Expected value of product of two uniformly distributed independent random variables

Question

What is E(XY), where X,Y ~ U(0,1)

i.e., Expected value of product of two uniformly distributed independent random variables? It is probably not 1/4? What is the exact value? how to derive the same? Thanks.

Answer 1

Since $X$ and $Y$ are independent, $$E(XY) = E(X) E(Y),$$ where $E(X) = E(Y)$ as $X$ and $Y$ share the same distribution. We have: \begin{align*} E(X) = \int_0^1 x f_X (x) \ dx = \int_0^1x \cdot 1 \; \ dx = \frac{x^2}{2} \Bigg|_0^1 = \frac{1}{2}. \end{align*} Hence, $$ E(XY) = \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{4}. $$ Alternatively, define the joint density, $$f_{X,Y} \begin{pmatrix} x \\ y \end{pmatrix}.$$ Since $X$ and $Y$ are independent, the densities multiply: $$f_{X,Y} \begin{pmatrix} x \\ y \end{pmatrix} = f_X (x) f_Y (y).$$ The densities of $X,Y \sim \text{Unif}(0,1)$ is $1$, so $$f_{X,Y} \begin{pmatrix} x \\ y \end{pmatrix} = 1.$$ Hence, by LOTUS: \begin{align*} E(XY) & = \int_0^1 \int_0^1 xy f_{X,Y} \begin{pmatrix} x \\ y \end{pmatrix} \ dx \ dy \\ & = \int_0^1 x \; dx \int_0^1 y \; dy \\ & = \frac{1}{2} \cdot \frac{1}{2} \\ & = \frac{1}{4}. \end{align*}