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??

  • $\begingroup$ The expectations should be double integrals.$$E(X)=\iint_{\Bbb R^2} x p(x,y) \, dx\,dy$$$$E(X^2)=\iint_{\Bbb R^2} x^2 p(x,y) \, dx\,dy$$ $\endgroup$
    – user170231
    Commented Apr 21, 2022 at 17:43
  • 1
    $\begingroup$ It should be zero. $E(Z) = 0$ and $E(XZ) = E(XE(Z \mid X)) = 0$ since $E(Z \mid X) = 0.$ $\endgroup$
    – William M.
    Commented Apr 21, 2022 at 17:43
  • $\begingroup$ @user170231 not when I got $P_X(X)$ right? Thats for joint density, but I already got $P_X(X)$ $\endgroup$ Commented Apr 21, 2022 at 17:46
  • $\begingroup$ @WilliamM. Oh I see!! thank you!! $\endgroup$ Commented Apr 21, 2022 at 17:47
  • 1
    $\begingroup$ 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. $\endgroup$
    – user170231
    Commented Apr 21, 2022 at 17:59

1 Answer 1


Here is a no calculation approach.

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.


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