A coin is tossed until the first time a head turns up. If this occurs at the nth step, then the player wins $\frac{2^n}{n}$ dollars if n is odd,and loses $\frac{2^n}{n}$ dollars if n is even. Let X be the amount the player wins. Find E(X) if it exists.
Im a little bit stucked at this question. I understand that X distribute G($\frac{1}{2}$) but I dont know how to refer to the fact that n can be n even or odd.
thanks for helping
Let's ignore $n$ being either even or odd for a moment and look at the expected value for any $n$. Let $X$ be the profit which can be negative if we lose. The expectation is given using (and assuming the coin is fair)
$$\mathbb{E}[X]=\sum_{n=1}^\infty\left((-1)^{n+1}\cdot\frac{2^n}{n}\right)\cdot\frac{1}{2^n}=\sum_{n=1}^\infty(-1)^{n+1}\cdot\frac{1}{n}.$$
This is the alternating harmonic series which converges to $\log(2)$. However the expectation only exists if it converges absolutely as well! Looking at
$$\sum_{n=1}^\infty\left|(-1)^{n+1}\cdot\frac{2^n}{n}\right|\cdot\frac{1}{2^n}=\sum_{n=1}^\infty\frac{1}{n}$$
we notice the harmonic series is divergent hence there is actually no expectation for $X$.
Just a sidenote if you're interested: The general harmonic series $\sum_{n=1}^\infty 1/n^\alpha$ does converge for $\alpha>1$; otherwise, it diverges. Therefore changing the game to have a profit of $2^n/n^2$ for example will yield $\mathbb E[X]\neq \infty$ but $\operatorname{Var}[X]=\infty$.
The probability that the first head occurs with the $n$th toss is $2^{-n}$. And in this case the player wins $w(n)=(-1)^{n+1}\cdot \frac {2^n}n$. Hence $$E(X)=\sum_{n=1}^\infty 2^{-n}\cdot w(n)= \sum_{n=1}^\infty \frac{(-1)^{n+1}}n.$$ You should ecognize the series on the right as some logarithm.