Let $(\Omega, \mathcal{F}, P)$ be a probability space and let $X$ be such that; $$P(X=(\frac{3}{2})^n) = 2^{-n}.$$
For $n \geq 1$, define $A_n = {\{X = (\frac{3}{2})^n)}\}$, and let $\mathcal{F_n}$ be the sigma field generated by $A_1, ..., A_n$.
Set $Z_n = E(X|\mathcal{F_n})$. It is easy to show this is a martingale with respect to the sigma field above.
Now, for $n \geq 1$ define $B_n = {\{X > (\frac{3}{2})^n)}\}$, then ${\{A_1, A_2, ..., A_n, B_n}\}$ forms a finite partition of $\Omega$ that generates $\mathcal{F_n}$.
I have been asked to show that $$Z_n = \sum_{k=1}^{n}(\frac{3}{2})^k\cdot I_{A_k} + 2(\frac{3}{2})^{n+1}\cdot I_{B_n}$$
Where $I_{A_k}, I_{B_n}$ are indicator functions.
I don't know where to start here. I don't really understand the variable $Z_n$, which is conditioned on the sigma field generated by ${\{A_1, A_2, ..., A_n, B_n}\}$.
Any help would be appreciated, thank you.
To find $Z_n$, it suffices to calculate $E(X|A_i)$ and $E(X|B_n)$. We can calculate $E(X|B_n)$, and, the others can be computed similarly. We set $B_{n,i}=\{X=\left (\frac{3}{2}\right)^{n+i}\}$ for $i\geq 1$.
\begin{align} E(X|B_n) &= \frac{E(X;B_n)}{\mathbb{P}(B_n)}\\ &=\frac{E(X1_{B_n})}{\mathbb{P}(B_n)}\\ &=\frac{\sum_{i\geq 1}E(X1_{B_{n,i}})}{\mathbb{P}(B_n)}\\ &=\frac{\sum_{i\geq 1}\left ( \frac{3}{2}\right )^{i}E(1_{B_{n,i}})}{\mathbb{P}(B_n)}\\ &=\frac{\sum_{i\geq 1}\left ( \frac{3}{2}\right )^{n+i}2^{-(n+i)}}{\sum_{i\geq 1}2^{-(n+i)} }\\ &=\frac{\sum_{i\geq 1}\left ( \frac{3}{2}\right )^{n+i}2^{-i}}{\sum_{i\geq 1}2^{-i} }\\ &=\frac{1}{2}\left ( \frac{3}{2}\right )^{n+1}\frac{\sum_{i\geq 0}\left ( \frac{3}{4}\right )^{i}}{\sum_{i\geq 1}2^{-i+1} }\\ &=2\left ( \frac{3}{2}\right )^{n+1} \end{align}