation to mess with people's heads

What about:

f(x) = 1 if x in T
f(x) = 0 if x not in T

where T is some non-measurable set? I forgot, is Q measurable?
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Old 18th June 2008, 04:30 PM   #19
Meridian
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Originally Posted by ddt View Post
Such a statement is just an invitation to mess with people's heads

What about:

f(x) = 1 if x in T
f(x) = 0 if x not in T

where T is some non-measurable set? I forgot, is Q measurable?
Yes - any set you can conceive of is measurable

In fact, I'm on pretty safe ground. It's impossible to explicitly construct a non (Lebesgue-)measurable set. To prove one exists you need to use the axiom of choice.
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Old 18th June 2008, 04:42 PM   #20
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Originally Posted by Meridian View Post
Yes - any set you can conceive of is measurable

In fact, I'm on pretty safe ground. It's impossible to explicitly construct a non (Lebesgue-)measurable set. To prove one exists you need to use the axiom of choice.
Thanks. Of course, I had to whip out my undergrad lecture notes to verify your answer. You can't trust just some guy on the internet, can you .

The style of your answer suggests you don't adhere to the axiom of choice? You're missing out then on a lot of mathematics .

I'm pro-choice
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Old 18th June 2008, 04:50 PM   #21
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No - I've nothing against the axiom of choice. But while I can prove non-measurable sets exist, needing AC to construct them makes the individual sets pretty hard to imagine!
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Old 18th June 2008, 05:39 PM   #22
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Originally Posted by Meridian View Post
No - I've nothing against the axiom of choice. But while I can prove non-measurable sets exist, needing AC to construct them makes the individual sets pretty hard to imagine!
But a true mathematician can perform magic uncountably many things in just the blink of an eye!

When it comes to imagining sets, try to imagine which points on a line are rational, which are algebraic, or try to imagine the Cantor set .
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Old 18th June 2008, 05:55 PM   #23
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Originally Posted by ddt View Post
But a true mathematician can perform magic uncountably many things in just the blink of an eye!

When it comes to imagining sets, try to imagine which points on a line are rational, which are algebraic, or try to imagine the Cantor set .
I just looked up "Cantor Set" on Wikipedia.

I'm sticking to engineering.
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Old 19th June 2008, 01:11 AM   #24
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Originally Posted by mijopaalmc View Post
In another thread, someone pointed out to me that functions such as f(X)=0X+c and f(X)=X^0+c, where X is a random variable and c is a real number, are not random variables because they yield the same value for ever value of the random variable X.
If I'm the person you're talking about then you're incorrectly characterizing what I said. You should look back through the transcript. I never said that the output of the function would not also be considered a random variable under some technical definition. What I maintained, which is what I had maintained throughout the discussion, was whether a system is modeled with random variables or not is independent of whether the system itself is random in any meaningful sense of the word.

The fact that constant functions can be functions of random variables proves the point. We have a function of a random variable and yet it models a deterministic system.

ETA: The whole discussion was in the context of your assertions that a system that took a random variable as an input could always have any output. This is clearly not the case for a constant function.

Last edited by zosima; 19th June 2008 at 01:27 AM. Reason: phrasing errors.
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Old 19th June 2008, 07:10 AM   #25
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Originally Posted by zosima View Post
If I'm the person you're talking about then you're incorrectly characterizing what I said. You should look back through the transcript. I never said that the output of the function would not also be considered a random variable under some technical definition. What I maintained, which is what I had maintained throughout the discussion, was whether a system is modeled with random variables or not is independent of whether the system itself is random in any meaningful sense of the word.

The fact that constant functions can be functions of random variables proves the point. We have a function of a random variable and yet it models a deterministic system.

ETA: The whole discussion was in the context of your assertions that a system that took a random variable as an input could always have any output. This is clearly not the case for a constant function.
The discussion on Borel-measurable function of random variables is posted below:

Originally Posted by mijopaalmc View Post
OK, I realize that there is a valid distinction between the process itself and the mathematical models that describe it and I agree that "random" and "non-random" may only meaningfully describe the mathematical models. However, if you are going to describe a mathematical model with random portions, then mathematically speaking to whole system is random. This is a non-negotiable point as we are describing a mathematical process and mathematically speaking any Borel-measurable function (which includes all elementary functions learned from primary school to elementary vector calculus) of a random variable is itself a random variable and the mathematical definition of a random process is a family of random variable defined of the same probability space. This is also why I am often frustrated with articulett when she insist that definition of random I am using makes algebra random, because the variables in the kind of algebra she is talking are not the random variables of probability and statistics, because they are merely values that you plug into a function and not functions in their own right as random variables are.
Originally Posted by zosima View Post
Wrong, but please do try again.

f(x) = 0*x+c

Originally Posted by mijopaalmc View Post
Obviously, you don't know what a trivial case is.
Originally Posted by zosima View Post
Sure I do, it is a counter-example to a bad definition, that takes no effort. Now if I had to think about it my counter-example would be pathological, but still acceptable. Just a post ago this was 'non-negotiable' , but now you make exceptions for what you term trivial examples. This is just another way of saying, "you're wrong".



As to my petulance. I can't have you just making up mathematics as you go along, and then try to trick people into believing it by trying to sound authoritative. As to all the other silliness you chose to make up, I'm not worrying about it. You've given up on trying to have a discussion, and I wouldn't want to be overdressed to the party.

