# Thread: independently and identically distributed random variables

1. ## independently and identically distributed random variables

Hi there,

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

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

Thanks
Casper

2. Check if they are uncorrelated. Uncorrelated normal random variables are independent of each other.

3. 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

4. $\displaystyle Y_1 + Y_2 - Y_3 - Y_4$ and $\displaystyle Y_1 - Y_2 + Y_3 -Y_4$ 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 $\displaystyle Cov(Y_1 + Y_2 - Y_3 - Y_4, Y_1 - Y_2 + Y_3 -Y_4)$ equal 0.

5. Originally Posted by cl85
$\displaystyle Y_1 + Y_2 - Y_3 - Y_4$ and $\displaystyle Y_1 - Y_2 + Y_3 -Y_4$ 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 $\displaystyle Cov(Y_1 + Y_2 - Y_3 - Y_4, Y_1 - Y_2 + Y_3 -Y_4)$ equal 0.
Hi there,
Before I continue, I got a stupid question to ask
is $\displaystyle Y_1 + Y_2 - Y_3 - Y_4 \sim N (0, 4 \sigma^2 )$ ? and

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

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

Thanks
Casper

6. let

$\displaystyle W_1 = Y_1 + Y_2 - Y_3 -Y_4 \sim N ( 0 , 4 \sigma^2 )$
$\displaystyle W_2 = Y_1 - Y_2 + Y_3 -Y_4 \sim N ( 0 , 4 \sigma^2 )$

when I calculate the

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

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

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

anyone else got an idea?

Thanks

Casper

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

$\displaystyle Y_1 - Y_2 + Y_3 -Y_4 \sim N (0, 4 \sigma^2 )$ ,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
$\displaystyle Cov(Y_1 + Y_2, Y_1 - Y_2)$
$\displaystyle = E[(Y_1 + Y_2 - E[Y_1 + Y_2])(Y_1 -Y_2 - E[Y_1 - Y_2])]$
$\displaystyle = E[\{(Y_1-E[Y_1]) + (Y_2 -E[Y_2])\}\{(Y_1-E[Y_1])-(Y_2-E[Y_2])\}]$
$\displaystyle =Var(Y_1) - Var(Y_2) - 2Cov(Y_1, Y_2)$
The trick here is to always rearrange terms so that the random variable is with its mean.

8. 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
$\displaystyle Cov(Y_1 + Y_2, Y_1 - Y_2)$
$\displaystyle = E[(Y_1 + Y_2 - E[Y_1 + Y_2])(Y_1 -Y_2 - E[Y_1 - Y_2])]$
$\displaystyle = E[\{(Y_1-E[Y_1]) + (Y_2 -E[Y_2])\}\{(Y_1-E[Y_1])-(Y_2-E[Y_2])\}]$
$\displaystyle =Var(Y_1) - Var(Y_2) - 2Cov(Y_1, Y_2)$
The trick here is to always rearrange terms so that the random variable is with its mean.
hmmmmmmmm
$\displaystyle Cov(Y_1 + Y_2, Y_1 - Y_2)=Var(Y_1) - Var(Y_2) - 2Cov(Y_1, Y_2)$

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

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

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

Casper

9. No! My example is not to give you a formula to use, it's to show you how to rearrange terms when doing covariance.