Find an expression for $X_n = E[Y | \mathcal{F}_n]$ with $\mathcal{F}_n = \sigma(\{1\}, \{2\}, \ldots, \{n-1\}, \{n, n+1, \ldots\})$

Question

Consider $\Omega = \mathbb{N}, \mathcal{F} = 2^{\mathbb{N}}$ and the probability measure $\mathbb{P}$ defined by:

\begin{align*} \mathbb{P}[k] &= 2^{-k}, \; \; k = 1,2, \ldots \\ \end{align*}

Let $Y$ be the random variable:

\begin{align*} Y(k) &= \frac{2^{k+1}}{k^2} \\ EY &= \int \frac{2^{k+1}}{k^2} dP = \sum\limits_{k=1}^\infty \frac{2}{k^2} = \frac{\pi^2}{3} \\ \end{align*}

Let $\mathcal{F}_n$ be the filtration:

\begin{align*} \mathcal{F}_n = \sigma(\{1\}, \{2\}, \ldots, \{n-1\}, \{n, n+1, \ldots\}) \\ \end{align*}

Find an expression for $X_n = E[Y | \mathcal{F}_n]$

Could someone give me a hint on how to approach this one? Thanks!

Answer 1

(1) $X_n$ needs to be $\mathcal F_n$-measurable, meaning only that it must be constant on $\{n,n+1,\ldots\}$.

(2) you need to have $E[X_n\cdot 1_B]=E[Y\cdot 1_B]$ for each $B\in\mathcal F_n$. For $B=\{n,n+1,\ldots\}$ this means the constant value of $X_n$ mentioned in (1) must be $$ b_n:={\sum_{k=n}^\infty 2k^{-2}\over \sum_{k=n}^\infty 2^{-k}}=2^n\sum_{k=n}^\infty k^{-2}. $$ For $B=\{k\}$ (where $k$ is an element of $\{1,2,\ldots,n-1\}$) this means $X_n(k) = 2^{k+1}k^{-2}$.

Answer 2

Conditional expectation is defined with the property such that for any $A \in \mathcal{F}_n$, $\int_A X_n \, dP = \int_A Y \, dP$.

Consider sets of the form $A = \{k \} \in \mathcal{F}_n$ which exist for $1 \le k \le n-1$. Each $k$ forms its own atom in $\mathcal{F}_n$ so $X_n(k)$ can have a distinct value.

\begin{align*} X_n^{-1}(Y(k)) &= \{k\} \\ \int_{\{k\}} X_n dP &= \int_{\{k\}} Y dP \\ X_n(k) 2^{-k} &= Y(k) 2^{-k} \\ X_n(k) &= Y(k) \\ \end{align*}

Consider the set of the form $A = \{ n, n+1, n+2, \ldots \} \in \mathcal{F}_n$ which exists for $n \le k$. All of these values of $k \ge n$ form one single atom in $\mathcal{F}_n$, so $X_n(k)$ must produce the same value for all such values of $k$.

\begin{align*} X_n^{-1}(Y(k)) &= \{ n, n+1, n+2, \ldots \} \\ \int_{\{ n, n+1, n+2, \ldots \}} X_n dP &= \int_{\{ n, n+1, n+2, \ldots \}} Y dP \\ \sum\limits_{k=n}^\infty X_n(k) 2^{-k} &= \sum\limits_{r=n}^\infty Y(r) 2^{-r} \\ X_n(k) 2^{1-n} &= \sum\limits_{r=n}^\infty Y(r) 2^{-r} \\ X_n(k) &= \sum\limits_{r=n}^\infty Y(r) 2^{n-r-1} \\ \end{align*}

Tying this together yields:

\begin{align*} X_n(k) &= \begin{cases} Y(k) & \text{if} \; 1 \le k \le n - 1 \\ \sum\limits_{r=n}^\infty 2^{n-r-1} Y(r) & \text{if} \; n \le k \\ \end{cases} \\ \end{align*}