0
$\begingroup$

Let $(X_1,X_2)$ be a pair of uniformly distributed random variables with $S=[-1,1]\times[0,1]$. Are $X_1$ and $X_2$ independent? Are $X_1$ and $X_2$ correlated.

My approach for independence:

For independence I think what was has to be shown is that $$f(x_1,x_2)dx_1dx_2=f(x_1)dx_1f(x_2)dx_2$$ where for $X_1$ the density is $$f(x_1)dx_1=\frac{1}{1-(-1)}dx_1=\frac{1}{2}dx_1$$ and for $X_2$ the density is $$f(x_2)dx_2\frac{1}{1-0}dx_2=dx_2.$$

I'm very confused on what $f(x_1,x_2)$ is.

My approach for correlation:

No approach yet. But I found out that two random variables are uncorrelated iff $cov[X_1, X_2]=0$. How can I show that for the above problem?

$\endgroup$
  • $\begingroup$ If you integrate the joint density it is 1. I think that the statement of the question is that they are jointly uniform, which means that the joint pdf is uniform, in this case the jpdf is 1/2 in the region. The phrase in the region is important, because you can write that as the product of two indicator functions, one that asks whether x is in the region. And one that asks if y is in the region. Thus it is seperable and independent. Independence implies uncorrelated. I believe you assumed they were each uniform and then you dont know about their relationship. But they are jointly uniform. $\endgroup$ – Mark Nov 30 '19 at 17:30
1
$\begingroup$

As far as I understand your question we have $f_{X_1, \, X_2}(x_1, x_2) = \mathbf{1}_{[-1, 1] \times [0, 1]}(x_1, x_2) \, \cdot \frac{1}{2}$. We can now get $f_{X_1}$ and $f_{X_2}$ up to a constant by integrating over the other random variable. $$f_{X_1} = \int_\mathbb{R} f_{X_1, X_2}(x_1, x_2) \, dx_2 = \int_\mathbb{R} \mathbf{1}_{[-1, 1] \times [0, 1]}(x_1, x_2) \, \cdot \frac{1}{2} \, dx_2 = \mathbf{1}_{[-1, 1]} \cdot \frac{1}{2} \int_0^1 \, dx_2 = \mathbf{1}_{[-1, 1]} \cdot \frac{1}{2}$$ You can do the same with $X_2$ and use your right idea about the splitting up of the density functions. Regarding correlation check again how independence and correlation are connected and how your result about independence might help you here already.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Yes! This was exactly what I was looking for. Took me 10 hours to get this integrating stuff but I'm there now! $\endgroup$ – Marc Dec 5 '19 at 14:40

Not the answer you're looking for? Browse other questions tagged or ask your own question.