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I have the following joint pdf:

$f(x,y)=0.5$ where $0 \leq|x|\leq|y|$, $0 \leq|y|\leq1$, and $0$ otherwise

The question is: are $X$ and $Y$ independent and uncorrelated?

I know that if $f(x)$*$f(y)$=$f(x,y)$, then $X$ and $Y$ are independent, and $Cov(x,y)$=$0$ means uncorrelated.

I found $f(x)$ using integral $0.5dy$ range($x$ to 1) + $0.5dy$ range(-1 to $x$) = 1 and

$f(y)$ using integral $0.5dx$ range($y$ to 0) + $0.5dx$ range(0 to $x$) = 0

So $1*0$ is not $0.5$, means $X$ and $Y$ are not independent and correlated.

But I have doubt in choosing the ranges of integral because of absolute value. Can anybody help with this?

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What is P(|x|>0.5 given |y|<0.3)? What is P(|x|>0.5)? – Henning Makholm Sep 28 '11 at 6:11
P(|x|>0.5)=1-F(0.5) and P(|x|>0.5 given |y|<0.3)=0? – Ab. Sep 28 '11 at 7:16
P(|x|>0.5)=1-F(0.5) and P(|x|>0.5 given |y|<0.3)=0? – Ab. Sep 28 '11 at 7:18
up vote 4 down vote accepted

First off, always draw a picture.


We certainly could do this problem by integration to get the marginal distributions $f_X$ and $f_Y$. Your integrals are not quite right, however, because of some sign ambiguities.

We have to set up the integrals to cover the right areas. It's easier to keep the limits in terms of the absolute values. For $y$ there is one interval where $x$ ranges from $-|y|$ to $|y|$. For $x$ there are two intervals, and hence two integrals. Abusing notation slightly, but hopefully in a clear way:

$$ \begin{align*} f_X(x) &= \int f(x,y) \text{ d}y = \int_{|x| < |y|} 0.5 \text{ d}y = \int_{y = -1}^{y = -|x|} 0.5 \text{ d}y + \int_{y = |x|}^{y = 1} 0.5 \text{ d}y= 1 - |x| \\ f_Y(y) &= \int f(x,y) \text{ d}x = \int_{|x| < |y|} 0.5 \text{ d}x = \int_{x=-|y|}^{x=|y|} 0.5 \text{ d}x = |y| \end{align*} $$

$f_X(x) f_Y(y) = (1 - |x|)|y| \neq f(x,y)$, so they are not independent.

However, we don't need to do the integrals to find this -- the condition that $f(x,y) = f_X(x) f_Y(y)$ means that slices of $f(x,y)$ at constant $x$ values must be scalar multiples of each other. By inspection this is not true: $f(0.5, y)$ is not a multiple of $f(0.6, y)$.

The covariance between x and y can be done by setting up integrals, but it can also readily be seen by symmetry to be 0.

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The system should allow one to give more than one vote for the initial comment (draw a picture). – André Nicolas Sep 28 '11 at 6:35

Once you realize that one should (always) draw a picture, the following might help:

Grand Principle 1: Include the conditions on the ranges in the densities as indicator functions.

In your case, the density $f_{(X,Y)}$ of $(X,Y)$ is defined for every $(x,y)$ in $\mathbb R\times\mathbb R$ by $$ f_{(X,Y)}(x,y)=\frac12\mathbf 1_{0\le |x|\le |y|}\mathbf 1_{0\le |y|\le 1}, $$ and the question is to know whether there exists some functions $g$ and $h$ defined on $\mathbb R$ such that, for every $(x,y)$ in $\mathbb R\times\mathbb R$, $$ f_{(X,Y)}(x,y)=g(x)h(y). $$ If this is so, you know that (1) $g$ is a multiple of the density of $X$, (2) $h$ is a multiple of the density of $Y$, and (3) $X$ and $Y$ are independent. Otherwise, $X$ and $Y$ are not independent.

Second remark:

Grand Principle 2: A way to disprove independence is to find a product $B_1\times B_2$ such that $f_{(X,Y)}$ is zero on $B_1\times B_2$ but neither zero on $B_1\times\mathbb R$ nor zero on $\mathbb R\times B_2$.

In other words, one is looking for a zone where $f_{(X,Y)}$ is zero but ought not to.

In your case, $B_1=(\frac12,1)$ and $B_2=(0,\frac12)$ would do. To see this, note that $f_{(X,Y)}$ is zero on $B_1\times B_2$ but that $B_1\times\mathbb R\supset(\frac12,\frac34)\times(\frac34,1)$ and $\mathbb R\times B_2\supset(0,\frac14)\times(\frac14,\frac12)$, which both have positive probability. Hence you are done.

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Very nice principles, showing both how to make it easier to set up the integrals, and how to make progress short of crassly doing the integrals. – wnoise Sep 28 '11 at 7:33
@wnoise, thanks. Grand Principle 1 also allows to prove independence automatically when independence there is. – Did Sep 28 '11 at 7:37
In our case, X and Y are not independent, thus should be correlated. But $Cov(x,y)=0$ (i.e uncorrelated), any illustrations for this case? Thanks – Ab. Sep 28 '11 at 7:39
@Ab. Is this another question? Then ask another question. Or better still, find the several pages on MSE where this question is already answered (and where it is explained why not independent, thus should be correlated is wrong). – Did Sep 28 '11 at 7:47
@Ab. If $(X, Y)$ is uniformly distributed on the unit disc, then $X$ and $Y$ are not independent (hint: Didier Piau's Grand Principle 2 can be used to prove this) but are uncorrelated since $E[X] = E[Y] = E[XY] = 0$ (hint: change to polar coordinates while evaluating the integrals). – Dilip Sarwate Sep 29 '11 at 3:17

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