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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 $[0, 64]$?

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    None of these is necessary, simply note that the distribution of $X$ is symmetric hence $$E(X^7)=E(X)=0$$ which implies that $$\mathrm{Cov}(X,Y)=E(XY)-E(X)E(Y)=E(X^7)-E(X)E(X^6)=0$$ which implies that $$\mathrm{Corr}(X,Y)=\frac{\mathrm{Cov}(X,Y)}{\sqrt{\mathrm{var}(X)\mathrm{var}(Y)}}=0$$ – 2017-02-15

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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.

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    How do I tell if $X$ and $Y$ are independent? – 2017-02-15
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    Independence is not needed. The COV formula is true for arbitrary r.v.'s. – 2017-02-15
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    By the way, if you knew that $X$ and $Y$ are independent, you would also know that they uncorrelated. Independence implies uncorrelatedness (but not vice versa). – 2017-02-15
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    How would I prove if $X$ and $Y$ are correlated or not? – 2017-02-15
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    Read first line of my answer. – 2017-02-15
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    Oh oops sorry I meant: How would I prove if $X$ and $Y$ are independent? – 2017-02-15
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    I have added this to my answer. – 2017-02-15
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    If X and Y are independent so are X^6 and Y which is not the case here! – 2017-02-15
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    @KaviRamaMurthy so essentially, $X$ and $Y$ are not independent because to find $Y$ you need to know $X$? – 2017-02-16
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$E(X) = \int_{-2}^2 x f_x(x)\, dx = 0$ 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*}

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    How did you get $E(Y)=64/7$? – 2017-02-15
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    $\int_{-2}^2 x^6 f_x(x) dx = \frac{1}{4}\left. \frac{x^7}{7} \right|_{-2}^2 = \frac{256}{28} = \frac{64}{7}$ – 2017-02-15