# Joint Density Question

So the problem is $$p(x,y) = 120xy(1-x-y)I \{x \geq0,y \geq0,x+y \leq 1 \}$$

Now that $$Z = Y - E(Y|X)$$ What is the correlation coefficient of $$Z$$ and $$X$$

So here I First tried to get $$E(Z)$$, which

$$E(Z) = E(Y) - E(E(Y|X)) = E(Y) - E(Y) = 0$$

Also $$E(Z^2) = E(Y^2 - 2YE(Y|X) - E(Y)^2) = E(Y^2) - E(Y)^2 = Var(Y)$$

Hence that

$$\sqrt{Var(Z)} = \sqrt{Var(Y)}$$

Also for covariance

$$Cov(X,Z) = E(XZ) - E(X)E(Z) = E(XY - XE(Y|X)) = E(XY) - E(X)E(Y) = Cov(X,Y)$$

Now that I got $$\rho_{XZ} = \rho_{XY} = \frac{Cov(X,Y)}{\sqrt{Var(X)}\sqrt{Var(Y)}}$$

Here now I tried to get $$E(X),E(Y),E(X^2)E(Y^2)$$

But the calculation gets really messy that I can't continue

$$P_X(X) = \int_{0}^{1-x}120xy(1-x-y)dy = 20(1-x)^3$$

$$E(X) = \int_{0}^{1-y}20x(1-x)^3dx = 10(1-y)^2-20(1-y)^3+15(1-y)^4-4(1-y)^5$$

$$E(X^2) = \int_{0}^{1-y}20x^2(1-x)^3dx = \frac{20}{3}(1-y)^3-15(1-y)^4+12(1-y)^5-\frac{20}{6}(1-y)^6$$

More it goes more messy it gets, thinking that I am doing something wrong, it will get more messy when I find $$E(X)^2$$

If you have any idea what I am doing wrong can you help??

• The expectations should be double integrals.$$E(X)=\iint_{\Bbb R^2} x p(x,y) \, dx\,dy$$$$E(X^2)=\iint_{\Bbb R^2} x^2 p(x,y) \, dx\,dy$$ Commented Apr 21, 2022 at 17:43
• It should be zero. $E(Z) = 0$ and $E(XZ) = E(XE(Z \mid X)) = 0$ since $E(Z \mid X) = 0.$ Commented Apr 21, 2022 at 17:43
• @user170231 not when I got $P_X(X)$ right? Thats for joint density, but I already got $P_X(X)$ Commented Apr 21, 2022 at 17:46
• @WilliamM. Oh I see!! thank you!! Commented Apr 21, 2022 at 17:47
• Using the PDF of $X$, your integral bounds are incorrect.$$E(X)=\int_0^1 P_X(x) \, dx$$$E(X)$ must be a constant. Commented Apr 21, 2022 at 17:59

Consider $$\langle X \rangle$$ the linear span of $$X$$ in $$\mathscr{L}^2,$$ and $$\mathrm{F}_X = \{\mathbf{E}(Z \mid X) \mid Z \in \mathscr{L}^2\},$$ which is also a vector space. Clearly $$\langle X \rangle \subset \mathrm{F}_X.$$ It is well known that $$Z \mapsto \mathbf{E}(Z \mid X)$$ is the orthogonal projection onto $$\mathrm{F}_X,$$ a fortiori $$Z - \mathbf{E}(Z \mid X) \perp \langle X \rangle.$$ The covariance between $$Z$$ and $$X$$ is just the inner product, so the covariance between $$Z - \mathbf{E}(Z \mid X)$$ and $$X$$ is zero.