Variance With Martingales Problem - Answered; Ignore the Bounty

Question

Let $(X_{j})_{j \geq 1}$ be random variables such that $X_{j}$ is $\mathcal{F}$-measurable for each $j$, where $(F_{j})_{j\geq 1}$ is an increasing sequence of $\sigma$-algebras. Assume $E(X_{j}|\mathcal{F}_{j-1})=0$ and $V_{j}=E(X_{j}^{2}|\mathcal{F_{j-1}})$ for $j \geq 2$. Finally, let $s_{n}^{2}=\sum_{j=1}^{n}V_{j}$ for $n \geq 2$, $\tilde{s}_{j}=\max(s_{j},1)$, and $Y_{n}=\sum_{j=2}^{n}\frac{X_{j}}{\tilde{s}_j}$ is a martingale for the sequence of $\sigma$-algebras $(\mathcal{F}_{j})_{j\geq1}$.

Show that $Var(Y_{n})=E(\sum_{j=2}^{n}\frac{V_{j}}{\tilde{s}_{j}^{2}})$

Answer 1

For every $n$, let $Z_n=(\tilde s_n)^{-1}X_n$. Here are the pieces you need to solve the problem:

• First, $s_n^2$ is $F_{n-1}$-measurable hence $\tilde s_n$ is $F_{n-1}$-measurable. Thus, $$ E[Z_n\mid F_{n-1}]=(\tilde s_n)^{-1}E[X_n\mid F_{n-1}]=0. $$ Integrating this, one gets $$ E[Z_n]=0. $$
• Second, for every $n\lt k$, $\tilde s_k$ and $Z_n$ are $F_{k-1}$-measurable hence $$ E[Z_nZ_k\mid F_{k-1}]=(\tilde s_k)^{-1}Z_nE[X_k\mid F_{k-1}]=0. $$ Integrating this, one gets $$ E[Z_nZ_k]=0. $$
• Third, for every $n$, $\tilde s_n$ is $F_{n-1}$-measurable hence $$ E[Z_n^2\mid F_{n-1}]=(\tilde s_n)^{-2}E[X_n^2\mid F_{n-1}]=(\tilde s_n)^{-2}V_n. $$ Integrating this, one gets $$ E[Z_n^2]=E[(\tilde s_n)^{-2}V_n]. $$