Exercise on martingale and sub/supermartingale

Question

I want to know whether I solved this exercise correct or not. It would be cool if someone takes the time to look over it:

Let $(Y_i)$ be i.i.d. with Rademacher distribution and $x\in\mathbb{Z}$. Consider the Filtration $\mathcal{F}_n=\sigma(Y_1,\dots,Y_n)$ and the random variable $$X_n:=x+Y_1+\cdots + Y_n$$ Decide whether the following are Martingales or not:

1. $X_n$

It's a martingale. Since $E(X_n)=x+\underbrace{1\cdot\frac{1}{2}-1\cdot\frac{1}{2}+\dots+1\cdot\frac{1}{2}-1\cdot\frac{1}{2}}_{n\ \text{times 0}}= x$ for ever $n\geq 0$

$$0=E(X_n)-E(X_{n+1})=\int\limits_{\Omega}X_n-E(X_{n+1}|\mathcal{F}_n) \mathrm{d}x$$ $$E(X_{n+1}|\mathcal{F}_n)=X_n\quad \text{a.s.}$$

2. $M_n:=\prod_{k=1}^n Y_k$

It's a martingale. Then since $(Y_i)$ are i.i.d. $$E(X_{n+1}|\mathcal{F}_n)=E(X_{n+1})=0$$ So same argument as in number $(1)$

3. $M_n:=X_n^2-n$

It's a submartingale. We already know that $X_n-n$ is a martingale and we now have a convext function $\phi(x,n)=x^2 -n$

$$E(X_n)-\phi (E(X_{n+1}|\mathcal{F}_n))\geq E(X_n) -E(\phi (X_{n+1})|\mathcal{F}_n) $$ $$X_n - \phi (E(X_{n+1}|\mathcal{F}_n)) \geq X_n - E(\phi(X_{n+1})|\mathcal{F}_n)$$ $$E(X_{n+1}^2-n+1|\mathcal{F_n})\geq X_n^2-n$$

4. $M_n^\lambda:=e^{\lambda X_n-a(\lambda)n}$ for every $\lambda \in \mathbb{R}$ where $a(\lambda)=\log \cosh \lambda$

It's a submartingal. Then $$e^{\lambda X_n-a(\lambda)n}=\frac{e^{\lambda X_n}}{\cosh(\lambda)n}$$

Now I would say that this is also a convex function and therefore I have a submartingal too.

Thank you very much.

Answer 1

1. Looks correct.

$$E[X_{n+1} | \mathscr F_n] = E[Y_{n+1}] + X_n = X_n$$

However, I think we're supposed to have

$$E(X_n)-E(X_{n+1})=\int\limits_{\Omega}X_n-E(X_{n+1}|\mathcal{F}_n) \mathrm{d}\color{red}{\mathbb P} \tag{*}$$

Note that $(*)$ implies $E[X_{n+1}] = E[X_n]$ and not vice-versa

2. Looks incorrect

$$E[M_{n+1} | \mathscr F_n] = E[Y_1Y_2 \cdots Y_nY_{n+1} | \mathscr F_n]$$ $$= Y_1Y_2 \cdots Y_nE[Y_{n+1} | \mathscr F_n]$$ $$= M_nE[Y_{n+1} | \mathscr F_n]$$ $$= M_nE[Y_{n+1}]$$ $$= M_n(0) = 0 \ne M_n \tag{**}$$

Note that $\{M_n = 0\} = \emptyset$

3. Looks incorrect

I think this is 1 of the 3 martingales of Brownian motion (which is the limit of a random walk)

$$E[X_{n+1}^2 - (n+1) | \mathscr F_n] = E[(X_{n+1} - X_n + X_n)^2 - (n+1) | \mathscr F_n]$$

$$ = E[(X_{n+1} - X_n + X_n)^2| \mathscr F_n] - (n+1) $$

$$ = E[(X_{n+1} - X_n)^2| \mathscr F_n] + 2E[(X_{n+1} - X_n)X_n | \mathscr F_n] + E[(X_n)^2| \mathscr F_n] - (n+1) $$

$$ = E[(Y_{n+1})^2| \mathscr F_n] + 2X_nE[(X_{n+1} - X_n) | \mathscr F_n] + (X_n)^2 - (n+1) $$

$$ = E[(Y_{n+1})^2] + 2X_n(0) + (X_n)^2 - (n+1) $$

$$ = 1 + 0 + (X_n)^2 - (n+1) $$

$$ = (X_n)^2 - n $$

Also, I don't think $X_n - n$ is a martingale:

$$E[X_{n+1} - (n+1) | \mathscr F_n] = X_n - (n+1) \ne X_n - n$$

However, your 3 inequalities look to be correct. Your last inequality is also true for an equality, I think.

4. Looks incorrect.

It's not exactly the 3rd martingale of Brownian motion, but similar is mentioned in Williams - Probability with Martingales (special case where x = 0):

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Pf:

$$E[\exp(\lambda X_{n+1} - a(\lambda)(n+1) | \mathscr F_{n+1}]$$

$$ = \exp(- a(\lambda)(n+1)) E[\exp(\lambda X_{n+1}) | \mathscr F_{n+1}]$$

$$ = \exp(- a(\lambda)(n)) \exp(- a(\lambda)) E[\exp(\lambda X_{n+1}) | \mathscr F_{n+1}]$$

$$ = \exp(- a(\lambda)(n)) \exp(- a(\lambda)) \exp(\lambda X_{n}) \frac{\exp(\lambda) + \exp(-\lambda)}{2}$$

$$ = \exp(- a(\lambda)(n)) \exp(\lambda X_{n})$$

QED