The question:
Fix $0<p=1-q<1$ reals, let $\Omega=\{0,1,2, \ldots\}$ be the non-negative integers, $\mathcal{F}=\mathcal{P}(\Omega)$ the power set, and $\mathbb{P}$ the probability measure that assigns $\mathbb{P}\{n\}=q^n p$ to $n \geq 0$. Define the function $X: \Omega \rightarrow\{0,1,2, \ldots\}$ to be the identity function. Notice that so far this is a way to describe a Pessimistic Geometric $(p)$ random variable (counting the failures before the first success). Define also $$ Y:=(X \bmod 2)=\mathbb{1}\{X \text { is odd }\}= \begin{cases}0, & \text { if } X \text { is even } \\ 1, & \text { if } X \text { is odd }\end{cases} $$ (a) What is the $\sigma$-algebra generated by $X$ ? (b) What is the $\sigma$-algebra generated by $Y$ ? (c) Use Kolmogorov's theorem on conditional expectations to calculate $\mathbb{E}(X \mid Y)$.
Would someone be able to let me know if I'm on the right path with my answer for part c). I think I'm suppose to use the fact that the sigma algebra generated by X is a subset of the algebra generated by Y.
Since $\sigma(Y) = \sigma(\{2,4,6,\dots,\},\{1,3,5,\dots,\})$ you get $$\mathbb{E}[X|Y]=1_{X \text{ odd}}\frac{\mathbb{E}[X1_{X \text{ odd}}]}{\mathbb{P}(X \text{ odd})}+1_{X \text{ even}}\frac{\mathbb{E}[X1_{X \text{ even}}]}{\mathbb{P}(X \text{ even})}$$ And each of these quantities can simply be calculated.