# X,Y Random Variables, X+Y , 1 are independent?

• Sep 30th 2009, 09:40 AM
noname
X,Y Random Variables, X+Y , 1 are independent?
Hi!

I take my statistics exam two days ago and I'm thankful for your help :)

Only one exercise gave me trouble. It was:

X and Y are random variables, check if random variables X+Y , 1 are independent.

The professor tell us that we should have used an appropriate example to check the independence of the variables.

Does anyone know a way to solve this exercise?

Thanks

• Sep 30th 2009, 10:56 AM
novice
Quote:

Originally Posted by noname
Hi!

I take my statistics exam two days ago and I'm thankful for your help :)

Only one exercise gave me trouble. It was:

X and Y are random variables, check if random variables X+Y , 1 are independent.

The professor tell us that we should have used an appropriate example to check the independence of the variables.

Does anyone know a way to solve this exercise?

Thanks

To prove X and Y being independent random variables:

If $\displaystyle h(xi,yj)$= sigma $\displaystyle f(xi)*g(yj)$, then X and Y are independent;
otherwise, X and Y are not independent random variables.

where h(xi,yj) is the joint distribution and i and j being the subscripts
where f(xi) is the probability distribution of X, and
g(yj) the probability distribution of Y.

The joint distribution of X+Y is the sum of h(xi,yj), where X=xi and Y=yj

I do not quite get your question about 1 being independednt of X+Y as I have not heard about it.
If 1 being a member in the probability distribution, then there must be only one random variable in the sample space, since 0< or = P and < or = 1.
If this is the case then P(X) = 1 and P(Y) =1, and P(X)*P(Y) = h(xi,yi) = 1.
• Sep 30th 2009, 10:59 AM
matheagle
by a constant, I interprete that as P(W=1)=1, and that is independent of any other random variable
• Sep 30th 2009, 01:18 PM
noname

Quote:

Originally Posted by novice
If 1 being a member in the probability distribution, then there must be only one random variable in the sample space, since 0< or = P and < or = 1.
If this is the case then P(X) = 1 and P(Y) =1, and P(X)*P(Y) = h(xi,yi) = 1.

Sorry but I can't understand this. Could you please explain it?

Quote:

Originally Posted by matheagle
by a constant, I interprete that as P(W=1)=1, and that is independent of any other random variable

Then i should have this, shouldn't I?

P(Z=z,W=1)=P(Z=z)P(W=1)=P(Z=z)*1

How can I prove that P(Z=z,W=1)=P(Z=z)?
• Sep 30th 2009, 02:15 PM
novice
Quote:

Originally Posted by noname

Sorry but I can't understand this. Could you please explain it?

Then i should have this, shouldn't I?

P(Z=z,W=1)=P(Z=z)P(W=1)=P(Z=z)*1

How can I prove that P(Z=z,W=1)=P(Z=z)?

The hardest part of answering this question is that we do not know whether "1" is a random variable or a probability in the distribution in the sample space, but I will try to simplify my thinking here:

Suppose "1" is in the distribution, then the sample space S must look like this: S={a}, let say P(X=1)= P({a}) = 1.
Then P(Y=1)=P({a})=1.

P(X=1,Y=1)=1 , which is the P(X=1)P(Y=1)=1, which also proved X and Y being independent.
In this case, P(X+Y) = P({$\displaystyle \emptyset$})

I have never heard of anyone looking for the probability of non-existence, i.e. P({$\displaystyle \emptyset$}), which render this nonsense.

Suppose "1" is one of the random variables, and
if P(X+Y)P(1)=P(X+Y,1), then they are independent, but You must know what P(X+Y) and P(1).
Say P(X+Y)=0.5 and P(1) =0.3, if P(X+Y)P(1)=0.15, they are independent.
• Sep 30th 2009, 11:20 PM
noname
Now it's more clear. Thank you

The last thing you said resembles the hint of the professor.
He said that we should use an example like:

$\displaystyle cosx^{2}+senx^{2}=1$

and calculate the derivatives (maybe are there some properties of indipendent random variables involving derivatives?)
• Oct 1st 2009, 02:32 PM
novice
Quote:

Originally Posted by noname
Now it's more clear. Thank you

The last thing you said resembles the hint of the professor.
He said that we should use an example like:

$\displaystyle cosx^{2}+sinx^{2}=1$

and calculate the derivatives (maybe are there some properties of indipendent random variables involving derivatives?)

I hope you don't mind that I am guessing. I am thinking as follows, and hope that I am not thinking wishfully. If not, you might be able to use same idea to approach the correct solution:

Since $\displaystyle cosx^{2}+senx^{2}=1$ ---(eq1)is given,we can use triangle as representation
where the base is X(x), height is Y(x)=1-X(x), $\displaystyle \sqrt{X(x)^2+Y(x)^2}$ is hypotenuse, and $\displaystyle 0\leq x \leq 1$

let

$\displaystyle P(Y(x))=1-P(X(x))$ where $\displaystyle P(X(x)) = \cos x$and
$\displaystyle P(Y(x)) = 1-P(X(x)) = \sin x$

The derivative of eq1 is

$\displaystyle 2 \sin \theta \cos \theta -\sin \theta \cos \theta =0$
Since $\displaystyle 0\leq P \leq 1, \theta\leq \pi x/2$

Let

$\displaystyle P(X=x) = \cos \theta, P(Y=y) = \sin \theta$

P(X,Y) = P(Y) P(Y)= $\displaystyle \\sin \theta \cos \theta$
which shows X and Y are independent.

$\displaystyle X+Y= 1$ always.

$\displaystyle P(X+Y,1) = P(X,Y)\cap P(1) = P(X=0,Y=1)\cap P(Y=1)$ P(X,Y) is a proper subset of P(X+Y+1)

or

$\displaystyle P(X=1,Y=0)\cap P(X=1)= P(X=1,Y=0) or P(X=0,Y=1)$; Both P(Y=1) and P(X=1) are proper subsets of P(X+Y,1).

$\displaystyle P(X=1)=0, P(X=0)=1$
$\displaystyle P(Y=1)=1, P(Y=0)=0$

$\displaystyle P(X+Y,1) =P(X=1,Y=0)=P(X=1)*P(Y=0) = 0$
$\displaystyle P(X+Y,1) =P(X=0,Y=1)=P(X=0)*P(Y=1) = 0$

Consequently, X, Y, X+Y, and 1 are independent.