Proof:Two definitions of independence are equivalent

From CS2800 wiki
If [math]\href{/cs2800/wiki/index.php/Pr}{Pr}(E_2) \neq 0 [/math] then [math]\href{/cs2800/wiki/index.php/Pr}{Pr}(E_1 \href{/cs2800/wiki/index.php/%5Ccap}{\cap} E_2) = \href{/cs2800/wiki/index.php/Pr}{Pr}(E_1)\href{/cs2800/wiki/index.php/Pr}{Pr}(E_2) [/math] if and only if [math]\href{/cs2800/wiki/index.php/Pr}{Pr}(E_1 \href{/cs2800/wiki/index.php/%5Cmid}{\mid} E_2) = \href{/cs2800/wiki/index.php/Pr}{Pr}(E_1) [/math]
Proof:
By the definition of conditional probability, we have [math]\href{/cs2800/wiki/index.php/Pr}{Pr}(E_1 \href{/cs2800/wiki/index.php/%5Cmid}{\mid} E_2) = \href{/cs2800/wiki/index.php/Pr}{Pr}(E_1 \href{/cs2800/wiki/index.php/%5Ccap}{\cap} E_2) / \href{/cs2800/wiki/index.php/Pr}{Pr}(E_2) [/math]. Multiplying both sides by [math]\href{/cs2800/wiki/index.php/Pr}{Pr}(E_2) [/math] gives


[math]\href{/cs2800/wiki/index.php/Pr}{Pr}(E_1 \href{/cs2800/wiki/index.php/%5Cmid}{\mid} E_2)\href{/cs2800/wiki/index.php/Pr}{Pr}(E_2) = \href{/cs2800/wiki/index.php/Pr}{Pr}(E_1 \href{/cs2800/wiki/index.php/%5Ccap}{\cap} E_2) [/math].


If [math]\href{/cs2800/wiki/index.php/Pr}{Pr}(E_1 \href{/cs2800/wiki/index.php/%5Cmid}{\mid} E_2) = \href{/cs2800/wiki/index.php/Pr}{Pr}(E_1) [/math], then the above equation becomes [math]\href{/cs2800/wiki/index.php/Pr}{Pr}(E_1)\href{/cs2800/wiki/index.php/Pr}{Pr}(E_2) = \href{/cs2800/wiki/index.php/Pr}{Pr}(E_1 \href{/cs2800/wiki/index.php/%5Ccap}{\cap} E_2) [/math].

Conversely, if we assume [math]\href{/cs2800/wiki/index.php/Pr}{Pr}(E_1)\href{/cs2800/wiki/index.php/Pr}{Pr}(E_2) = \href{/cs2800/wiki/index.php/Pr}{Pr}(E_1 \href{/cs2800/wiki/index.php/%5Ccap}{\cap} E_2) [/math], then dividing both sides by [math]\href{/cs2800/wiki/index.php/Pr}{Pr}(E_2) [/math] gives [math]\href{/cs2800/wiki/index.php/Pr}{Pr}(E_1) = \href{/cs2800/wiki/index.php/Pr}{Pr}(E_1 \href{/cs2800/wiki/index.php/%5Cmid}{\mid} E_2) [/math].