Before going to find E( X-3 | X > 3 ) we need to understand the corresponding conditional probability needed which is : P(X-3 | X > 3)
Lets think about the possible values of X under the given condition ...So X = 4 , 5, 6 ..............
So X - 3 will take any value > 0..
So E(X - 3 | X > 3) = E(X = 1) or E(X = 2)................
= 1 . p(1) + 2 . p(2) ....+ k . p(k) .............
= 1 . (3/4)0 . (1/4) + 2 . (3/4)1 . (1/4) + 3 . (3/4)2 . (1/4) .......................
= 1/4 . [ 1.1 + 2.(3/4) + 3.(3/4)2 + .................. ]
Now the expression within the brackets is an infinite arithmetico geometric progression..
The sum of an infinite arithmetico-geometric sequence is , where is the common difference of (arithmetic progression part) and is the common ratio of (geometric progression part) and x1 = a1 g1
So according to the series in the brackets , we have :
d = 1 , g2 = 3/4 , r = 3/4 , x1 = a1 . g1 = 1..
Hence substituting these values in the expression mentioned in quotes , we get
sum =
= (1 . 3/4) / (1 - 3/4)2 + 1 / (1 - 3/4)
= (3/4) / (1/16) + 4
= 12 + 4
= 16
So E( X - 3 | X > 3 ) = 1/4 * sum
= 1/4 * 16
= 4
Hence required expectation value = 4