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Sorry if this question is so basic, it hurts. I feel like not understanding this topic well enough is holding me back. If any of this language is wrong in some fundamental way, please correct me! Moving on..

Say I make a probabilistic statement about a random variable $X$ drawn from some unknown continuous distribution:

$$Pr(X > t) < t$$

Assume that statement holds for all $t > 0$.

Now let's say I draw a sample $Y$ from the distribution of $X$, but I don't tell you what $Y$ is. Is there any real difference between reasoning about $Y$ and reasoning about $X$? Is $Y$ a random variable? Does $X = Y$? Does the following implicitly hold?

$$Pr(Y > t) < t$$

Now let's say we do know the value of $Y$, e.g. $Y = 2$. Would probabilistic statements about random variable $Y$ still apply? Or would they go out the window with knowing the variable? Specifically, would the following hold?

$$Pr(2 > t) < t$$

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when you wrote the letter t twice in your first equation, was that intentional? for example: Pr(x > 5) < 5 if t = 5? – frogeyedpeas Jul 17 '13 at 2:56
Yes, that was intentional. – Sebastian Goodman Jul 17 '13 at 2:57
Yes it does imply that, but that conclusion is not really important to the questions I asked. – Sebastian Goodman Jul 17 '13 at 14:33
up vote 0 down vote accepted

Okay so if we select X randomly from a set Q such that any randomly selected X from Q generally follows the statistical rule:

Probability(X > t) < t

Then your question is: if we select some Y randomly from the set Q does Y also obey the same law?

Yes, because selecting an X and selecting a Y is the same thing if both of them are randomly selected and have no difference in how they are selected.

Now if you did know the value of Y ahead of time would it change things? Most certainly!

The way the formula works is as follows:

you have the statement Probability(X > t) < t

You first give it a value of t (Ex: t = 1/2)

And then you get some useful expression:

Probability(X > 1/2) < 1/2

If you now define the value of X...

Probability(X$_{defined}$ > 1/2) < 1/2

The statement above is meaningless since you can just compare the two values and get a definite yes or no whether X is greater than or less than t.

On the other hand let us say t is undefined and x is defined:

Probability(X$_{defined}$ > t) < t

You have a new probability law all together which concerns a new variable t and takes a constant value X

So I guess it depends on how you interpret the formula you gave to be honest.

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hopefully this makes sense! please let me know if it doesn't :) – frogeyedpeas Jul 17 '13 at 3:04
if your going to downvote please at least explain why... even if it is a one word clue – frogeyedpeas Jul 17 '13 at 3:52
Yes, your answer does make sense! Thank you! BTW, I was assuming that $t$ was just a number, but didn't mention that in my question. When you say that $Pr(X_{defined} > .5) < .5$ is meaningless, is it nonsense or is it just not useful? Also, I'd still like to know if it's semantically correct to say $X = Y$ if they are random variables drawn from the same distribution? – Sebastian Goodman Jul 17 '13 at 4:41
that particular expression is not useful since we already know what $x_{defined}$ is and rather than having probabilistic data we can do a single comparison to figure for sure if indeed it is greater than or less than .5 – frogeyedpeas Jul 20 '13 at 1:57
As far as X = Y, that really depends. If your doing a problem involving X and Y are two separate variables (though they "behave" similarly" then you need to keep the separate – frogeyedpeas Jul 20 '13 at 1:58

First, the continuous distribution that you have defined is strange. $Pr(X>t)<t$ for $t>0$ means that the random variable is always $0$ or less. There is no problem with this but it is unusual and not particularly helpful for the question that you have asked. It is equivalent to $Pr(X>0)=0$.

You speak of taking a "sample $Y$" from $X$. If $X$ is a random variable then $Y$ is $X$. If $X$ is a collection of iid random variables $\{X_1,\dots,X_n\}$ then what is true of the "family" is true of $Y$.

If you know that $Y=2$, then $Y$ is not a random variable except in the trivial sense that $Pr(Y=2)=1$ and $Pr(Y\ne2)=0$.

If $Y$ is $X$ then the same is true of $X$, as Julius Caesar said "Alea iacta est" - "The die is cast" - once a random event has occurred it is no longer described by a random variable. Like a quantum wave function, it collapses when observed.

If $Y$ is an instance of $X$ then there are some very limited observations that you can make about $X$, pretty much confined to $Y=2$ is in the domain of $X$, i.e. it is an outcome with a non-zero probability but you have no knowledge of the value of that probability, it could be $1$ it could be $0.5$ or $10^{-4567}$ right down to $\lim_{\epsilon\to 0}\epsilon$.

As you gather more observations, so that $Y$ is now a random variable that consists of a sample of $X$ then there is a very real difference between reasoning about $Y$ and reasoning about $X$. I refer you to the Central Limit Theorem. Without reexplaining what the Wikipedia article says, the distribution of $Y$ will be related to but very different from $X$. For example, if the prerequisites of the Classical CLT are satisfied, $Y$ will be normally distributed irrespective of the distribution of $X$.

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You are right that the statement I gave is strange and a more compact way of saying it is that $Pr(X > 0) = 0$. I didn't want anyone to get distracted by that, though. My questions have to do with the nature of random variables and not with that probabilistic statement. – Sebastian Goodman Jul 17 '13 at 14:32
Also, I thought the CLT was a convergence result having to do with sampling statistics like the sample mean over $iid$ random variables following a normal distribution as the sample size goes to $\infty$. How does this relate to $X$ and $Y$? – Sebastian Goodman Jul 17 '13 at 14:35
I believe that I was confused by your referring to $Y$ as a "sample". I will edit my answer. – Dale M Jul 18 '13 at 2:01

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