91Ó°ÊÓ

Probability simply refers to the likelihood of something occurring. We may talk about the probabilities of particular outcomes—how likely they are—when we're unclear about the result of an event. Statistics is the study of occurrences guided by probability.

With pdf \({\rm{f(x)}}\), the Expected Value (mean value) of a continuous random variable \({\rm{X}}\) is

\(E(X) = {\mu _X} = \int_{ - \infty }^\infty x \cdot f(x)dx.\)

By the definition we have

\(\begin{aligned}E(X) \cdot E(Y) &= \int_{ - \infty }^\infty x \cdot {f_X}(x)dx \cdot \int_{ - \infty }^\infty y \cdot {f_Y}(y)dx\\ &= \int_{ - \infty }^\infty {\int_{ - \infty }^\infty x } \cdot {f_X}(x) \cdot y \cdot {f_Y}(x)dxdy\\ &= \int_{ - \infty }^\infty {\int_{ - \infty }^\infty x } y \cdot {f_X}(x){f_Y}(x)dxdy\\\mathop &= \limits^{(1)} \int_{ - \infty }^\infty {\int_{ - \infty }^\infty x } y \cdot f(x,y)dxdy\\ &= E(XY)\end{aligned}\)

(1): the random variables are independent (see the proposition below).

Two random variables\({\rm{X}}\)and\({\rm{Y}}\)are independent if and only if

1., \({\rm{p(x,y) = }}{{\rm{p}}_{\rm{X}}}{\rm{(x) \times }}{{\rm{p}}_{\rm{Y}}}{\rm{(y)}}\)

for every \({\rm{(x,y)}}\) and when \({\rm{X}}\) and \({\rm{Y}}\) discrete rv's,

2. \({\rm{f(x,y) = }}{{\rm{f}}_{\rm{X}}}{\rm{(x) \times }}{{\rm{f}}_{\rm{Y}}}{\rm{(y)}}\),

for every \({\rm{(x,y)}}\) and when \({\rm{X}}\) and \({\rm{Y}}\) continuous rv's,

otherwise, they are dependent.

If you substitute integral with the sum, you get the proof for the discrete case.

The expected value of uniformly distributed independent random variables \({\rm{X}}\) and \({\rm{Y}}\) is\({\rm{L}}\), as stated in the answer to exercise\({\rm{25}}\).

\({\rm{E(XY) = E(X) \times E(Y) = L \times L = }}{{\rm{L}}^{\rm{2}}}\)

Therefore, the solution is \({\rm{E(XY) = }}{{\rm{L}}^{\rm{2}}}\).