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- #1

- Thread starter oyth94
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- Thread starter
- #1

- May 12, 2013

- 84

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.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??

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?

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- #3

Yes that is what I meant! I tried the question for X+Y and got 0,1,2So 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?

For X-Y I got -1,0,1

Why do I do next?

- May 12, 2013

- 84

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 $X-Y=b$.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, 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?