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

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

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

## Answers and Replies

Homework Helper
Dearly Missed
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
$\begin{array}{|c|c|c|c|} \hline x+y & 0 & 1 & 2 \\ \hline p(x+y) & .49 & .42 & .09 \\ \hline \end{array}$

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?

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.)

Homework Helper
Dearly Missed
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!

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

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

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
$\begin{array}{|c|c|c|c|} \hline x+y & 0 & 1 & 2 \\ \hline p(x+y) & .5 & .4 & .1 \\ \hline \end{array}$

not

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

Homework Helper
Dearly Missed
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).

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

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

Gahh I'm a moron. I think this is what it will be:

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

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

does this seem right now

Gold Member
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.

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Homework Helper
Gold Member
Gahh I'm a moron. I think this is what it will be:

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

Correct

x-y:
$\begin{array}{|c|c|c|c|} \hline x+y & -1 & 0 & 1 \\ \hline p(x+y) & .6 & .4 & 0 \\ \hline \end{array}$
Check your second one. Should start like:
$\begin{array}{|c|c|c|c|} \hline x\color{red} - y & -1 & 0 & 1 \\ \hline p(x\color{red} -y) & ? & ? & ? \\ \hline \end{array}$
and you need to check the probabilities.

Correct

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

Ahh sorry, meant to write:

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

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

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