Are (X+Y) and (X-Y) independent?

[r0bHadz]

Homework Statement

joint distribution p(x,y):
\begin{array}{|c|c|c|c|}
\hline & \frac xy & 0 & 1 \\
\hline & 0 & .5 & .2 \\
\hline & 1 & .2 & .1 \\
\hline
\end{array}

Relevant Equations

compute the marginal pmf from joint pmf: for x sum all of the probablities in one column
for y: sum all probabilities in one row

test for independence: if x(n)y(n) = x,y(n) then for all n then it is independent

I have x(0) = .7 x(1) = .3, y(0)=.7 y(1) = .3

since x,y(0) = .5 =/= x(0)y(0), x(0)y(0) = .49, x and y are dependent

now I need to determine whether x and y are independent or not

for x+y
$\begin{array}{|c|c|c|c|}<br /> \hline x+y & 0 & 1 & 2 \\<br /> \hline p(x+y) & .49 & .42 & .09 \\<br /> <br /> \hline<br /> \end{array}$

but how can I possibly determine x-y? since the domain will be 0 1 and -1 how can I determine -1?

 

[Ray Vickson]

I have x(0) = .7 x(1) = .3, y(0)=.7 y(1) = .3

since x,y(0) = .5 =/= x(0)y(0), x(0)y(0) = .49, x and y are dependent

now I need to determine whether x and y are independent or not

for x+y
$\begin{array}{|c|c|c|c|}<br /> \hline x+y & 0 & 1 & 2 \\<br /> \hline p(x+y) & .49 & .42 & .09 \\<br /> <br /> \hline<br /> \end{array}$

but how can I possibly determine x-y? since the domain will be 0 1 and -1 how can I determine -1?

What is stopping you from computing 0-1?

 

[r0bHadz]

What is stopping you from computing 0-1?

I'm guessing I just take the negative of 1? I'm not sure, it seems to me, like for example, if this was rolling dice, that you can't have a -1. I'm probably just not understanding the concept though..

and how do I prove that X+Y and X-Y are independent anyways? I can't construct a joint distribution from marginal distributions (which X+Y and X-Y would be.)

 

[Ray Vickson]

I'm guessing I just take the negative of 1? I'm not sure, it seems to me, like for example, if this was rolling dice, that you can't have a -1. I'm probably just not understanding the concept though..

and how do I prove that X+Y and X-Y are independent anyways? I can't construct a joint distribution from marginal distributions (which X+Y and X-Y would be.)

You are over-thinking it. Just perform X-Y and let X and Y be whatever they want to be. If X > Y, X-Y will b e >0; if X < Y, X-Y will be < 0. That's all there is to it!

 

[r0bHadz]

Hmm

When computing X+Y, I have a domain of: 0 1 2

I can compute 0 by the product of Px(0)Py(0) = .7*.7 = .49 = X+Y(0)
I can compute 1 by Px(0)Py(1) + Px(1)Py(0) = 42/100 = .42 = X+Y(1)
I can compute 2 by Px(1)Py(1) = .3*.3 = .09 = X+Y(2)

Now X-Y has domain: -1, 0,1

How do I compute -1 given that my marginal probabilities are: Px(0) = .7, Px(1) = .3, Py(0) = .7, Py(1) = .3

There is no way to get -1 from 0+0, 0+1 or 1+1

 

[StoneTemplePython]

edit: I misread the post. Follow along with what Ray said.

 

[r0bHadz]

Hmm I think what I was doing wrong is computing X+Y and X-Y from the marginal distribution, when I should be computing it from the joint distribution

so for
:x+y
$\begin{array}{|c|c|c|c|}<br /> \hline x+y & 0 & 1 & 2 \\<br /> \hline p(x+y) & .5 & .4 & .1 \\<br /> <br /> \hline<br /> \end{array}$

not

for x+y
$\begin{array}{|c|c|c|c|}<br /> \hline x+y & 0 & 1 & 2 \\<br /> \hline p(x+y) & .49 & .42 & .09 \\<br /> <br /> \hline<br /> \end{array}$

