Question

A discrete random variable has a probability mass function of P(X=n) = 0.5*n.
Let Y = 1 for x even and -1 for x odd ...
What is the expected value of 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

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

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

$$E(Y)=-{2\over 3}+{1\over 3}=-{1\over 3}$$