TITLE: independent, identically distributed (IID) random variables
QUESTION [12 upvotes]: I am having trouble understanding IID random variables. I've tried reading http://scipp.ucsc.edu/~haber/ph116C/iid.pdf, http://www.math.ntu.edu.tw/~hchen/teaching/StatInference/notes/lecture32.pdf, and http://www-inst.eecs.berkeley.edu/%7Ecs70/sp13/notes/n17.sp13.pdf but I don't get it.
Would someone explain in simple terms what IID random variables are and give me an example?

REPLY [5 votes]: "Independent" means for any $x_i \in X$, $P(x_0, x_1,..., x_i) = \prod_0^i P(x_i)$ 
For example, toss 2 dice. Let $X_1$ be event of the first being {1, 2}, $X_2$ the first being {2, 3, 4} and $X_3$ the second being {6} and $X_4$ the second being {4, 6}. It's intuitive to conclude that $P(X_1 \cap X_2, X_3) = P(X_1 \cap X_2) P(X_3) = P(X_1=2) P(X_3)= P(X_2=2) P(X_3) = \frac{1}{6} \cdot \frac{1}{6}$, so are other combinations. 
However, synonyms of "identical" include "alike" and "equal". That's, the probability of every variable should be equal, or identical. In the above example $P(X_1) \neq P(X_2)$ or $\frac{2}{6} \neq \frac{3}{6}$. To make $X_1$ and $X_2$ be indentical variables, we can let $X_2$ be event of the second die being {1, 6} or the first being {4, 2} or other similar posibilities you like. Then we get $P(X_1) = P(X_2)=\frac{1}{3}$. 
"Independent but not identical" as shown in the above two instances, like any combinations of evens between two dice: $P(X_1)$ and $P(X_3)$ or $P(X_2)$ and $P(X_4)$ and etc. 
"Identical and independent", any event combinations between the two dice that have the same probability. For example $P(X_1=1)$ and $P(X_3=6)$ and etc.
"Identical but not independent", any two events for one die. For instance, $P(X_1=1)$ and $P(X_2=3)$. It is not independent since if the die shows you 1 you can not see a 3 at the same time, and hence the probability of seeing 1 and 3 at the same time while rolling one die is 0. 
"Not identical and not independent", any two events for one die with unequal probability, for instance $P(X_1)$ and $P(X_2)$ or $P(X_3)$ and $P(X_4)$, or $P(X_2=4)$ and $P(X_1)$ and etc.