# Homogeneous Martingales and BDG Inequality

### Part I

1. Let $$f(x,y):\mathbb{R}^2 \rightarrow \mathbb{R}$$ be a twice differentiable function in both $$x$$ and $$y$$. Let $$M(t)$$ be defined by $M(t)=\int_0^t \sigma(s,\omega) dB(s,\omega)$. Assume that $$\sigma(t,\omega)$$ is adapted and that $$\mathbb{E} M^2 < \infty$$ for all $$t$$ a.s. .(Here $$B(t)$$ is standard Brownian Motion.) Let $$\langle M \rangle(t)$$ be the quadratic variation process of $$M(t)$$. What equation does $$f$$ have to satisfy so that $$Y(t)=f(M(t),\langle M \rangle(t))$$ is again a martingale if we assume that $$\mathbf E\int_0^t \sigma(s,\omega)^2 ds < \infty$$.
2. Set
\begin{align*}
f_n(x,y) = \sum_{0 \leq m \leq \lfloor n/2 \rfloor} C_{n,m} x^{n-2m}y^m
\end{align*}
here $$\lfloor n/2 \rfloor$$ is the largest integer less than or equal to $$n/2$$. Set $$C_{n,0}=1$$ for all $$n$$. Then find a recurrence relation for $$C_{n,m+1}$$ in terms of $$C_{n,m}$$, so that $$Y(t)=f_n(B(t),t)$$ will be a martingale.Write out explicitly $$f_1(B(t),), \cdots, f_4(B(t),t)$$ as defined in the previous item.

### Part II

Now consider $$I(t)$$ defined by $I(t)=\int_0^t \sigma(s,\omega)dB(s,\omega)$ where $$\sigma$$ is adapted and $$|\sigma(t,\omega)| \leq K$$ for all $$t$$ with probability one. In light of the above let us set
\begin{align*}Y(t,\omega)=I(t)^4 – 6 I(t)^2\langle I \rangle(t) + 3 \langle I \rangle(t)^2 \ .\end{align*}

1. Quote  the problem “Ito Moments” to show that $$\mathbb{E}\{ |Y(t)|^2\} < \infty$$ for all $$t$$. Then use the first part of this problem to conclude that $$Y$$ is a martingale.
2. Show that $\mathbb{E}\{ I(t)^4 \} \leq 6 \mathbb{E} \big\{ \{I(t)^2\langle I \rangle(t) \big\}$
3. Recall the Cauchy-Schwartz inequality. In our language it states that
\begin{align*}
\mathbb{E} \{AB\} \leq (\mathbb{E}\{A^2\})^{1/2} (\mathbb{E}\{B^2\})^{1/2}
\end{align*}
Combine this with the previous inequality to show that\begin{align*}\mathbb{E}\{ I(t)^4 \} \leq 36 \mathbb{E} \big\{\langle I \rangle(t)^2 \big\} \end{align*}
4. As discussed in class $$I^4$$ is a submartingale (because $$x \mapsto x^4$$ is convex). Use the Kolmogorov-Doob inequality and all that we have just derived to show that
\begin{align*}
\mathbb{P}\left\{ \sup_{0\leq s \leq T}|I(s)|^4 \geq \lambda \right\} \leq ( \text{const}) \frac{ \mathbb{E}\left( \int_0^T \sigma(s,\omega)^2 ds\right)^2 }{\lambda}
\end{align*}