4

Let $X_1, X_2,\cdots$ be i.i.d. random variable on $(\Omega, \mathbb F)$ with $EX_1=0$ and $VX_1 = \sigma^2$. Consider a filtration $\mathbb F_n = \mathbb F(X_1,\cdots, X_n)$.

I want to show that $Y_n = \left( \sum_{k=1}^n X_k \right)^2$ is a submartingale.

This is my attempt:

\begin{eqnarray} E[Y_{n+1} | \mathbb F_n] &= E \left[ \left( \sum_{k=1}^{n+1} X_k \right)^2 \,\,\big|\,\, \mathbb F_n \right] \\ &= E \left[ \sum_{i,j=1}^{n+1} X_iX_j \,\,\big|\,\, \mathbb F_n \right] \\ &= \sum_{i,j=1}^{n+1} E \left[ X_i \,\,\big|\,\, \mathbb F_n \right] E \left[ X_j \,\,\big|\,\, \mathbb F_n \right] \\ &= \sum_{i,j=1}^{n} X_i X_j + 2 \sum_{i=1}^{n+1} X_i E \left[ X_{n+1} \,\,\big|\,\, \mathbb F_n \right] \\ &= Y_n + 2 \sum_{i=1}^{n+1} X_i E \left[ X_{n+1} \,\,\big|\,\, \mathbb F_n \right] \end{eqnarray}

How can I proceed from here? Of course if I knew $X_i\geq 0$ I would be done, but as is I don't know what to do next.

Moreover, I'm not sure how to show that $E |Y_n| < \infty$ when we don't know that $X_i\geq 0$.

Any help is MUCH appreciated.

CRYPTONEWBIE
  • 429
  • 2
    It's starting to go wrong in the third line of your calculation. (There is nothing wrong about pulling the sum outside the conditional expectation, but $$\mathbb{E}(X_i X_j \mid \mathbb{F}_n) \neq \mathbb{E}(X_i \mid \mathbb{F}_n) \mathbb{E}(X_j \mid \mathbb{F}_n).$$ – saz Jan 01 '18 at 14:14
  • Doesn't that follow from independence though? @saz – CRYPTONEWBIE Jan 01 '18 at 14:53
  • 1
    No, it doesn't (and for $i=j$ the random variables are not independent). You might want to take a look at this answer: https://math.stackexchange.com/a/2562920/36150 – saz Jan 01 '18 at 15:06
  • Right, thank you @saz Do you have a hint for the solution? – CRYPTONEWBIE Jan 02 '18 at 10:10
  • 1
    Well, did you take a look at the answer I linked in my previous comment? It should be hint enough. – saz Jan 02 '18 at 10:47

1 Answers1

1

$\newcommand{\E}{\mathbb E}\newcommand{\F}{\mathcal F}\newcommand{\Var}{\operatorname{Var}}$First let's do the integrability part of $Y_n$: \begin{align} \E|Y_n|=\E\left[\left(\sum_{k=1}^n X_k\right)^2\right]=\Var\left(\sum_{k=1}^n X_k \right)+\left(\E\left[ \sum_{k=1}^n X_k \right]\right)^2 <\infty \end{align} Where we have used $\E[Z^2]=\Var(Z)+\E[Z]^2$. Now let's show the rest: \begin{align} \E[Y_{n+1}|\F_n]&=\E\left[\left(\sum_{k=1}^{n+1} X_k\right)^2\ \Bigg |\ \F_n\right]\\ &=\E\left[\sum_{k=1}^{n+1}X_k^2+2\sum_{j=1}^{n+1}\sum_{k=1}^{j-1}X_kX_j \ \Bigg |\ \F_n \right]\\ \end{align} where the formula in this post is used. By linearity we have: \begin{align} \E[Y_{n+1}|\F_n]&=\sum_{k=1}^n X_k^2+\E[X_{n+1}^2]+2\sum_{j=1}^{n+1}\sum_{k=1}^{j-1}X_k\E[X_j|\F_n]\\ &=\sum_{k=1}^n X_k^2+\E[X_{n+1}^2]+2\sum_{j=1}^{n}\sum_{k=1}^{j-1}X_k\E[X_j|\F_n]+2\sum_{k=1}^{n}X_k\E[X_{n+1}|\F_n]\\ \end{align} where we have used the pull-out property of "known" variables and we will use it once again. The most right term is $\E[X_{n+1}|\F_n]=\E[X_{n+1}]=0$ by assumption hence: \begin{align} \E[Y_{n+1}|\F_n]&=\sum_{k=1}^n X_k^2+\E[X_{n+1}^2]+2\sum_{j=1}^{n}\sum_{k=1}^{j-1}X_kX_j\\ &\geq \sum_{k=1}^n X_k^2+2\sum_{j=1}^{n}\sum_{k=1}^{j-1}X_kX_j\\ &= \left(\sum_{k=1}^n X_k \right)^2\\ &= Y_n \end{align} So $(Y_n,\F_n)$ is a submartingale.

Shashi
  • 8,738