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

1. ## 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

2. 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.

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

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?

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)?

5. 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.

6. 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?)

7. 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.