Independent and Idetically Distributed...

[Sujeeva Fernando]

What is the real meanning of the following statement?
"X and Y are independent and identically distributed"
If one can explain by taking an examle, I really appreciate.

[its_me]

For 2 variables X & Y, it means X & Y are independant of each other
( ie.cov(X,Y)=0 ) And X & Y have the same pdf.
If X & Y are iid N(mu, sigma)
It means X & Y are mutually indept and X~N(mu,sigma) and Y~N(mu,sigma)

An example could be rolliing of an fair die & observing the outcomes

[Macroman]

Independent means that one outcome is independent of other outcomes. If we roll a die or toss a coin the probability of successive outcomes is independent of previous outcomes.

For a fair coin suppose 2 sequences of 5 tosses:

HTHHT

HHHHH

In each case the probability that the next toss is heads is 1/2, independent of the previous outcomes. Also the distribution of each toss is the same as all the others, each time we have 50% probability of heads and 50% probability of tails, that's identically distributed.

[Sujeeva Fernando]

Thank You very much...
If X and Y are iid, can one say P(X>Y)=P(X<=Y)=1/2?

If one can justify it is great.

[Gandalf]

If X and Y are independent and identically distributed, then P(X>Y) = P(X<Y) by symmetry.
If X and Y are continuous random variables, then P(X=Y) = 0.
Therefore, if X and Y are continuous random variables which are independent and identically distributed, then P(X>Y) = P(X<=Y) = 1/2 [since P(X<Y) + P(X>Y) + P(X=Y) = 1; P(X>Y) = P(X<Y); and P(X=Y) = 0]

if X and Y are discrete random variables which are independent and identically distributed, then P(X>Y) = P(X<=Y) <> 1/2. E.g., in its_me's dice example with two dice, it is equally likely which is larger when they are unequal, but they are equal with probability 1/6, so P(X>Y) = 5/12.

[jason.]

Originally Posted by its_me

For 2 variables X & Y, it means X & Y are independant of each other
( ie.cov(X,Y)=0 ) And X & Y have the same pdf.

Be careful here: It is true that if and are independent then but the converse is not true! For example, say that is with probability and with probability and set . Then but and are not independent.

[Kazodev]

This might be worth a read http://en.wikipedia.org/wiki/Statistical_independence (as is this http://en.wikipedia.org/wiki/Indepen...ndom_variables)

[Sujeeva Fernando]

Thank you all.....

Can any one prove that why the following statement is true?
(symmetrycity fact is not clear)

"If X and Y (continuous random variables) are independent and identically distributed, then P(X>Y) = P(X<Y) by symmetry".

[jason.]

Originally Posted by Sujeeva Fernando

Thank you all.....

Can any one prove that why the following statement is true?
(symmetrycity fact is not clear)

"If X and Y (continuous random variables) are independent and identically distributed, then P(X>Y) = P(X<Y) by symmetry".

and have the same distributions because and do too. If this isn't immediately obvious to you, you can compute the distribution of and using convolutions and you'll see that they are the same.

[Gandalf]

1/2 by symmetry just means you are doing to identical random experiment twice, independently. If the experiments are really independent and identical, half the time the first experiment will "win"; half the time the second will "win". Assuming they can't tie, isn't that obvious?

Mathematically, let f(x,y) be the joint density, which since independent must equal g(x)g(y) where g is the (identical) marginal density.

Assuming a and b are the lower and upper limits of the outcomes (which might be -infinity and infinity)

Pr(X>Y) =

Pr(Y>X) =

Either expression could equally well be written as
, so they must be equal.

We have P(X>Y) + P(Y>X) + P(X=Y) = 1; all possible outcomes have combined probability 1.
If (as is the case for continuous random variables) P(X=Y) = 0,
then P(X>Y) + P(Y>X) = 1.
Since, per above, they are equal, we have
P(X>Y) + P(X>Y) = 1,
so P(X>Y) = 1/2.
Since equal, P(Y>X) = 1/2.