But just as a friendly warning I would recommend you avoid straying towards mathematics. You do remember that business with 'almost surely' where you demonstrated you didn't understand the cardinality of a set don't you?
Originally Posted by zosima View Post
Not only is it useless, it is also wrong. Mijo conceded my counter-example; going from this point being 'non-negotiable' to negotiable. Mijo's statement only holds true insofar as all the examples that disprove his definition are ignored.
Then again that is not so different than the strategy Mijo applies to all the points he makes. So I suppose it is not surprising.
Originally Posted by zosima View Post
Originally Posted by mijopaalmc View Post
I did not actually concede your "counter-example"; I just never adequately responded to it. [...]
#1 As far as I am concerned running away from a strong objection to a fallacious claim is concession, or perhaps somewhat worse than concession. This is particularly true for you insofar as you have a habit of showing complete amnesia to the points that 'sink your battleship'.

[...]

Originally Posted by mijopaalmc View Post
Your example is in fact a random variable, just as all measurable functions of random variables are. In essence, just because one event has a probability of 1 doesn't make the distribution non-random because other events still exist they just happen with a probability of zero.
#1 This is the 'almost surely' business... we went over this before and you still seem to be missing the point. This claim about the possibility existing with a vanishingly small probability is only true if the function I provided( f(x)=0*x+c) involves a random variable defined over a field with an uncountably large number of elements(like the reals or the complex numbers) if the distribution is defined over a finite field or a countably infinite field(like the rationals or the natural numbers) then the probability of the outcome is exactly 0. Thus my counter example is definitely correct if defined over a finite field, like the non-negative integers modulo 7.

#2 Since evolution involves a discrete number of individuals, it will never correspond to the idealized models of the reals or the complex numbers. It will always exhibit behavior over a finite field. So your objection above, specifically doesn't apply to evolution.

#3 This ignores the fact that a probability distribution over the reals with a dirac delta in it, is not random at all, but simply not-deterministic or not probable. If a model involves a significant random variable the term that is often used technically is stochastic. Stochastic is not freely interchangeable with random. Models with random components may be Stochastic, but not all Stochastic models have random components. The components may be probabilistic, but behave in ways that deviate significantly from random.

[...]
Could you show me where I made the claim (with in that discussion or the entire thread) "that a system that took a random variable as an input could always have any output"?

Note: I have edited the posts to keep them as on-topic for this thread as possible. I urge anyone who is interested to review the entire thread for the appropriate context.
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Old 19th June 2008, 10:17 AM   #26
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I should note in the quotes Mijo presented, what you saw me doing is talking a lot of smack. Which is generally what one does when they've made no progress in an argument for 700+ posts. Mijo seems to infer from my claim that 'he is wrong', that I made sort of specific mathematical claim.

Originally Posted by mijopaalmc View Post
Could you show me where I made the claim (with in that discussion or the entire thread) "that a system that took a random variable as an input could always have any output"?

Note: I have edited the posts to keep them as on-topic for this thread as possible. I urge anyone who is interested to review the entire thread for the appropriate context.
I thought you'd never ask.
Here are the highlights:
Originally Posted by mijopaalmc View Post
That is a distinction without a difference. Something that is non-random is deterministic and vice versa.
Originally Posted by mijopaalmc View Post
You, like sol invictus, are relying on uncertainty in the initial conditions to declare that my definition is meaningless. However, my point is that, in the case of a stochastic process, each possible value in the distribution of initial condition yields at least two distinct outcome, not just one, as would be the case with a deterministic system.
Originally Posted by mijopaalmc View Post
No, you didn't actually read what I wrote or what I referred to (which seems to be a huge problem with those who argue that evolution is non-random.

A sure event (e.g., hitting the dart board universe) is deterministic, because one, and only one, outcome exists). An almost sure event (e.g., hitting a specific point or line the dart board universe) is random, because strictly more than one outcome exists. You can never not hit the dart board universe (deterministic event), but you could get really lucky and hit a specific point or line on the dart board universe (random event)

Do try and actually read before you post.
Note that in the example above, when Mijo says 'strictly more than one event' he is including events that have exactly 0 probability in his conception.

Originally Posted by mijopaalmc View Post
zosima-
The problem is that statistical hypothesis testing (even in maximum parsimony methods, especially when the number of taxa is above 8) used in evolutionary biology has a much wider usage of random that the aforementioned one, and it is therefore inconsistent to state that evolution by natural selection is non-random while making the assumptions of randomness necessary to perform any number of statistical hypothesis tests.
Originally Posted by mijopaalmc View Post
Uh....the other definitions by which evolutionary biologists claim evolution is not random are not consistent with the understanding of randomness needed to meaningfully practice statistics and to use statistical analysis to demonstrate that evolution does occur.
The problem is that you are assuming that there is only one correct usage of 'random variable' and 'random process', that mathematical definitions are applicable to non-mathematical objects, and that anyone who uses the the word 'random' outside of a mathematical context is somehow being inconsistent. Something can be a random process in one sense and not a random process in another, with no contradiction. Moreover, something can be a random process and yet still not random. This confusion results from conflation of the sense of the term from one field to a field where it does not directly apply.

So it seems like we've finally resolved the 1000+ post evolution thread, in a completely different thread. The definition of random process you apply is only meaningful in the context of mathematics. Insofar as evolution is not a mathematical entity* you were incorrect to apply that definition to evolution. That probably explains why 30-40 people told you exactly that 30-40 times each. So I guess my statements that you were incorrect were not incorrect after all. I wouldn't actually recommend anyone spend that much time reading the evolution thread. It is an awful hive of skum and trollery, much like the Mos Eisely spaceport. (Unless you really feel like verifying our statements. You'll find I'm understating its awfulness)

*even though some aspects of it may involve mathematics.

Last edited by zosima; 19th June 2008 at 10:19 AM. Reason: punctuation
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Old 20th June 2008, 10:02 AM   #27
mijopaalmc
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wrong forum

Last edited by mijopaalmc; 20th June 2008 at 10:10 AM. Reason: decided question belonged in its own thread in Forum Management
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