Show that $X$ and $Y$ are uncorrelated

Question

The question I'm given is: Suppose that $X$ has the uniform distribution on the interval $[-2,2]$ and $Y=X^6$. Show that $X$ and $Y$ are uncorrelated. Also are $X$ and $Y$ independent?

I know that $f_x(x)=1/4$, $E(X)=2$, $Var(X)=4/3$ and that $X=Y^(1/6)$.

What do I do from here? I'm thrown by the fact that $Y(-2)=64$ and $Y(2)=64$. Does this mean that Y is in $[0, 64]$?

Answer 1

$E(X) = \int_{-2}^2 x f_x(x)\, dx = 0$ and $E(Y) = E(X^6) = \int_{-2}^{2} x^6 f_x(x)\, dx = 64/7$. Thus \begin{align*} Cov(X,Y) &= E((X-E(X))(Y-E(Y))\\ &= E(X(X^6-\frac{64}{7})) \\ &= E(X^7 - \frac{64}{7}X)\\ &= \int_{-2}^2 \left(x^7 - \frac{64}{7}x\right)f_x(x) \, dx \\ &= 0 \end{align*}

Answer 2

Two r.v.'s $X,Y$ are uncorrelated if their covariance $COV (X,Y) = 0$.

Recall that $COV (XY) = E(XY) - E(X)E(Y)$. Moreover, when $X$ is uniform on $[-2,2]$ (and thus symmetric around $0$, then $E(X^k) = 0$ when $k$ is an odd integer.

Now you are ready to compute $$COV (XY) = E(XY) - E(X)E(Y) = E(X \cdot X^6) - E(X)E(X^6) = E(X^7)- E(X)E(X^6) =0$$ because $E(X)=E(X^7) = 0$.

[ADDED AFTER YOUR REQUEST in comments]

Two r.v.'s $X,Y$ are independent if $P(A_X \cap B_Y) = P(A_X) P(B_Y)$ for any event $A_X$ concerning $X$ and any event $B_Y$ concerning $Y$.

Let $A_X = \{ X \ge 1 \}$ and $B_Y = \{ Y \ge 1 \}$. Then $$P(A_X \cap B_Y) = P (X \ge 1, Y \ge 1) = P (X \ge 1) = \frac{1}{4}$$ but $$P(A_X) P (B_Y) = P (X \ge 1) P(Y \ge 1) = \frac{1}{4} \frac{1}{2} = \frac{1}{8}$$ Thus, $X$ and $Y$ are not independent.