Joint dist of X+Y and X-Y

[oyth94]

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??

 

[TheBigBadBen]

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?

 

[oyth94]

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?

 

[TheBigBadBen]

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?

 

[CellCoree]

The joint distribution of X+Y and X-Y can be found by first determining the values of X+Y and X-Y for all possible combinations of u1 and u2. Since X and Y are defined as indicator functions based on the values of u1 and u2, we can determine that X+Y and X-Y can take on the following values:

X+Y = 2 if u1 <= 1/2 and u2 <= 1/3
X+Y = 1 if u1 <= 1/2 and u2 > 1/3
X+Y = 1 if u1 > 1/2 and u2 <= 1/3
X+Y = 0 if u1 > 1/2 and u2 > 1/3

X-Y = 0 if u1 <= 1/2 and u2 <= 1/3
X-Y = 1 if u1 <= 1/2 and u2 > 1/3
X-Y = -1 if u1 > 1/2 and u2 <= 1/3
X-Y = 0 if u1 > 1/2 and u2 > 1/3

Since u1 and u2 are both uniformly distributed on the interval (0,1), we can see that the joint distribution of X+Y and X-Y is a discrete distribution, as it takes on a finite number of values.

To fully determine the joint distribution, we would need to know the probabilities associated with each of these values. This can be calculated by considering the probabilities of each event (u1 <= 1/2 and u2 <= 1/3, u1 <= 1/2 and u2 > 1/3, etc.) and using the properties of the uniform distribution.

In terms of whether the joint distribution is jointly discrete or jointly absolutely continuous, it would depend on the specific values of u1 and u2. If they are both randomly and independently sampled from the interval (0,1), then the joint distribution would be jointly absolutely continuous. However, if there is some relationship or dependency between u1 and u2, then the joint distribution may be jointly discrete.