# Expectation problem

Let $X$ and $Y$ be continuous random variables with joint PDF $f(x,y) = kxy^2$ and support set $0<x$, $0<y$, $x+y<=1$.

a) Find the constant $k$. $k = 60$? This is what I got.

b) Find the marginal PDFs of $X$ and $Y$. (I know for $X$ you take the integral from $0$ to $\infty$ of $f(x,y) dy$ and vice versa for $Y$.)

c) Find $\mathbf E(X|Y = y)$ and $\mathbf E(Y|X =x)$.

d) Find the correlation coefficient, which is $\frac{\mathbf E(XY) - \mathbf E(X) \mathbf E(Y)}{\sqrt{\mathbf E(X^2)-\mathbf E(X)}\sqrt{\mathbf E(Y^2)-\mathbf E(Y)}}$.

e) Find $\mathbf P(Y<4X/5+2/3)$

I know how to do most of this but I don't know how to find the constant $k$? I got $k = 60$ but I'm not sure its correct? Help? Thank you in advance.

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Presumably your density is $kxy^2$ on some set, and $0$ elsewhere. If your post had specified where the density is $kxy^2$, the $k$ would not be hard to find. – André Nicolas Dec 15 '12 at 2:28

For the constant, we want the integral of the density over our region to be $1$. The region is the triangle with sides the axes and the line $x+y=1$.
So $x$ should go from $0$ to $1-y$, and then $y$ go from $0$ to $1$. Or else we want $y$ to go from $0$ to $1-x$, and $x$ then go from $0$ to $1$. It doesn't really matter all that much in this case. I'll do the first way, you can do it the second way. So we want $$\int_{y=0}^1\left(\int_{x=0}^{1-y} kxy^2\,dx\right)\,dy=1.$$ The first integral gives us $\frac{k}{2}(1`-y)^2y^2$. The second integration gives $\dfrac{k}{60}$. So to make the integral equal to $1$ we need $k=60$.