independently and identically distributed random variables

• Dec 3rd 2008, 08:41 AM
casperyc
independently and identically distributed random variables
Hi there,

Assume that $Y1, Y2, Y3$ and $Y4$ are independently and identically distributed $N(\mu, \sigma^2)$ random variables.

how do I show that $Y1 + Y2 - Y3 - Y4$and $Y1 - Y2 + Y3 - Y4$ are independent.

Thanks
Casper
• Dec 3rd 2008, 09:13 AM
cl85
Check if they are uncorrelated. Uncorrelated normal random variables are independent of each other.
• Dec 3rd 2008, 10:07 AM
casperyc
Quote:

Originally Posted by cl85
Check if they are uncorrelated. Uncorrelated normal random variables are independent of each other.

could you explain a bit more?

thanks
• Dec 3rd 2008, 10:21 AM
cl85
$Y_1 + Y_2 - Y_3 - Y_4$ and $Y_1 - Y_2 + Y_3 -Y_4$ are both linear combination of independent normal random variables and are hence normal. To check if two normal random variables are independent, we just need to check if they are uncorrelated, ie, their covariance $Cov(Y_1 + Y_2 - Y_3 - Y_4, Y_1 - Y_2 + Y_3 -Y_4)$ equal 0.
• Dec 3rd 2008, 11:28 AM
casperyc
Quote:

Originally Posted by cl85
$Y_1 + Y_2 - Y_3 - Y_4$ and $Y_1 - Y_2 + Y_3 -Y_4$ are both linear combination of independent normal random variables and are hence normal. To check if two normal random variables are independent, we just need to check if they are uncorrelated, ie, their covariance $Cov(Y_1 + Y_2 - Y_3 - Y_4, Y_1 - Y_2 + Y_3 -Y_4)$ equal 0.

Hi there,
Before I continue, I got a stupid question to ask
is $Y_1 + Y_2 - Y_3 - Y_4 \sim N (0, 4 \sigma^2 )$ ? and

$Y_1 - Y_2 + Y_3 -Y_4 \sim N (0, 4 \sigma^2 )$ ,too?

I know how to sum normal distribution but I forgot how to take the difference...

Thanks
Casper
• Dec 3rd 2008, 01:04 PM
casperyc
let

$
W_1 = Y_1 + Y_2 - Y_3 -Y_4 \sim N ( 0 , 4 \sigma^2 )
$

$
W_2 = Y_1 - Y_2 + Y_3 -Y_4 \sim N ( 0 , 4 \sigma^2 )
$

when I calculate the

$
Cov( W_1 , W_2 ) = E( W_1, W_2 ) - E( W_1 ) E ( W _ 2)
$

I cant find the $E( W_1, W_2 )$

how should I proceed? I dont think it's a good way to proceed....

anyone else got an idea?

Thanks

Casper
• Dec 3rd 2008, 02:13 PM
cl85
Quote:

Originally Posted by casperyc
Hi there,
Before I continue, I got a stupid question to ask
is $Y_1 + Y_2 - Y_3 - Y_4 \sim N (0, 4 \sigma^2 )$ ? and

$Y_1 - Y_2 + Y_3 -Y_4 \sim N (0, 4 \sigma^2 )$ ,too?

I know how to sum normal distribution but I forgot how to take the difference...

Thanks
Casper

Yes and Yes, but you do not need to know it to solve for the question.

I'll show through an example how to proceed
$Cov(Y_1 + Y_2, Y_1 - Y_2)$
$= E[(Y_1 + Y_2 - E[Y_1 + Y_2])(Y_1 -Y_2 - E[Y_1 - Y_2])]$
$= E[\{(Y_1-E[Y_1]) + (Y_2 -E[Y_2])\}\{(Y_1-E[Y_1])-(Y_2-E[Y_2])\}]$
$=Var(Y_1) - Var(Y_2) - 2Cov(Y_1, Y_2)$
The trick here is to always rearrange terms so that the random variable is with its mean.
• Dec 3rd 2008, 02:23 PM
casperyc
Quote:

Originally Posted by cl85
Yes and Yes, but you do not need to know it to solve for the question.

I'll show through an example how to proceed
$Cov(Y_1 + Y_2, Y_1 - Y_2)$
$= E[(Y_1 + Y_2 - E[Y_1 + Y_2])(Y_1 -Y_2 - E[Y_1 - Y_2])]$
$= E[\{(Y_1-E[Y_1]) + (Y_2 -E[Y_2])\}\{(Y_1-E[Y_1])-(Y_2-E[Y_2])\}]$
$=Var(Y_1) - Var(Y_2) - 2Cov(Y_1, Y_2)$
The trick here is to always rearrange terms so that the random variable is with its mean.

hmmmmmmmm
$Cov(Y_1 + Y_2, Y_1 - Y_2)=Var(Y_1) - Var(Y_2) - 2Cov(Y_1, Y_2)$

so in my case
$Cov(W_1 + W_2, W_1 - W_2)=Var(W_1) - Var(W_2) - 2Cov(W_1, W_2)$
to find $Cov(W_1, W_2) = \dfrac{Var(W_1) - Var(W_2) - Cov(W_1 + W_2, W_1 - W_2) }{2}$

and now i have to find $Cov(W_1 + W_2, W_1 - W_2)$ ??

dose that make it more complicated or I have misunderstood it again?

Casper
• Dec 3rd 2008, 02:41 PM
cl85
No! My example is not to give you a formula to use, it's to show you how to rearrange terms when doing covariance.