Re: Probability expectations question

Posted by Amit Goyal on Apr 13, 2014; 1:03am
URL: http://discussion-forum.276.s1.nabble.com/Probability-expectations-question-tp7586690p7586761.html

Since, E(X^2)= E(X) = 1. This implies variance of X, V(X) = E(X^2) - (E(X))^2 = 0.
Thus we have E(X) = 1, V(X) = 0
which together implies
Pr(X= 1) =  1.
Hence,
E(X^100) = 1.