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#1
06-15-2007, 12:13 AM
 Sujeeva Fernando Member Join Date: Jun 2007 Posts: 141
Independent and Idetically Distributed...

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.
#2
06-15-2007, 12:31 AM
 its_me Member Join Date: Jan 2003 Location: AO Posts: 1,770

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
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#3
06-15-2007, 12:33 AM
 Macroman Member Join Date: Sep 2001 Location: you know where Favorite beer: Mountain Dew Posts: 3,692

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.
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#4
06-15-2007, 12:41 AM
 Sujeeva Fernando Member Join Date: Jun 2007 Posts: 141
Further...

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.
#5
06-15-2007, 06:46 AM
 Gandalf Site Supporter Site Supporter SOA Join Date: Nov 2001 Location: Middle Earth Posts: 26,452

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.
#6
06-15-2007, 09:05 AM
 jason. Member Join Date: Feb 2007 Studying for CFA II Favorite beer: Samuel Smith Old Brewery Pale Ale Posts: 888

Quote:
 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 $X$ and $Y$ are independent then ${\rm Cov}[X,Y] = 0$ but the converse is not true! For example, say that $X$ is $-1$ with probability $1/2$ and $+1$ with probability $1/2$ and set $Y=X^2$. Then ${\rm Cov}[X,Y] = 0$ but $X$ and $Y$ are not independent.
#7
06-15-2007, 09:10 AM
 Kazodev Member SOA Join Date: May 2004 Posts: 3,393

This might be worth a read http://en.wikipedia.org/wiki/Statistical_independence (as is this http://en.wikipedia.org/wiki/Indepen...ndom_variables)
#8
06-16-2007, 10:11 AM
 Sujeeva Fernando Member Join Date: Jun 2007 Posts: 141
Symmetry..

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".
#9
06-16-2007, 10:18 AM
 jason. Member Join Date: Feb 2007 Studying for CFA II Favorite beer: Samuel Smith Old Brewery Pale Ale Posts: 888

Quote:
 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
$X-Y$ and $Y-X$ have the same distributions because $X$ and $Y$ do too. If this isn't immediately obvious to you, you can compute the distribution of $X-Y$ and $Y-X$ using convolutions and you'll see that they are the same.
#10
06-16-2007, 10:43 AM
 Gandalf Site Supporter Site Supporter SOA Join Date: Nov 2001 Location: Middle Earth Posts: 26,452

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) = $\int_a^b \int_a^x g(x)g(y) dy dx$

Pr(Y>X) = $\int_a^b \int_a^y g(x)g(y) dx dy$

Either expression could equally well be written as
$\int_a^b \int_a^s g(s)g(t) dt ds$, 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.

Last edited by Gandalf; 06-16-2007 at 10:50 AM.. Reason: Changed "distribution" to "density" to be more precise

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