Answer in Statistics and Probability for Kyei William Frimpong #119625
January 23rd, 2023
“E[Y[=-1({1over 2})+1({1over 2^2})-1({1over 2^3})+1({1over 2^4})-1({1over 2^5})+…=”
“=-({1over 2}+{1over 2}cdot{1over 4}+{1over 2}cdot({1over 4})^2+… )+”
“+({1over 4}+{1over 4}cdot{1over 4}+{1over 4}cdot({1over 4})^2+… )”
We have an infinite series that is geometric
“{1over 2}+{1over 2}cdot{1over 4}+{1over 2}cdot({1over 4})^2+… =dfrac{{1over 2}}{1-{1over 4}}={2over 3}”
“{1over 4}+{1over 4}cdot{1over 4}+{1over 4}cdot({1over 4})^2+… =dfrac{{1over 4}}{1-{1over 4}}={1over 3}”
“E[Y]=-{2over 3}+{1over 3}=-{1over 3}”
