00:01
Okay, so we're gonna first assume that x, y are independent, and then prove that covariance of x, y is equal to 0.
00:14
Okay, so we can express the covariance as the expectation of x times y minus expectation of x times expectation of y.
00:33
Okay, so let's calculate e of x, y first.
00:43
This is equal to x1, y1, p1, p2 plus x1, y2, p1 times 1 minus p2 plus x2, y1, 1 minus p1 times p2 plus, let's put it here, x2, y2, 1 minus p1 times 1 minus p2.
01:11
Okay, now let's calculate e of x, which is x1 p1 plus x2 1 -p1, and e of y is equal to y1 p2 plus y2 1 -p2.
01:34
So multiplying out e and e gives the four terms that is exactly the same as in e , y, plus x2 y1, 1 -p1, p2, plus x2 y2, 1 -p1, 1 -p2, so that this is equal to e , and therefore covariance of x, y which is equal to e minus e times e is equal to 0.
02:31
So okay, now we're gonna assume that covariance of is equal to 0, and then we're gonna show that that implies that x and y are independent...