00:01
In this problem, here xn is equal to summation of y n, where y i is equal to 0 or 1.
00:11
Probability of y is equal to 0 is q and probability of y is equal to 1 is p.
00:23
Now, in a part e x n plus 1, divided by en.
00:33
Here, e n includes y1, y2, yn.
00:45
That is, it is e summation of y, i, y n plus 1, divided by yn, y2, yn, that is e of y n plus 1, and this implies e of yn plus 1, xn.
01:05
And this implies e of xn plus 1 and it is fn, that is x, n plus e of y n plus 1.
01:21
Here, e of y of y, i is summation of y, p, y, that is 1, that is 1p plus 0, which is equal to p.
01:36
Then this implies we can say e of xn plus 1, fn is equal to xn plus p.
01:46
Now in second part, mn is equal to xn minus hn...