0
$\begingroup$

We're learning about independent random variables in the context of multivariate probability distributions and I just need some help with this one question.

If $f(y_1, y_2)=6y_1^2y_2$ when $0\leq y_1 \leq y_2, y_1+y_2\leq 2$ and $0$ elsewhere

Show that $Y_1$ and $Y_2$ are dependent random variables.

The real problem I'm having with this question I've realized, is that I don't really understand how to get the marginal densities of Y1 and Y2. If someone could walk me through that, it would be greatly appreciated!

$\endgroup$
  • $\begingroup$ Consider the last two paragraphs of this answer of mine which can be used to assert that $Y_1$ and $Y_2$ are dependent just by looking at the shape of the region $0\leq y_1 \leq y_2, y_1+y_2\leq 2$. $\endgroup$ – Dilip Sarwate Aug 18 '14 at 21:47
0
$\begingroup$

A general approach is to find the marginal distributions $$f_{Y_1}(y_1) = \int f(y_1,y_2)\mathop{dy_2},\quad f_{Y_2}(y_2) = \int f(y_1,y_2)\mathop{dy_1},$$ and show that $f_{Y_1}$ is not the same as the conditional distribution $$f_{Y_1 \mid Y_2=y_2}(y_1) = \frac{f(y_1,y_2)}{f_{Y_2}(y_2)}.$$

There might be a faster approach though...

$\endgroup$
  • 1
    $\begingroup$ A faster approach is to say that the support of the joint density is not a product set (simplest case: rectangle with sides parallel to the axes) and so the random variables are dependent. This is essentially Robert Israel's hint reduced to visual inspection. $\endgroup$ – Dilip Sarwate Aug 18 '14 at 22:22
0
$\begingroup$

Hint: For example, $P(Y_1 > 1 \ \text{ and}\ Y_2 < 1) = 0$ but $P(Y_1 > 1) P(Y_2 < 1)$ is not.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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