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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 $\frac{dg_2}{(1-r)^2}+\frac{x_1}{1-r}$, where $d$ is the common difference of $a_n$(arithmetic progression part) and $r$ is the common ratio of $g_n$(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  =     $\frac{dg_2}{(1-r)^2}+\frac{x_1}{1-r}$

        =      (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 

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