"P(X=n)=(1\/2)^n"
"Y=\\begin{cases}\n1, \\ \\text{if x is even}\n\\\\\n-1, \\ \\text{if x is odd}\n\\end{cases}"
"P(Y=1)=\\sum \\limits_{n=1}^{\\infin}P(X=2n)= \\sum \\limits_{n=1}^{\\infin} (1\/2)^{2n}= \\sum \\limits_{n=1}^{\\infin} (1\/4)^{n}=\\frac{1\/4}{1-1\/4}=1\/3"
"P(Y=-1)=1-P(Y=1)=1-1\/3=2\/3"
"E[Y]=1\\times P(Y=1)+(-1)\\times P(Y=-1)=1\/3-2\/3=-1\/3"
Answer: the expected value of "Y" is -1/3.
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