Finding a Doob decomposition

Question

Let $(M_n)_{n\in \mathbb{N}}$ be a square integrable martingale and $(X_k)_{k\in \mathbb{N}}$ iid square integrable random variables with $\mathbb{E}X_1=1$. Show $M_n := \Pi_{k=1}^n X_k$ is a square integrable martingale and determine its quadratic variation.

I tried $$E(M_{n+1}|\mathcal{A}_n)=E(\Pi_{k=1}^{n+1} X_k|\mathcal{A}_n)=\Pi_{k=1}^nX_k E(X_{n+1}|\mathcal{A}_n)=M_n\cdot1$$ and $$EM_n^2=E((\Pi_{k=1}^{n}X_k)^2)=\Pi_{k=1}^{n}E(X_k^2)<\infty$$ So $M_n$ is a square integrable martingale and $M_n^2$ a submartingale. I wanted to use Doob decomposition: $$\langle M_n\rangle=\sum_{i=1}^n(E(M_i^2|\mathcal{A_{i-1}})-M^2_{i-1})=\sum_{i=1}^n(E(\Pi_{k=1}^{i}X_i^2|\mathcal{A_{i-1}})-\Pi_{k=1}^{i-1}X_k^2)= ...$$

I do not know how to continue... what can I do now? Thanks for any help!

Answer 1

You've made a good start. To finish, by independence, $$ E(\Pi_{k=1}^{i}X_i^2|\mathcal{A_{i-1}})=\Pi_{k=1}^{i-1}X_i^2\cdot E(X_i^2|\mathcal A_{i-1}) = \Pi_{k=1}^{i-1}X_i^2\cdot E(X_i^2)=\Pi_{k=1}^{i-1}X_i^2\cdot (\sigma^2+1), $$ so $$ E(\Pi_{k=1}^{i}X_i^2|\mathcal{A_{i-1}})-\Pi_{k=1}^{i-1}X_k^2=\sigma^2\cdot \Pi_{k=1}^{i-1}X_k^2=\sigma^2M_{i-1}^2. $$ where $\sigma^2:=E(X_1^2)-1$. Therefore $\langle M\rangle_n=\sigma^2\sum_{k=0}^{n-1}M_k^2$.

Answer 2

I thank Snoop for a comment and John Dawkins for his other answer which led me to elaborate a bit to show what can go wrong (and in fact was going wrong in my earlier versions):

Since the $X_k$ are i.i.d. \begin{align}\tag{1} \mathbb E[X_k^2]&={\rm Var}[X_1]+1=:\sigma^2+1\,. \end{align} The quadratic variation is the unique increasing predictable process $\langle M\rangle_n$ such that $M_n^2-\langle M\rangle_n$ is a martingale. For a discrete process predictable means that at $n$ it has to be ${\cal A}_{n-1}$-measurable.

Notation: \begin{align} A_n&=\sum_{k=1}^{n-1}\sigma^2M_k^2\,,\tag{2}\\ B_n&:=\sum_{k=1}^n(M_k-M_{k-1})^2\,,\tag{3}\\ C_n&:=\mathbb E\Big[B_n\Big|{\cal A}_{n-1}\Big]\tag{4} \end{align}

• When $\sigma>0$ only the two processes $M_n^2-A_n\,,M_n^2-B_n$ are martingales, and only $A_n$ is the predictable quadratic variation $\langle M\rangle_n\,$.

• $B_n$ is not predictable.

• $C_n$ is predictable but not necessarily increasing and does not make $M_n^2-C_n$ a martingale.

Proof. Using the i.i.d properties of the $X_i\,,$ \begin{align} \mathbb E[M_{n+1}^2|{\cal A}_n]&=\mathbb E\Big[\prod_{k=1}^{n+1}X_k^2\,\Big|{\cal A}_n\Big]= \mathbb E[X_{n+1}^2]\prod_{k=1}^nX_k^2=\sigma^2M^2_n+M^2_n\,.\tag{5} \end{align} Since taking the conditional expectation twice does not change the result, \begin{align} \mathbb E\Big[C_{n+1}\,\Big|{\cal A}_n\Big]&=\mathbb E\Big[B_{n+1}\,\Big|{\cal A}_n\Big]=\underbrace{\mathbb E\Big[(M_{n+1}-M_n)^2\,\Big|{\cal A}_n\Big]}_{(*)}+B_n\,.\tag{6} \end{align} Because $M_n$ is a martingale we can further develop the (*) term in (6): \begin{align} (*)&=\mathbb E\Big[M_{n+1}^2 \,\Big|{\cal A}_n\Big]-2M_n^2+M_n^2 \stackrel{(5)}{=}\sigma^2M^2_n\,.\tag{7} \end{align} Therefore, by (5) and (7), $$ \mathbb E\Big[M_{n+1}^2-C_{n+1}\,\Big|{\cal A}_n\Big]=M^2_n-\color{red}{B_n}\,,\tag{8} $$ that is, $M_n^2-C_n$ is $\color{red}{\rm not}$ a martingale.

The proof that $M_n^2-B_n$ is a martingale is the same. Use (5)-(7).

To see that $M_n^2-A_n$ is a martingale we only have to observe that, since $A_{n+1}$ is ${\cal A}_n$-measurable by definition, \begin{align} \mathbb E\Big[A_{n+1}\,\Big|{\cal A}_n\Big]&=\sigma ^2M_n^2+ A_n\tag{9} \end{align} trivially holds. Combining (5) with (9) shows that $M_n^2-A_n$ is a martingale. $$\tag*{$\Box$} \quad $$ Remark. The situation in the discrete case is a bit analogous to discontinuous processes. We know that for the Poisson process $N$ the predictable quadratic variation is $\langle N\rangle_t=\lambda t\,.$ But there is also the quadratic variation $$ [N]_t=\sum_{s\le t}(N_s-N_{s-})^2=\sum_{s\le t}(\Delta N_s)^2=\sum_{s\le t}\Delta N_s=N_t\tag{10} $$ which makes $N_t-[N]_t$ a martingale. But $[N]_t$ is not predictable.