# Joint Distribution, expected value correlation of a graphed triangle

Let X and Y be the coordinates of a point chosen uniformly at random from the triangle that joins the points (0,1), (0,0), and (1,0).

1) Find the joint distribution of X and Y.

2) Determine the expected value of X and the expected value of Y (these are the expected coordinates of a point chosen at random).

3) Find the correlation between X and Y.

4) If the original units were measured in inches, would there be a different correlation if the units were changed to centimeters? Justify your answer mathematically?

For 1, I got $f_{X,Y}(x,y) = 2$ and 2 I got $f_X(x) = 2-2y$ and $f_Y(y) = 2-2x$. Then for correlation in 3 I got $\frac{1}{12} - (4-4x-4y+4xy)$. I feel like I am forgetting something however and this should be different. For 4, I know this is unchanged but I am not sure how to show this. Any help is appreciated.

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Starting with 1, one cannot declare that the density is $f(x,y) = 2$. What you probably mean is that $f(x,y)=2\mathbf 1_T(x,y)$ for some well chosen domain $T$. –  Did Dec 13 '12 at 11:21

$1.$ For the joint density, you must specify it everywhere. This is not just a matter of fussiness, you are likely to get wrong answers if you don't. So say $f(x,y)=2$ on the triangle, and $f(x,y)$ is $0$ elsewhere.

$2.$ $E(x)=\iint_{\mathbb{R}^2} xf(x,y)\,dx\,dy$. Now to do the actual integration, use the fact that the density is $2$ on the triangle, and $0$ elsewhere. So $E(X)=\iint_T 2x\,dx\,dy$, where $T$ is our triangle.

For the actual integration, integrate first with respect to $y$, $y=0$ to $1-x$. Then integrate with respect to $x$, $x=0$ to $x=1$.

Of course by symmetry $E(Y)=E(X)$.

$3.$ You will need the covariance, which is $E(XY)-E(X)E(Y)$. For $E(XY)$, integrate $xyf(x,y)$ over the plane, so effectively $2xy$ over the triangle. You will also need the variance of $X$ and of $Y$. For the variance of $X$ you will need $E(X^2)$ as part of the calculation. Another integration!

$4.$ If the original units were inches, and they are switched to cm, we are dealing with new random variables $U$ and $V$, with $U=aX$, $V=aY$, where $a$ is about $2.54$.

Argue that the covariance of $U$ and $V$ is $a^2$ times the covariance of $X$ and $Y$. This is easy, just substitute in the formula for covariance. Argue in a similar way that $\sigma_U=a\sigma_X$, and similarly for $V$. When we divide to get the correlation, the $a^2$ cancel.

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