 

[Ray Vickson]

Hmm

When computing X+Y, I have a domain of: 0 1 2

I can compute 0 by the product of Px(0)Py(0) = .7*.7 = .49 = X+Y(0)
I can compute 1 by Px(0)Py(1) + Px(1)Py(0) = 42/100 = .42 = X+Y(1)
I can compute 2 by Px(1)Py(1) = .3*.3 = .09 = X+Y(2)

Now X-Y has domain: -1, 0,1

How do I compute -1 given that my marginal probabilities are: Px(0) = .7, Px(1) = .3, Py(0) = .7, Py(1) = .3

There is no way to get -1 from 0+0, 0+1 or 1+1

Right: for X+Y you can never get -1. However, that was not the issue: you want to compute X-Y, and for that you can certainly get -1 (but you can never get 2).

 

[r0bHadz]

I see. So the table for X-Y will be:

for x+y
$\begin{array}{|c|c|c|c|}<br /> \hline x-y & -1 & 0 & 1 \\<br /> \hline p(x-y) & .2 & .6 & .2 \\<br /> <br /> \hline<br /> \end{array}$

 

[r0bHadz]

Gahh I'm a moron. I think this is what it will be:x+y:
$\begin{array}{|c|c|c|c|}<br /> \hline x+y & 0 & 1 & 2 \\<br /> \hline p(x+y) & .5 & .4 & .1 \\<br /> <br /> \hline<br /> \end{array}$

x-y:
$\begin{array}{|c|c|c|c|}<br /> \hline x+y & -1 & 0 & 1 \\<br /> \hline p(x+y) & .6 & .4 & 0 \\<br /> <br /> \hline<br /> \end{array}$

does this seem right now

 

[WWGD]

I'm guessing I just take the negative of 1? I'm not sure, it seems to me, like for example, if this was rolling dice, that you can't have a -1. /QUOTE]
No, but you can have the difference in the values of the faces equal -1. Or -2. The sums, differences are not modeling throws of dice.

 

[LCKurtz]

Gahh I'm a moron. I think this is what it will be:x+y:
$\begin{array}{|c|c|c|c|}<br /> \hline x+y & 0 & 1 & 2 \\<br /> \hline p(x+y) & .5 & .4 & .1 \\<br /> <br /> \hline<br /> \end{array}$

Correct

x-y:
$\begin{array}{|c|c|c|c|}<br /> \hline x+y & -1 & 0 & 1 \\<br /> \hline p(x+y) & .6 & .4 & 0 \\<br /> <br /> \hline<br /> \end{array}$

Check your second one. Should start like:
$\begin{array}{|c|c|c|c|}<br /> \hline x\color{red} - y & -1 & 0 & 1 \\<br /> \hline p(x\color{red} -y) & ? & ? & ? \\<br /> \hline<br /> \end{array}$
and you need to check the probabilities.

 

[r0bHadz]

CorrectCheck your second one. Should start like:
$\begin{array}{|c|c|c|c|}<br /> \hline x\color{red} - y & -1 & 0 & 1 \\<br /> \hline p(x\color{red} -y) & ? & ? & ? \\<br /> \hline<br /> \end{array}$
and you need to check the probabilities.

Ahh sorry, meant to write:

x-y:
$\begin{array}{|c|c|c|c|}<br /> \hline x-y & -1 & 0 & 1 \\<br /> \hline p(x-y) & .2 & .6 & .2 \\<br /> <br /> \hline<br /> \end{array}$

The probabilities add up to 1 so I think its correct? Does this seem right to you?

 

[r0bHadz]

Hmm interesting. Didn't see it from that perspective!

 

[FactChecker]

It's very important to understand the distribution of X-Y, but that is not really necessary for answering the question of the independence of X+Y and X-Y. Consider P(X+Y=1) and P(X+Y=1 | X-Y=0). What do their values tell you?
To prove independence, you need to consider all cases, but to prove dependence you only need to find one case where one variable changes the probability of the other.