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I guess it is $\frac12$ as $P(0) = P(1) = 0.5$. While I was told it is $\frac13$. How to solve this type of problem?

Thank you.

Edit by advices:

$f(n) = \left\{\begin{aligned}{0, \text{n is odd natural number}\\1, \text{n is even natural number or 0}} \end{aligned}\right.$

What is the E[f(n)]?

user3059627
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2 Answers2

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The expected value is the value of that term! $$\lambda + \mu \text{ mod }2 $$ where $$\lambda = \left\{\begin{aligned}{0, \text{first sequence}\\1, \text{second sequence}} \end{aligned}\right. $$ $$\mu = \left\{\begin{aligned}{0, \text{odd term}\\1, \text{even term}} \end{aligned}\right. $$

Now, if you have either sequence, and stop it at some "random stopping point" the expected value is $1/2$. Or, if you pick the sequence at random and stop it at $n$, the expected value is $1/2$. Or, if you pick both the sequence and the stopping point at random, the expected value is $1/2$.

mjw
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This is a reverse-engineered response that finds a way of interpreting the problem by adding a plausible but omitted fact.

Expected value of $\{0, 1, 0, 1, 0, 1, \cdots \} =$ ?

Unfounded Assumption
The chance of selecting the $k$-th element in the sequence is $2^{(-k)}.$

Then, the expected value is

$$\sum_{k=1}^n 4^{(-k)} = \frac{(1/4)}{1 - [(1/4)]} = \frac{1}{3}.$$

user2661923
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