Hard expected value problem

Question

Given a series of discrete random variables $Y_2, Y_3...Y_n$, such that for all $Y_i$ :

$P(Y_i = e^i) = \frac{1}{i}$, $P(Y_i = 3) = \frac{1}{3}$, $P(Y_i = X) = \frac{2}{3} - \frac{1}{i}$,

and $X$ is a random variable such that $P(X = 1) = \frac{2}{3}$, $P(x = -1) = \frac{1}{3}$. find $\mathbb{E}(Y_i)$ in terms of $i$.

I tried quite a lot of approaches to solve this but no success yet. First, I acknowledged that $P(Y_i = -1) + P(Y_i = 1) = P(Y_i =X)$, but I can't find another equation and so I can't find all of the probabilities for some $Y_i$.

Answer 1

Use the Law of Total Expectation: $\begin{align}\Bbb E(Y_i) = \Bbb E(\Bbb E(Y_i\mid X)) \end{align}$

For clarrity, let us define $\mathsf P_{i}(y):=\Pr(Y_i{=}y)$

$$\begin{align}\Bbb E(Y_i) &= \Bbb E(\Bbb E(Y_i\mid X))\\[1ex]&= \Bbb E\big(e^i\mathsf P_{i}(e^i)+3\mathsf P_{i}(3)+X\mathsf P_{i}(X)\big)\\[1ex]&= e^i\mathsf P_{i}(e^i)+3\mathsf P_{i}(3)+\Bbb E\big(X\mathsf P_{i}(X)\big)\\[1ex]&~~\vdots \end{align}$$

Answer 2

We remove $X$ from this by saying that $P(Y_i=X)= \frac23-\frac1i$ is the same as saying $P(Y_i=1)= \frac23(\frac23-\frac1i)$ and $P(Y_i=-1)= \frac13(\frac23-\frac1i)$

Then, $E[Y_i]=y_iP(Y_i=y_i)=e^i(\frac1i)+3(\frac13)+1(\frac23(\frac23-\frac1i))-1(\frac13(\frac23-\frac1i))=\frac{e^i}i+1+\frac49-\frac2{3i}-\frac29+\frac1{3i}=\frac{e^i}i+\frac{11}9-\frac1{3i}$