MHB How to Determine the Joint Distribution of X+Y and X-Y?

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To determine the joint distribution of X+Y and X-Y, where X and Y are defined based on uniform samples u1 and u2, the possible values for X+Y are 0, 1, and 2, while X-Y can take values -1, 0, and 1. The joint probability mass function (pmf) needs to be calculated for these combinations, such as the probability of X-Y being -1 and X+Y being 1, which is found to be 1/6. The discussion emphasizes the importance of clearly defining the variables and calculating the probabilities for each case to complete the joint distribution. Overall, the focus is on deriving the pmf for the specified combinations of X+Y and X-Y.
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Suppose u1 and U2 are a sample from Unif(0,1).
X= 1{u1<=1/2} , Y=1{u2<=1/3}
Get the joint dist of X+Y and X-Y
Is it jointly discrete or jointly abs cont?

How do info about finding the joint dist??
 
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oyth94 said:
Suppose u1 and U2 are a sample from Unif(0,1).
X= 1{u1<=1/2} , Y=1{u2<=1/3}
Get the joint dist of X+Y and X-Y
Is it jointly discrete or jointly abs cont?

How do info about finding the joint dist??

So first of all, you weren't very clear with your notation, so first I'd like to check that I understand what you mean.

We say that
$$
X = \begin{cases} 1 &\mbox{if } u_1 \leq \frac12 \\
0 & \mbox{otherwise } \end{cases}
$$
and
$$
Y = \begin{cases} 1 &\mbox{if } u_2 \leq \frac13 \\
0 & \mbox{otherwise } \end{cases}
$$
Where $u_1$ and $u_2$ are independently sampled from a uniform distribution over $(0,1)$. Your goal is to find the joint distribution of $X+Y$ and $X-Y$. Is that correct?
 
TheBigBadBen said:
So first of all, you weren't very clear with your notation, so first I'd like to check that I understand what you mean.

We say that
$$
X = \begin{cases} 1 &\mbox{if } u_1 \leq \frac12 \\
0 & \mbox{otherwise } \end{cases}
$$
and
$$
Y = \begin{cases} 1 &\mbox{if } u_2 \leq \frac13 \\
0 & \mbox{otherwise } \end{cases}
$$
Where $u_1$ and $u_2$ are independently sampled from a uniform distribution over $(0,1)$. Your goal is to find the joint distribution of $X+Y$ and $X-Y$. Is that correct?

Yes that is what I meant! I tried the question for X+Y and got 0,1,2
For X-Y I got -1,0,1
Why do I do next?
 
oyth94 said:
Yes that is what I meant! I tried the question for X+Y and got 0,1,2
For X-Y I got -1,0,1
Why do I do next?

So as you rightly stated, $X+Y$ can be 0,1, or 2, and $X-Y$ can be -1,0,1. What they're asking for now is a probability mass function (pmf) on these two variables, which I'll call $f(x,y)$. That is, for a given $(a,b)$, we need to be able to figure out the probability that $X+Y=a$ and $X-Y=b$.

So, for example: the probability that $X-Y=-1$ and $X+Y=1$ is $\frac12 \cdot \frac13 = \frac 16$, since this will only happen when $X=0$ (which has probability $\frac12$) and $Y=1$ (which has probability $\frac13$). So, we would say that for our joint pmf $f(x,y)$, we have $f(-1,1)=\frac16$. The value of our joint probability distribution at $(-1,1)$ is $\frac16$.

Does that help? Can you finish the problem from there?
 
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