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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!

Justin
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  • 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 $0\leq y_1 \leq y_2, y_1+y_2\leq 2$. – Dilip Sarwate Aug 18 '14 at 21:47

2 Answers2

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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...

angryavian
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    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. – Dilip Sarwate Aug 18 '14 at 22:22
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Hint: For example, $P(Y_1 > 1 \ \text{ and}\ Y_2 < 1) = 0$ but $P(Y_1 > 1) P(Y_2 < 1)$ is not.

Robert Israel
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