Functional Expectation Problem: Let  $X$  be distributed over the set  $\Bbb N$  of non-negative integers, with pmf: $$P(X=i)=\frac{\alpha}{2^i}$$

*

*$\alpha$

*$E[X]$
For $Y=X$ mod $3$, find:

*

*$P(Y=1)$

*$E[Y]$


*

*I will assume $\alpha=1$ for now to set the stage althought is there a way to find this value rather than assume?


*Assuming this is simlar to an infinite expectation that starts with Basel problem: $$\sum_{i=0}^\infty \frac{1}{i^2}=\frac{\pi^2}{6}$$
$$\frac{6}{\pi^2}\sum_{i=0}^\infty \frac{1}{i^2}=1$$
$$p_i = \frac{6}{\pi^2}\cdot\frac{1}{i^2}$$
$$E[X]=\sum_{i=0}^\infty i \cdot p_i=\frac{6}{\pi^2}\sum_{i=0}^\infty \frac{1}{i}=\infty$$
However, just not certain.
For the "$Y=X$ mod $3$" part I don't even know what that means.  Does that mean $Y=X+3$?
Thus I don't know how to answer the rest of them without understanding what "mod $3$" means.
 A: 

*

In order to guarantee that $p_i=\frac{\alpha}{2^i}$ is a distribution, we have that:
$$\sum_{i=0}^{+\infty} p_i = 1 \Rightarrow \\\alpha \sum_{i=0}^{+\infty}\frac{1}{2^i} = \alpha \frac{1}{1 - \frac{1}{2}} = 2 \alpha = 1 \Rightarrow \alpha = \frac{1}{2}.$$
1.
$$P(Y=1) = P(X ~\text{mod}~ 3 = 1) = \sum_{i : i ~\text{mod}~ 3 = 1}\frac{\alpha}{2^i}.$$
But all $i$ such that $i ~\text{mod}~ 3 = 1$ can be rewritten as:
$$i = 3j + 1,$$
for $j \geq 0.$ Hence:
$$P(Y=1)  = \sum_{j=0}^{+\infty} \frac{\alpha}{2^{3j + 1}} = \frac{\alpha}{2}\sum_{j=0}^{+\infty}\frac{1}{8^{j} } = \frac{\alpha}{2}\frac{1}{1 - \frac{1}{8}} = \frac{4}{7}\alpha = \frac{2}{7} .$$
2.
$$\mathbb{E}[Y=1] = \sum_{j=0}^{+\infty} (3j + 1)\frac{\alpha}{2^{3j + 1}} = \sum_{j=0}^{+\infty} \frac{3j\alpha}{2^{3j + 1}} + \sum_{j=0}^{+\infty} \frac{\alpha}{2^{3j + 1}} = \\
=\frac{3\alpha}{2}\sum_{j=0}^{+\infty} j\left(\frac{1}{8}\right)^j + \frac{4}{7}\alpha = \frac{3\alpha}{16}\sum_{j=0}^{+\infty} j\left(\frac{1}{8}\right)^{j-1} + \frac{4}{7}\alpha =\\
 = \frac{3\alpha}{16}\frac{1}{\left(1 - \frac{1}{8}\right)^2}  + \frac{4}{7}\alpha   = \frac{3\alpha}{16}\frac{64}{49}  + \frac{4}{7}\alpha = \frac{12\alpha}{49} + \frac{28 \alpha}{49} = \frac{40}{49}\alpha = \frac{20}{49}.
  $$
