Reasoning about random variables and instances 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?
$$Pr(2 > t) < t$$
 A: 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. 
A: 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 $X$.
