Question

A discrete random variable has probability mass function, P(X = n) = =
1
2

n
.
Let Y =

1, for x even
n1, for x odd
Find the expected value of Y ; (E[y]).

Answer 1

$$E[Y[=-1({1\over 2})+1({1\over 2^2})-1({1\over 2^3})+1({1\over 2^4})-1({1\over 2^5})+...=$$

$$=-({1\over 2}+{1\over 2}\cdot{1\over 4}+{1\over 2}\cdot({1\over 4})^2+... )+$$

$$+({1\over 4}+{1\over 4}\cdot{1\over 4}+{1\over 4}\cdot({1\over 4})^2+... )$$

We have an infinite series that is geometric

$${1\over 2}+{1\over 2}\cdot{1\over 4}+{1\over 2}\cdot({1\over 4})^2+... =\dfrac{{1\over 2}}{1-{1\over 4}}={2\over 3}$$

$${1\over 4}+{1\over 4}\cdot{1\over 4}+{1\over 4}\cdot({1\over 4})^2+... =\dfrac{{1\over 4}}{1-{1\over 4}}={1\over 3}$$

$$E[Y]=-{2\over 3}+{1\over 3}=-{1\over 3}$$