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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.

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closed as off-topic by BCLC, Namaste, Leucippus, JonMark Perry, Shailesh Jun 1 '18 at 1:42

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  • $\begingroup$ The sigma field generated by ${\{A_1, A_2, ..., A_n, B_n}\}$ is $\mathcal{F}_n$ since $B_n = (\bigcup_{i=1}^nA_i)^{\text{c}}$ $\endgroup$ – clark Jan 15 '18 at 18:15
  • $\begingroup$ @clark Ok thank you. I don't see how this helps solve the problem, though? $\endgroup$ – the man Jan 15 '18 at 18:22
  • $\begingroup$ Since $\cup A_i \cup B_n$ forms partition of the sigma field $\mathcal{F}_n$ is suffices to verify the desired equality for each of the events. To show that for each event you can use $E(X|A)=E(X1_A)/\mathbb{P}(A)$ where $A$ is an event. $\endgroup$ – clark Jan 15 '18 at 18:23
  • $\begingroup$ @clark So $Z_n = E(X|A_1) + ... + E(X|A_n) + E(X|B_n)$? $\endgroup$ – the man Jan 15 '18 at 18:31
  • $\begingroup$ Not quite, notice that your RHS, as is, is a number not a random variable, we should write it as $E(X|A_1)1_{A_1}+\ldots$ . More specifically, on the event $A_i$ $Z_n= E(X|F_n)= E(X|A_i)$, which you can compute as mentioned in the above comments. $\endgroup$ – clark Jan 15 '18 at 18:38
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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}

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    $\begingroup$ Thank you for your help, I appreciate it. $\endgroup$ – the man Jan 15 '18 at 20:37
  • $\begingroup$ If $T =$ min${\{n \geq 1 : Z_n = X}\}$ is a stopping time, could you possibly give me any hints on how to show that $T = \sum_{n=1}^{\infty} n \cdot I_{A_n}?$ My thoughts are that $X = E[X|\mathcal{F_n}]$, so $X$ is independent of the sigma field, but this isn't getting me anywhere! $\endgroup$ – the man Jan 15 '18 at 20:43
  • $\begingroup$ @theman I think it would be easier to address this in a different post. As, there would be the option to give feedback in form of an answer, and others would also be able to contribute as well. $\endgroup$ – clark Jan 15 '18 at 21:00

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