Uniform Random Variable: Correlation and Independence

Question

Let X be a uniform random variable defined on the interval $(0,1)$. If $Y = 6X^2−6X+1$, compute the correlation of X and Y . Are X and Y independent? Are X and Y uncorrelated?

So my work is.

$F(X) = \int{1} dx = x$

$E(X) = \frac{a+b}{2} = \frac{1}{2}$

$F(Y) = \int 6X^2-6X+1 dx = 2x^3-3x^2+x$

So is $E(XY) = \int\int f(x)f(y)dxdy$?

Answer 1

$X$ is a uniform continuous random variable on the support $(0;1)$.

Some features of this are that it has a cummulative distribution function $F_X(x) = x\;\raise{0.25ex}\chi_{x\in(0;1)}$, a probability mass function of $f_X(x)=1\,\raise{0.25ex}\chi_{x\in(0;1)}$, an expectation $\mathsf E(X)= \frac 1 2$ and variance: $\mathsf E(X^2)-\mathsf E(X)^2 = \frac 1 {12}$

Indeed $\mathsf E(X^k) = \int_0^1 x^k\,f_X(x)\operatorname d x = \frac{1}{k+1}$ for all $\Re (k)>-1$

Use this.

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$$\begin{align} \mathsf E(X^k) & =\frac{1}{k+1} \\[2ex] \mathsf E(XY) & = \mathsf E(6X^3-6X^2+X) \\[1ex] & = 6\mathsf E(X^3)-6\mathsf E(X^2)+\mathsf E(X) \\[1ex] & = \\[2ex] \mathsf {Cov}(X,Y) & = \mathsf E(XY)-\mathsf E(X)\mathsf E(Y) \\[1ex] & = \mathsf E(6X^3-6X^2+X)-\mathsf E(X)\mathsf E(6X^2-6X+1) \\[1ex] & = 6\mathsf E(X^3)-6\mathsf E(X^2)+\mathsf E(X)-\mathsf E(X)\Big(6\mathsf E(X^2)-6\mathsf E(X)+\mathsf E(1)\Big) \\[1ex] & = \end{align}$$

Answer 2

That's too much work, you would rather do $$E[XY] = E[X(6X^2-6X+1)].$$ Then use the properties of expectation.

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For the covariance, I would proceed as follows: \begin{align*} \text{Cov}(X,Y) &= \text{Cov}(X,6X^2-6X+1)\\ &=6\text{Cov}(X,X^2)-6\text{Cov}(X,X)+\text{Cov}(X,1)\\ &=6\left[E[X^3]-E[X]E[X^2]\right]-6\text{Var}(X)+0\\ &=6\left[\int_0^1x\cdot x^3\,dx-\frac{1}{2}\left(\frac{1}{12}+\frac{1}{4}\right)\right]-\frac{1}{2}\\ &=6\left[\frac{1}{5}\cdot 1 -\frac{1}{2}\cdot \frac{4}{12}\right] -\frac{1}{2}\\ &=-\frac{3}{10}. \end{align*}