Using X and Y Independence to prove expected value and covariance properties

Question

I am new to stats, and I am taking a high paced course for the summer, so I know this may seem trivial, but can someone explain to me or prove to me why

$$E(XY) = E(X)E(Y)$$

when X and Y are independent and why independence also implies that

$$ Cov(X,Y) = 0 $$

I just want to make sure that I have a foundational understanding of these.

Answer 1

Let us consider the particular case where the random variables $X$ and $Y$ are independent and absolutely continuous. Then we have that \begin{align*} \textbf{E}(XY) & = \iint_{\mathbb{R}^{2}}xyf_{X,Y}(x,y)\mathrm{d}x\mathrm{d}y\\\\ & = \iint_{\mathbb{R}^{2}}xyf_{X}(x)f_{Y}(y)\mathrm{d}x\mathrm{d}y\\\\ & = \int_{\mathbb{R}}xf_{X}(x)\mathrm{d}x\int_{\mathbb{R}}yf_{Y}(y)\mathrm{d}y\\\\ & = \textbf{E}(X)\textbf{E}(Y) \end{align*}

Moreover, according to the definition of Covariance, we also have that \begin{align*} \textbf{Cov}(X,Y) & = \textbf{E}[(X - \textbf{E}(X))(Y - \textbf{E}(Y))]\\\\ & = \textbf{E}(XY - X\textbf{E}(Y) - Y\textbf{E}(X) + \textbf{E}(X)\textbf{E}(Y))\\\\ & = \textbf{E}(XY) - \textbf{E}(X)\textbf{E}(Y) \end{align*}

Consequently, based on the assumption of independence, we conclude the covariance equals zero.

Hopefully this helps!